Jean-Xavier Rampon

for their patience and support. Their suggestions clarified the presentation of this text and improved the accessibility of the proofs. Merci encore !

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1. If M is a strong module of σ, then M P is a strong module of σ P .

2. Suppose that all the blocks of P are strong modules of σ. If Q is a strong module of σ P , then ∪Q is a strong module of σ.

Proof. First, let M be a strong module of σ. By Lemma 2.10, M P is a module of σ P . Consider a module Q of σ P such that Q ∩ (M P ) ≠ ∅.

Consequently, M P is a strong module of σ P . Second, let Q be a strong module of σ P . Consider a module M of σ such that there exists x ∈ (∪Q) ∩ M . Denote by X the block of P containing x. Clearly X ∈ Q∩(M P ). By Lemma 2.10, M P is a module of σ P .

Since Q is a strong module of σ P and X ∈ Q ∩ (M P ), we have Q ⊆ M P or M P ⊆ Q. It follows that ∪Q ⊆ M or M ⊆ ∪Q. Consequently, ∪Q is a strong module of σ.

Preface

My purpose is to present the main results on prime 2-structures. We consider primality in terms of the usual modular decomposition. We are mainly interested in the downward hereditary properties of primality. The first six chapters are devoted to finite prime 2-structures whereas the last three ones are to infinite prime 2-structures. The main focus is to establish results from the literature proven for graphs, digraphs, binary relational structures, etc., in the setting of 2-structures.

In Chapter 1, we provide the definition of a 2-structure. The 2-structures are the suitable generalizations of usual structures in Graph Theory, such as graphs and digraphs, to study the modular decomposition. In a 2-structure, the link between two vertices is not an edge or an arc, but a type of links, that is, an equivalence class of ordered pairs of distinct vertices. In this manner, a 2-structure is defined as an equivalence relation on the set of ordered pairs of distinct vertices. This equivalence relation is sufficient to define the notion of a module.

In Chapter 2, we define different types of connectedness for 2-structures. They generalize known connectedness for graphs and tournaments. We examine the components which are generated by these different types of connectedness. This examination leads us to introduce the notions of a module, of a modular cut, and of a strong module. These three notions induce three different types of primality. We study these three types of primality, and we conclude with Gallai's decomposition theorem.

In Chapter 3, we examine the prime 2-substructures in a prime 2-structure. First, we prove that every vertex is covered by prime 2-substructures of size 3, 4 or 5. Second, we introduce the outside partition associated with a prime 2-substructure. The outside partition allows us to build from a prime 2-substructure a new prime 2-substructure by adding two vertices. The first downward hereditary property of primality follows: A prime 2-structure admits prime 2substructures obtained by removing one or two vertices.

In Chapter 4, we characterize the critical 2-structures, that is, the prime 2-structures with the property that all the 2-substructures obtained by removing one vertex are decomposable. We introduce the primality graph associated with every prime 2-structures. Its edges are the unordered pairs whose removal provides a prime 2-substrucure. We examine the neighbourhoods of the primali PREFACE ity graph of a critical graph. We deduce that the primality graph of a critical graph is a path, a cycle of odd length or a path of odd length together with one isolated vertex. For each of these four types, we characterize the corresponding critical 2-structures. The characterization of critical 2-structures constitutes an important step in the study of prime 2-structures.

In Chapter 5, we demonstrate the Schmerl-Trotter theorem: a prime 2structure, with at least seven vertices, admits an unordered pair whose removal provides a prime 2-substructure. In other words, the primality graph of a prime graph, with at lest seven vertices, is nonempty. The Schmerl-Trotter theorem is the first substantial theorem in the study of prime 2-structures. It is an important downward hereditary property of primality. We prove also different refinements of the Schmerl-Trotter theorem.

In Chapter 6, we characterize the prime 2-structures that are minimal for a singleton or an unordered pair. Precisely, a prime 2-structure is minimal for a vertex subset if every proper induced 2-substructure with at least three vertices containing this vertex subset is not prime. We mainly characterize the prime 2-structures with at least six vertices that are minimal for an unordered pair. This characterization allows us to provide a concise proof of the Schmerl-Trotter theorem.

Chapter 7 is devoted to the following compactness theorem on infinite prime 2-structures. An infinite 2-structure is prime if and only if every finite vertex subset is contained in a finite vertex subset which induces a prime 2substructure.

Chapter 8 is the analogue of Chapter 4 for infinite 2-structures. Precisely, we characterize the infinite prime 2-structures, all the 2-substructures of which obtained by removing one vertex are decomposable, and that admit at least a prime 2-substructure obtained by removing finitely many vertices.

In Chapter 9, we characterize finite or infinite partially critical 2-structures. A prime 2-structure is partially critical whenever the removal of every vertex outside a given proper and prime 2-substructure provide a decomposable 2substructure. As in Chapter 3, we associate with the prime 2-substructure an outside partition. We also associate with it an outside graph which plays an important role in our characterization.

Finally, in Chapter 10, we provide a downward hereditary property of primality in the case of infinite 2-structures. Precisely, we prove that an infinite prime 2-structure admits a proper vertex subset equipotent to the vertex set which induces a prime 2-substructure.

Chapter 1

2-structures

A 2-structure [START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF] σ consists of a vertex set V (σ) , and of an equivalence 2-structure vertex set relation ≡ σ defined on (V (σ) × V (σ)) ∖ {(v, v) ∶ v ∈ V (σ)}. The cardinality of V (σ) is denoted by v(σ). A vertex subset of σ is a subset of V (σ). The set vertex subset of the equivalence classes of ≡ σ is denoted by E(σ) . Given a 2-structure σ, if E(σ) admits a unique element e, then σ is said to be constant or e-constant. constant, e-constant Warning. Unless indicated to the contrary, we consider 2-structures to be finite.

Notation 1.1. Let σ be a 2-structure. Given distinct v, w ∈ V (σ), the equivalence class of ≡ σ to which (v, w) belongs is denoted by (v, w) σ . Moreover, set [v, w] σ = ((v, w) σ , (w, v) σ ), and ≺v, w ≻ σ = {(v, w) σ , (w, v) σ }.

Let σ be a 2-structure. With each W ⊆ V (σ) associate the 2-substructure 2-substructure σ[W ] of σ induced by W defined on V (σ[W ]) = W such that

≡ σ[W ] = (≡ σ ) ↾(W ×W )∖{(w,w)∶w∈W } .
Given W ⊆ V (σ), σ[V (σ)∖W ] is denoted by σ-W , and by σ-w when W = {w}.

We use the next notation.

Notation 1.2. Let S be a set. Given W ⊆ S ×S, set W ⋆ = {(v, w) ∶ (w, v) ∈ W }.

We associate with a 2-structure σ its dual σ ⋆ defined on V (σ ⋆ ) = V (σ) dual as follows. Given x, y, v, w ∈ V (σ ⋆ ), with x ≠ y and v ≠ w, (x, y) ≡ σ ⋆ (v, w) if (y, x) ≡ σ (w, v). Hence E(σ ⋆ ) = {e ⋆ ∶ e ∈ E(σ)}. A 2-structure σ is reversible reversible 1 if σ = σ ⋆ . Hence, a 2-structure σ is reversible if and only if for each e ∈ E(σ), e ⋆ ∈ E(σ). Let σ be a reversible 2-structure. For each e ∈ E(σ), we have e ⋆ ∈ E(σ), so e = e ⋆ or e ∩ e ⋆ = ∅. A 2-structure σ is symmetric if for each symmetric e ∈ E(σ), e = e ⋆ . On the other hand, it is asymmetric if for each e ∈ E(σ), asymmetric e ∩ e ⋆ = ∅1 .

Isomorphism

Given 2-structures σ and τ , an isomorphism from σ onto τ is a bijection from isomorphism V (σ) onto V (τ ) satisfying for x, y, v, w ∈ V (σ), with x ≠ y and v ≠ w, (x, y) ≡ σ (v, w) if and only if (f (x), f (y)) ≡ τ (f (v), f (w)). Therefore, given a bijection f ∶ V (σ) → V (τ ), f is an isomorphism from σ onto τ if and only if f induces a bijection f ∶ E(σ) → E(τ ) satisfying for any v, w ∈ V (σ), with v ≠ w, we have

(f (v), f (w)) τ = f ((v, w) σ ).
Two 2-structures are isomorphic if there exists an isomorphism from one onto isomorphic the other.

Let σ be a 2-structure. An automorphism of σ is an isomorphism from σ automorphism onto itself. For example, the identity function Id V (σ) ∶ V (σ) → V (σ), defined identity function by Id V (σ) (v) = v for every v ∈ V (σ), is an automorphism of σ. The family of the automorphisms of σ, endowed with composition, is the automorphism group of automorphism group σ. It is denoted by Aut(σ). A 2-structure σ is rigid if Aut(σ) = {Id V (σ) }. On rigid the other hand, it is vertex-transitive if for any v, w ∈ V (σ), there is f ∈ Aut(σ) vertex-transitive such that f (v) = w. Lastly, given 2-structures σ and τ , σ embeds into τ if σ is isomorphic to a embedding 2-substructure of τ .

Graphs

A (simple) graph G is defined by a vertex set V (G) and an edge set E(G), graph vertex set edge set where an edge of G is an unordered pair of distinct vertices of G. Such a graph is denoted by (V (G), E(G)). For instance, given a nonempty set S, K S = (S, S 2 ) is the complete graph on S whereas (S, ∅) is the empty graph. With each graph complete graph empty graph G we associate its complement G = (V (G), V (G)

2 ∖ E(G)). complement A graph G is multipartite with a partition P of V (G) if the subgraph G[X]
multipartite of G induced by X is empty for each X ∈ P . It is bipartite when P = 2.

bipartite Given n ≥ 2, the path P n is the graph defined on V (P n ) = {0, . . . , n -1} as path follows. Given v, w ∈ {0, . . . , n -1}, with v ≠ w, {v, w} ∈ E(P n ) if vw = 1 (see Figure 1.1). The length of the path P n is n -1. length

0 • 1 • ......... n -2 • n -1 • Figure 1.1:
The path P n

Given n ≥ 3, the cycle C n is the graph defined on V (C n ) = {0, . . . , n -1} cycle obtained from P n by adding the edge {0, n -1}. The length of C n is n.

Let G be a graph. Given a vertex v of G, a neighbour of v is a vertex w neighbour of G such that {v, w} ∈ E(G). The neighbourhood of v is the set N G (v) of its neighbourhood neighbours, and d G (v) = N G (v) is its degree. Given a nonempty subset X of degree V (G), G[X] is connected if for any x, y ∈ X, with x ≠ y, there are elements connected x 0 , . . . , x n of X such that x 0 = x, x n = y, and {x m , x m+1 } ∈ E(G) for every 0 ≤ m ≤ n -1. Given a nonempty subset

X of V (G), G[X] is a component of G component if G[X] is connected, and for any x ∈ X and v ∈ V (G) ∖ X, {x, v} ∈ E(G). A vertex v of a graph G is isolated if G[{v}] is a component of G.
isolated Let G and H be graphs such that V (G)∩V (H) = ∅. The disjoint union of G disjoint union and H is the graph G⊕H = (V (G)∪V (H), E(G)∪E(H)). If V (G)∩V (H) ≠ ∅, then we can define G ⊕ H up to isomorphism by considering a graph H ′ such that H ≃ H ′ , and

V (G) ∩ V (H ′ ) = ∅.
A graph G is identified with the symmetric 2-structure σ(G) defined on V (σ(G)) = V (G) as follows. Given x, y, v, w ∈ V (σ(G)), with x ≠ y and v ≠ w, Given a graph G, observe that σ(G) = σ(G). A graph is self-complementary if it is isomorphic to its complement. Consider a self-complementary graph self-complementary G. Since σ(G) = σ(G), an isomorphism from G onto G is an automorphism of σ(G). Given a nonempty set S of integers, the usual linear order on S is denoted by L S . Given m ≥ 1, L {0,...,m-1} is also denoted by L m . Given n ≥ 1, we consider the tournament T 2n+1 defined on V (T 2n+1 ) = {0, . . . , 2n} by Consider a reversible 2-structure σ. Given e ∈ E(σ), the 2-structure σ is linear or e-linear if (V (σ), e) is a linear order. linear, e-linear Remark 1.3. Consider an e-linear 2-structure σ, where e ∈ E(σ). We have (V (σ), e) is a linear order. Clearly, (V (σ), e ⋆ ) is a linear order as well. Since σ is reversible, e ⋆ ∈ E(σ). Thus, σ is (e ⋆ )-linear, and E(σ) = {e, e ⋆ }. Moreover, we have σ = σ((V (σ), e)) = σ((V (σ), e ⋆ )).

Digraphs

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ T 2n+1 -(2n) = L 2n , (2n, 2m) ∈ A(T 2n+1 ) for 0 ≤ m ≤ n -1, and
(2m + 1, 2n) ∈ A(T 2n+1 ) for 0 ≤ m ≤ n -1 (see Figure 1.2). 0 • 1 • 2i • . . . 2i+1 • . . . 2n -2 • 2n -1 • E E E B B 2n • A Q ¢ ¢ ¢ ¢ ¢ ¢
Chapter 2

Connectedness and modules

We use the following notation. Notation 2.1. Let σ be a 2-structure. For W, W ′ ⊆ V (σ), with W ∩ W ′ = ∅, W ←→ σ W ′ signifies that (v, v ′ ) ≡ σ (w, w ′ ) and (v ′ , v) ≡ σ (w ′ , w) for any v, w ∈ W and v ′ , w ′ ∈ W ′ . The negation is denoted by W ←→ σ W ′ . Given v ∈ V (σ) and W ⊆ V (σ) ∖ {v}, {v} ←→ σ W is also denoted by v ←→ σ W . The negation is denoted by v ←→ σ W .

Given W, W ′ ⊆ V (σ) such that W ←→ σ W ′ , (W, W ′ ) σ denotes the equivalence class (w, w ′ ) σ of (w, w ′ ), where w ∈ W and w ′ ∈ W ′ . Furthermore, set [W, W ′ ] σ = ((W, W ′ ) σ , (W ′ , W ) σ ).

Lastly, given v ∈ V (σ) and W ⊆ V (σ) ∖ {v} such that {v} ←→ σ W , ({v}, W ) σ is also denoted by (v, W ) σ , (W, {v}) σ is denoted by (W, v) σ , and [{v}, W ] σ is denoted by [v, W ] σ .

Different types of connectedness

Let σ be a 2-structure. With each (e, f ) ∈ E(σ) × E(σ), we associate a type of connectedness. Given (e, f ) ∈ E(σ)×E(σ), we require that if σ is not connected in terms of the type associated with (e, f ), then the ordered pairs of vertices that are not in the same component, belong to e or f . Given a 2-structure σ, consider e, f ∈ E(σ). We define on V (σ) the equivalence relation ≈ (e,f ) in the following way. Given v, w ∈ V (σ), v ≈ (e,f ) w if v = w or v ≠ w and there exist sequences v 0 , . . . , v m and w 0 , . . . , w n of vertices of σ satisfying

• v 0 = v and v m = w; • for 0 ≤ i ≤ m -1, [v i , v i+1 ] σ ≠ (e, f );
• w 0 = w and w n = v;

• for 0 ≤ j ≤ n -1, [w j , w j+1 ] σ ≠ (e, f ).

Note that we do not need the second sequence w 0 , . . . , w p when e = f . Moreover, for 0 ≤ i ≤ m -1, [v i+1 , v i ] σ ≠ (f, e), and for 0 ≤ j ≤ n -1, [w j+1 , w j ] σ ≠ (f, e). By considering the sequences v = w n , . . . , w 0 = w and w = v m , . . . , v 0 = v, we obtain v ≈ (f,e) w. Consequently, for any e, f ∈ E(σ) and v, w ∈ V (σ), we have v ≈ (e,f ) w if and only if w ≈ (f,e) v. Definition 2.2. Let σ be a 2-structure. Consider e, f ∈ E(σ). The equivalence classes of ≈ (e,f ) are called the {e, f }-components of σ. The family of the {e, f }-{e, f }-components components of σ is denoted by C {e,f } (σ). Lastly, we say that the 2-structure σ is {e, f }-connected if it admits a unique {e, f }-component. Moreover, the {e, f }-connected 2-structure σ is connected if σ is {e, f }-connected for all e, f ∈ E(σ). connected We have E(σ(G)) = {e 0 , e 1 }.

The {e 0 }-components of σ(G) are exactly the components of G, whereas the {e 1 }-components of σ(G) are exactly the components of G. Since σ(G) is symmetric, σ(G) is {e 0 , e 1 }-connected.

Second, consider a tournament T . We have E(σ(T )) = {A(T ), A(T ) ⋆ }.

The {A(T ), A(T ) ⋆ }-components of σ(T ) are exactly the strongly connected components of T . Since σ(T ) is asymmetric, σ(T ) is {A(T )}-connected, and {A(T ) ⋆ }-connected.

The following lemma is established in [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF] for binary structures, that is, labeled 2-structures [START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF].

Lemma 2.4. Given a 2-structure σ, consider e, f ∈ E(σ). Let X be an {e, f }component of σ. For each v ∈ V (σ) ∖ X, we have v ←→ σ X. Precisely, for each y ∈ V (σ) ∖ X, we have [v, X] σ = (e, f ) or (f, e).

Proof. Let v ∈ V (σ) ∖ X. Consider x ∈ X. Since v ≈ (e,f ) x, we have [x, v] σ = (e, f ) or (f, e).

(2.1)

For a contradiction, suppose that there exist x, y ∈ X such that [x, v] σ ≠ [y, v] σ . It follows from (2.1) that e ≠ f . Morever, by interchanging x and y if necessary, we can assume that [x, v] σ = (e, f ) and [y, v] σ = (f, e). Hence [v, x] σ ≠ (e, f ) and [y, v] σ ≠ (e, f ). Since x ≈ (e,f ) y, there exists a sequence x 0 , . . . , x m satisfying • x 0 = x and x m = y;

• for 0 ≤ i ≤ m -1, [x i , x i+1 ] σ ≠ (e, f ).

By considering the sequences x 0 , . . . , x m , v and v, x, we obtain x ≈ (e,f ) v, which contradicts v ∈ X. Therefore, [x, v] σ = [y, v] σ for any x, y ∈ X. It follows from (2.1) that [v, X] σ = (e, f ) or (f, e).

Modules and quotient

Given Lemma 2.4, we introduce the following definition. Given a 2-structure σ, a subset M of V (σ) is a module 1 of σ if for each v ∈ V (σ) ∖ M , we have module v ←→ σ M . The classical properties of modules follow.

Proposition 2.5. Let σ be a 2-structure.

(M1) ∅, V (σ) and {v}, where v ∈ V (σ), are modules of σ.

(M2) Given W ⊆ V (σ), if M is a module of σ, then M ∩W is a module of σ[W ].
(M3) Let M be a module of σ. For every N ⊆ M , N is a module of σ[M ] if and only if N is a module of σ.

(M4) For any modules M and N of σ, M ∩ N is a module of σ.

(M5) Given modules M and N of σ, if M ∩ N ≠ ∅, then M ∪ N is a module of σ.

(M6) Given modules M and N of σ, if M ∖ N ≠ ∅, then N ∖ M is a module of σ.

(M7) Given modules M and N of σ, if M ∩ N = ∅, then M ←→ σ N .

Proof. It is easy to verify that the first assertion holds. For the second one, consider a subset W of V (σ), and a module M of σ. Let v ∈ W ∖ M . Clearly, v ∈ V (σ) ∖ M . Since M is a module of σ, we have v ←→ σ M , so v ←→ σ M ∩ W . For the third assertion, consider a module M of σ and a subset N of M . By the preceding assertion, if N is a module of σ, then M ∩ N = N is a module of σ[M ]. Conversely, suppose that N is a module of σ[M ], and consider v ∈ V (σ) ∖ N . We have v ∈ V (σ) ∖ M or v ∈ M ∖ N . In the first instance, since M is a module of σ, v ←→ σ M , and hence v ←→ σ N . In the second instance,

v ←→ σ N because N is a module of σ[M ].
Now, let M and N be modules of σ.

To verify that M ∩N is a module of σ, consider v ∈ V (B)∖(M ∩N ). We have v ∈ (V (σ) ∖ M ) ∪ (V (σ) ∖ N ). By interchanging M and N if necessary, assume that v ∈ V (σ) ∖ M . As M is a module of σ, v ←→ σ M , and hence v ←→ σ M ∩ N .

To show that M ∪ N is a module of σ, suppose that there exists x ∈ M ∩ N . Let v ∈ V (σ) ∖ (M ∪ N ). Since M is a module of σ, x ∈ M and v ∈ V (σ) ∖ M , we have

[v, M ] σ = [v, x] σ . Similarly, we have [v, N ] σ = [v, x] σ . It follows that [v, M ∪ N ] σ = [v, x] σ . Thus v ←→ σ M ∪ N .
Lastly, to prove that N ∖ M is a module of σ, suppose that there exists x ∈ M ∖ N . Let v ∈ V (σ) ∖ (N ∖ M ). Clearly, v ∈ (V (σ) ∖ N ) ∪ (M ∩ N ). First, suppose that v ∈ V (σ)∖N . Since N is a module of σ, v ←→ σ N , so v ←→ σ N ∖M . Second, suppose that v ∈ M ∩ N . Consider u, u ′ ∈ N ∖ M . We have to verify that v ←→ σ {u, u ′ }. Since M is a module of σ, x, v ∈ M and u ∈ V (σ) ∖ M , we have [v, u]

σ = [x, u] σ . Similarly, [v, u ′ ] σ = [x, u ′ ] σ . Moreover, we have [x, u] σ = [x, u ′ ] σ because N is a module of σ with u, u ′ ∈ N and x ∈ V (σ) ∖ N . It follows that [v, u] σ = [v, u ′ ] σ , so v ←→ σ {u, u ′ }.
Finally, let M and N be nonempty modules of σ such that M ∩ N = ∅. Consider x ∈ M and y ∈ N . For any v ∈ M and w ∈ N , we have [v, w] σ = [x, w] σ because M is a module of σ with x, v ∈ M and w ∈ V (σ) ∖ M . Furthermore, [x, w] σ = [x, y] σ because N is a module of σ with y, w ∈ N and x ∈ V (σ) ∖ N . Therefore, [v, w] σ = [x, y] σ for any v ∈ M and w ∈ N . Thus M ←→ σ N .

Let σ be a 2-structure. Following Assertion (M1) of Proposition 2.5, the modules ∅, V (σ) and {v}, where v ∈ V (σ), are called trivial modules. A 2-trivial module structure is indecomposable if all its modules are trivial 2 . Otherwise, it is indecomposable decomposable. Observe that a 2-structure, with at most two vertices, is indedecomposable composable. This leads us to the following notion. A 2-structure σ is prime if prime σ is indecomposable, with v(σ) ≥ 3. For instance, if σ is a constant 2-structure, then all the subsets of V (σ) are modules of σ. Hence, a constant 2-structure σ is decomposable if v(σ) ≥ 3. The same holds for linear 2-structures. Instead, consider a linear 2-structure σ such that v(σ) ≥ 3. By Remark 1.3, there exists a linear order L such that σ = σ(L). As above mentioned, the intervals of L are modules of σ. By denoting by v and w the first two vertices of L, we obtain that {v, w} is an interval of L. Thus, {v, w} is a nontrivial module of σ, so σ is decomposable. Fact 2.6. For n ≥ 4, the path P n (see Figure 1.1) is prime.

Proof. Let M be a module of P n with M ≥ 2. We have to show that M = {0, . . . , n -1}. Consider p, q ∈ M such that 0 < p < q. Since {p -1, p} ∈ E(P n ) and {p -1, q} ∈ E(P n ), we have p -1 ∈ M . In the same manner, if 0 < p -1, then p -2 ∈ M . It follows that {0, . . . , p} ⊆ M . Similarly {q, . . . , n -1} ⊆ M . Therefore {0, . . . , p} ∪ {q, . . . , n -1} ⊆ M. Now, consider p, q ∈ M such that p < q and M ∩ {p, . . . , q} = {p, q}. Suppose for a contradiction that p < q -1. Since {p, p + 1} ∈ E(P n ) and p + 1 ∈ M , we have {x, p + 1} ∈ E(P n ) for every x ∈ M . Therefore M ⊆ {p, p + 2}. Since {0, . . . , p} ∪ {q, . . . , n -1} ⊆ M , we obtain p = 0, q = n -1 = 2. Since n ≥ 4, we have p = q -1. Thus M = {0, . . . , n -1}.

Fact 2.7. For n ≥ 1, the tournament T 2n+1 (see Figure 1.2) is prime.

Proof. Consider a module M of T 2n+1 such that M ≥ 2. We have to show that M = {0, . . . , 2n}. By Proposition 2.5, M ∩ {0, . . . , 2n -1} is a module of T 2n+1 [{0, . . . , 2n -1}] = L 2n . Since M ∩ {0, . . . , 2n -1} ≠ ∅, there exist p, q ∈ {0, . . . , 2n -1} such that p ≤ q and M ∩ {0, . . . , 2n -1} = {p, . . . , q}. If p = q, then 2n ∈ X because M ≥ 2. If p < q, then 2n ∈ X because 2n ←→ σ {2m, 2m+1} for 0 ≤ m ≤ n -1. Thus 2n ∈ M . Since (2n, 0) ∈ A(T 2n+1 ) and (0, r) ∈ A(T 2n+1 ) for 1 ≤ r ≤ 2n -1, we have 0 ∈ M . Since (2n -1, 2n) ∈ A(T 2n+1 ) and (r, 2n -1) ∈ A(T 2n+1 ) for 0 ≤ r ≤ 2n -2, we have 2n -1 ∈ M . Consequently, p = 0, q = 2n -1 and M = {0, . . . , 2n}.

Let σ be a 2-structure. For any e, f ∈ E(σ), the {e, f }-components of σ are modules of σ by Lemma 2.4. Hence, the family C {e,f } (σ) (see Definition 2.2) realizes a partition of V (σ) in modules of σ. Generally, we introduce the following definition. A partition P of V (σ) is a modular partition of σ if all the modular partition blocks of P are modules of σ. Given a modular partition P of σ, it follows from Assertion (M7) of Proposition 2.5 that for distinct X, Y ∈ P , we have X ←→ σ Y . Hence, the blocks of P can be considered as the vertices of a new 2-structure defined in the following manner. With each modular partition P of σ, we associate the quotient σ P of σ by P defined on V (σ P ) = P as follows. Given quotient X, X ′ , Y, Y ′ ∈ V (σ P ), with X ≠ X ′ and Y ≠ Y ′ , (X, X ′ ) ≡ (σ P ) (Y, Y ′ ) if (x, x ′ ) ≡ σ (y, y ′ ), where x ∈ X, x ′ ∈ X ′ , y ∈ Y , and y ′ ∈ Y ′ .

Let σ be a 2-structure. Given e, f ∈ E(σ), C {e,f } (σ) is a modular partition of σ as mentioned above. We characterize the quotient σ C {e,f } (σ) as follows.

Proposition 2.8 (Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF]). Let σ be a 2-structure. For every e ∈ E(σ), σ C {e} (σ) is constant. Moreover, for distinct e, f ∈ E(σ), σ C {e,f } (σ) is linear.

Proof. To begin, consider e ∈ E(σ). Given distinct X, Y ∈ C {e} (σ), it follows from Lemma 2.4 that (X, Y ) σ = e.

(2.2)

Consider X, X ′ , Y, Y ′ ∈ C {e} (σ), with X ≠ X ′ and Y ≠ Y ′ . Let x ∈ X, x ′ ∈ X ′ , y ∈ Y , and y ′ ∈ Y ′ . It follows from (2.2) that (x, x ′ ) σ = e and (y, y ′ ) σ = e, so (x, x ′ ) ≡ σ (y, y ′ ). By the definition of quotient, we have (X, X ′ ) ≡ (σ C {e} (σ)) (Y, Y ′ ). Hence σ C {e} (σ) is constant. Now, consider distinct e, f ∈ E(σ). Given distinct X, Y ∈ C {e,f } (σ), it follows from Lemma 2.4 and Assertion (M7) of Proposition 2.5 that [X, Y ] σ = (e, f ) or (f, e).

(2.3)

Set ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ e C {e,f } (σ) = {(X, Y ) ∈ C {e,f } (σ) × C {e,f } (σ) ∶ X ≠ Y, (X, Y ) σ = e} and f C {e,f } (σ) = {(X, Y ) ∈ C {e,f } (σ) × C {e,f } (σ) ∶ X ≠ Y, (X, Y ) σ = f }.
We prove that

E(σ C {e,f } (σ)) = {e C {e,f } (σ), f C {e,f } (σ)}. (2.4)
Consider X, X ′ , Y, Y ′ ∈ C {e,f } (σ), with X ≠ X ′ and Y ≠ Y ′ . Let x ∈ X, x ′ ∈ X ′ , y ∈ Y , and y ′ ∈ Y ′ . First, suppose that (X, X ′ ) ≡ (σ C {e} (σ)) (Y, Y ′ ). By the definition of quotient, we have (x, x ′ ) ≡ σ (y, y ′ ), so (x, x ′ ) σ = (y, y ′ ) σ . By (2.3), (x, x ′ ) σ , (y, y ′ ) σ ∈ {e, f }. Thus, either (x, x ′ ) σ = (y, y ′ ) σ = e or (x, x ′ ) σ = (y, y ′ ) σ = f . In the first instance, we obtain (X, X ′ ) σ = (Y, Y ′ ) σ = e, and hence (X, X ′ ), (Y, Y ′ ) ∈ e C {e,f } (σ). In the second one, we have (X, X ′ ), (Y, Y ′ ) ∈ f C {e,f } (σ). Second, suppose that (X, X ′ ), (Y, Y ′ ) ∈ e C {e,f } (σ). We have (X, X ′ ) σ = (Y, Y ′ ) σ = e. Thus (x, x ′ ) σ = (y, y ′ ) σ = e, so (x, x ′ ) ≡ σ (y, y ′ ). By the definition of quotient, we have (X, X ′ ) ≡ (σ C {e} (σ)) (Y, Y ′ ). Similarly, we have (X, X ′ ) ≡ (σ C {e} (σ)) (Y, Y ′ ) when (X, X ′ ), (Y, Y ′ ) ∈ f C {e,f } (σ). Consequently (2.4) holds.

We continue by showing that (e C {e,f } (σ)) ⋆ = f C {e,f } (σ).

(2.5)

Consider distinct X, Y ∈ C {e,f } (σ). Suppose that (X, Y ) ∈ (e C {e,f } (σ)) ⋆ . We have (Y, X) ∈ e C {e,f } (σ), so (Y, X) σ = e. By (2.2), [X, Y ] σ = (e, f ) or (f, e). Since (Y, X) σ = e, we obtain (X, Y ) σ = f , so (X, Y ) ∈ f C {e,f } (σ). Conversely, suppose that (X, Y ) ∈ f C {e,f } (σ). We have (X, Y ) σ = f . By (2.2), [X, Y ] σ = (e, f ) or (f, e). Hence (Y, X) σ = e, so (Y, X) ∈ e C {e,f } (σ), that is, (X, Y ) ∈ (e C {e,f } (σ)) ⋆ . Consequently (2.5) holds.

To conclude, we have to prove that (C {e,f } (σ), e C {e,f } (σ)) is a linear order. It follows from (2.4) and (2.5) that (C {e,f } (σ), e C {e,f } (σ)) is a tournament. Thus, we have to verify that (C {e,f } (σ), e C {e,f } (σ)) is transitive. Consider X, Y, Z ∈ C {e,f } (σ) such that (X, Y ), (Y, Z) ∈ e C {e,f } (σ). Since (X, Y ), (Y, Z) ∈ e C {e,f } (σ), we have (X, Y ) σ = (Y, Z) σ = e. It follows from (2.3) that [X, Y ] σ = [Y, Z] σ = (e, f ). Since e ≠ f , we have X ≠ Z. Let x ∈ X, y ∈ Y , and z ∈ Z. We obtain [z, y] σ = [y, x] σ = (f, e). Since e ≠ f , we have [z, y] σ ≠ (e, f ) and [y, x] σ ≠ (e, f ). Since X ≠ Z, we have x ≈ (e,f ) z. It follows that [x, z] σ = (e, f ). By (2.3), [X, Z] σ = (e, f ). Consequently, (C {e,f } (σ), e C {e,f } (σ)) is transitive. Notation 2.9. Given a 2-structure σ, consider a partition P of V (σ). With W ⊆ V (σ), we associate the set W P of the blocks X of P such that X ∩ W ≠ ∅. Moreover, with Q ⊆ P , we associate the union ∪Q of the elements of Q.

In the following result, we compare the modules of a 2-structure with those of its quotients. Lemma 2.10. Given a 2-structure σ, consider a modular partition P of σ.

1. If M is a module of σ, then M P is a module of σ P .

2. If Q is a module of σ P , then ∪Q a module of σ.

Proof. First, we consider a module M of σ. Consider X ∈ P ∖ (M P ), and Y, Z ∈ M P . Let x ∈ X. Since Y, Z ∈ M P , there exist y, z ∈ M such that y ∈ Y ∩ M and z ∈ Z ∩ M . Since M is a module of σ, we have (x, y) ≡ σ (x, z) and (y, x) ≡ σ (z, x). By the definition of quotient, (X, Y ) ≡ (σ P ) (X, Z) and (Y, X) ≡ (σ P ) (Z, X). Thus, M P is a module of σ P .

Second, let Q be a module of σ P . Consider v ∈ V (σ)∖(∪Q), and y, z ∈ (∪Q). Since v ∈ V (σ) ∖ (∪Q), there exist X ∈ P ∖ Q such that v ∈ X. Furthermore, since y, z ∈ (∪Q), there exist Y, Z ∈ Q such that y ∈ Y and z ∈ Z. Since Q is a module of σ P , Y, Z ∈ Q and X ∈ P ∖ Q, we have (X, Y ) ≡ σ P (X, Z) and (Y, X) ≡ σ P (Z, X). It follows from the definition of quotient that (v, y) ≡ σ (v, z) and (y, v) ≡ σ (z, v). Therefore, ∪Q is a module of σ.

Modular cuts

Given a 2-structure σ, we continue the examination of the properties of the {e, f }-components of σ, where e, f ∈ E(σ). The next result is a consequence of Proposition 2.8. Corollary 2.11 (Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF]). Given a 2-structure σ, consider e, f ∈ E(σ). If σ is not {e, f }-connected, then there exists X ∈ C {e,f } (σ) such that [X, V (σ) ∖ X] σ = (e, f ).

Proof. If e = f , then it follows from Proposition 2.8 that [X, V (σ) ∖ X] σ = (e, e) for every X ∈ C {e} (σ). Suppose that e ≠ f . By Proposition 2.8 and Remark 1.3, there exists a linear order L defined on V (L) = C {e,f } (σ) such that σ C {e,f } (σ) = σ(L). The least vertex Y of L satisfies [Y, V (σ) ∖ Y ] σ = (e, f ) or (f, e). Similarly, the greatest vertex Z of L satisfies [Z, V (σ) ∖ Z] σ = (e, f )

or (f, e). Since [Y, Z] σ ≠ [Z, Y ] σ , we have [Y, V (σ) ∖ Y ] σ ≠ [Z, V (σ) ∖ Z] σ .
Therefore, there exists X ∈ {Y, Z} such that [X, V (σ) ∖ X] σ = (e, f ).

In Corollary 2.11, observe that X and V (σ) ∖ X are modules of σ. This leads us to the following definition. Given a 2-structure σ, a subset X of V (σ) is a modular cut3 of σ if X and V (σ) ∖ X are modules of σ. For instance, ∅ modular cut and V (σ) are modular cuts of σ, called trivial modular cuts. A 2-structure is trivial modular cut uncuttable if all its modular cuts are trivial, otherwise it is cuttable 4 . The fol-uncuttable cuttable lowing characterization of uncuttable 2-structures follows from Assertion (M7) of Proposition 2.5 and from Corollary 2.11. Proposition 2.12 (Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF]). A 2-structure is uncuttable if and only if it is connected.

Proof. Let σ be a 2-structure. To begin, suppose that σ is not connected. There exist e, f ∈ E(σ) such that σ is not {e, f }-connected. By Corollary 2.11, σ admits a nontrivial modular cut among its {e, f }-components. Hence σ is cuttable.

Conversely, suppose that σ is cuttable, and consider a nontrivial modular cut X of σ. Since X is a nontrivial modular cut of σ, X and V (σ) ∖ X are nonempty modules of σ. It follows from Assertion (M7) of Proposition 2.5 that there exist e, f ∈ E(σ) such that [X, V (σ) ∖ X] σ = (e, f ). Consequently, there is no sequence x 0 , . . . , x n satisfying x 0 ∈ X, x n ∈ V (σ)∖X, and [x m , x m+1 ] σ ≠ (e, f ) for m ∈ {0, . . . , n -1}. Thus σ is not {e, f }-connected, so σ is not connected.

Strong modules and Gallai's decomposition

Let σ be a 2-structure. If σ is prime, then {V (σ)} and {{v} ∶ v ∈ V (σ)} are the only modular partitions of σ. On the other hand, if σ is constant, then every partition of V (σ) is a modular partition of σ. Hence, in order to obtain a successful modular decomposition process, we have to associate intrinsically a modular partition with each 2-structure and to characterize the corresponding quotient. Furthermore, for the efficiency of the process, we inquire that if we repeat the process a second time, we would get an isomorphic quotient. For instance, consider a binary structure σ, and suppose that σ is not {e}-connected, where e ∈ E(σ). First, given Lemma 2.4, we can associate with σ the modular partition C {e} (σ). By Proposition 2.8, the corresponding quotient σ C {e} (σ) is constant. Set τ = σ C {e} (σ),

and ε = (C {e} (σ) × C {e} (σ)) ∖ {(X, X) ∶ X ∈ C {e} (σ)}.
Since τ is constant, we have E(τ ) = {ε}. Moreover, C {e} (σ) ≥ 2 because σ is not {e}-connected. Thus τ is not {ε}-connected. Second, associate with τ the family C {ε} (τ ) of its {ε}-components. Since E(τ ) = {ε}, the {ε}-components of τ are reduced to singletons. Therefore, the quotient of τ C {ε} (τ ) is isomorphic to τ . To proceed for any 2-structure σ, we return to the examination of the properties of the {e, f }-components of σ, where e, f ∈ E(σ).

Lemma 2.13 (Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF]). Given a 2-structure σ, consider an {e, f }-component X of σ, where e, f ∈ E(σ). For every module M of σ, if X ∩M ≠ ∅, then X ⊆ M or M ⊆ X.

Proof. Let M be a module of σ such that X ∩ M ≠ ∅ and X ∖ M ≠ ∅. We have to show that M ⊆ X. Consider x ∈ X ∖ M and y ∈ X ∩ M . Since X is an {e, f }component of σ containing x and y, there exist sequences x = x 0 , . . . , x p = y and y = y 0 , . . . , y q = x of elements of X such that for 0 ≤ m ≤ p -1, [x m , x m+1 ] σ ≠ (e, f ), and for 0 ≤ m ≤ q -1, [y m , y m+1 ] σ ≠ (e, f ). Since x 0 ∈ M and x p ∈ M , there exists m ∈ {0, . . . , p-1} such that x m ∈ X ∖M and x m+1 ∈ X ∩M . Similarly, since y 0 ∈ M and y q ∈ M , there exists n ∈ {0, . . . , q -1} such that y n ∈ X ∩ M and y n+1 ∈ X ∖ M . Now, let v ∈ M . Since M is a module of σ, x m+1 , v ∈ M and x m ∈ M , we have [x m , x m+1 ] σ = [x m , v] σ . Hence [x m , v] σ ≠ (e, f ). Since y n , v ∈ M and y n+1 ∈ M , [y n , y n+1 ] σ = [v, y n+1 ] σ . Thus [v, y n+1 ] σ ≠ (e, f ). By considering the sequences x = x 0 , . . . , x m , v and v, y n+1 , . . . , y q = x, we obtain x ≈ (e,f ) v. It follows that v ∈ X. Therefore M ⊆ X.

This result leads us to introduce the following definition. Given a 2-structure σ, a subset M of V (σ) is a strong module 5 of σ provided that M is a module strong module of σ, and for every module N of σ, we have

if M ∩ N ≠ ∅, then M ⊆ N or N ⊆ M .
Given e, f ∈ E(σ), it follows from Lemma 2.13 that each {e, f }-component of σ is a strong module of σ. As for modules, ∅, V (σ) and {v}, v ∈ V (σ), are strong modules of σ, called trivial strong modules. A 2-structure is primitive 6 if all primitive its strong modules are trivial. Three types of primitive 2-structures occur.

Lemma 2.14. Given a 2-structure σ, if σ is prime, constant or linear, then σ is primitive.

Proof. If σ is indecomposable, then all its modules are trivial, and hence all its strong modules are also. Therefore, if σ is prime, then σ is primitive. Now, suppose that σ is constant or linear. Recall that a 2-structure with at most 2 vertices is indecomposable. Hence, suppose also that v(σ) ≥ 3. To show that σ is primitive, it suffices to verify that every nontrivial module M of σ is not strong, that is, there exists a module N of σ such that M ∩ N ≠ ∅, M ∖ N ≠ ∅ and N ∖ M ≠ ∅.

Suppose that σ is constant. As previously observed, any subset of V (σ) is a module of σ. Consider distinct x, y ∈ M , and v ∈ V (σ) ∖ M . The module {x, v} of σ satisfies x ∈ M ∩ {x, v}, y ∈ M ∖ {x, v} and v ∈ {x, v} ∖ M .

Lastly, suppose that σ is linear. By Remark 1.3, there exists a linear order L defined on V (L) = V (σ) such that σ = σ(L). Recall that the modules of σ are exactly the intervals of L. Hence, M is a nontrivial interval of L. Up to isomorphism, we can assume that L = L n , where n ≥ 3. Since M is a nontrivial interval of L, M = [p, q], where 0 ≤ p < q ≤ n -1 and (p, q) ≠ (0, n -1). Observe that σ = σ(L ⋆ ) as well. Thus, by considering L ⋆ instead of L if necessary, we can assume that p ≥ 1. To conclude, it suffices to consider for N the interval [0, p] of L.

The analogue of Lemma 2.10 for strong modules follows.

Lemma 2.15. Given a 2-structure σ, consider a modular partition P of σ. 5 Also called prime module in [START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF] for 2-structures, and strong interval for digraphs. 6 Also called special in [START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF] .

In the second assertion of Lemma 2.15, the hypothesis that all the blocks of P are strong modules of σ is necessary. Indeed, for each X ∈ P , {X} is a strong module of σ P , so we must have ∪{X} = X is a strong module of σ.

The following property of the {e, f }-components of a 2-structure σ, where e, f ∈ E(σ), completes our examination. Lemma 2.16 (Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF]). Given a 2-structure σ, consider an {e, f }-component X of σ, where e, f ∈ E(σ). For every strong module M of σ, if X ⊆ M , then M = X or M = V (σ).

Proof. Let M be a strong module of σ such that X ⊊ M ⊆ V (σ). We have to show that M = V (σ). It follows from Lemma 2.4 that C {e,f } (σ) is a modular partition of σ. Furthermore, each block of C {e,f } (σ) is a strong module of σ by Lemma 2.13. Since M is a strong module of σ, it follows from Lemma 2.15 that

M C {e,f } (σ) is a strong module of σ C {e,f } (σ). By Proposition 2.8, σ C {e,f } (σ) is constant or linear. Thus, σ C {e,f } (σ) is primitive by Lemma 2.14. There- fore, M C {e,f } (σ) is a trivial strong module of σ C {e,f } (σ). Since X ⊊ M , M C {e,f } (σ) ≥ 2, so M C {e,f } (σ) = C {e,f } (σ). Lastly, consider Y ∈ C {e,f } (σ). Since M C {e,f } (σ) = C {e,f } (σ), we have Y ∩ M ≠ ∅ and M ∖ Y ≠ ∅. Since M is a strong module of σ, we obtain Y ⊆ M . It follows that M = V (σ).
Notation 2.17. Let σ be a 2-structure. Suppose that σ is not {e, f }-connected, where e, f ∈ E(σ). It follows from Lemma 2.16 that C {e,f } (σ) is the set of the strong modules of σ that are maximal under inclusion among the proper strong modules of σ. In a general way, we associate with each 2-structure σ the set Π(σ) of the strong modules of σ that are maximal under inclusion among the proper strong modules of σ. (Note that Π(σ) can be empty when σ is infinite.) Proposition 2.18. Let σ be a 2-structure such that v(σ) ≥ 2. The set Π(σ) constitutes a modular partition of σ7 , and the quotient σ Π(σ) is primitive.

Proof. To begin, consider X, Y ∈ Π(σ) such that X ∩ Y ≠ ∅. Since X is a strong module of σ, we have X ⊆ Y or Y ⊆ X.
It follows from the maximality of X and Y that X = Y . Moreover, consider v ∈ V (σ). As previously mentioned, {v} is a strong module of σ. Denote by S v the set of the proper strong modules of

σ containing v. Since v(σ) ≥ 2, {v} ∈ S v . Let M, N ∈ S v . We have v ∈ M ∩ N . Since M is a strong module of σ, we obtain M ⊆ N or N ⊆ M . Therefore, (S v , ⊊) is a linear order. Since V (σ) is finite, (S v , ⊊) admits a greatest element M v . Clearly, M v ∈ Π(σ). Consequently, we have ∪Π(σ) = V (σ). It follows that Π(σ) is a modular partition of σ. Now, we prove that σ Π(σ) is primitive. Consider a strong module Q of σ Π(σ) such that Q ≥ 2.
We have to show that Q = Π(σ). Since all the blocks of Π(σ) are strong modules of σ, it follows from Lemma 2.15 that ∪Q is a strong module of σ. Given X ∈ Q, we have X ⊊ (∪Q) because Q ≥ 2. By the maximality of X, ∪Q = V (σ), and hence Q = Π(σ).

The characterization of primitive 2-structures is an easy consequence of Lemma 2.14, and of the following two propositions. Proposition 2.19. Given a primitive 2-structure σ such that v(σ) ≥ 3, σ is prime if and only if σ is uncuttable. Proof. To begin, suppose that σ is cuttable, and consider a nontrivial modular cut X of σ. Since v(σ) ≥ 3, X or V (σ)∖X are nontrivial modules of σ. Therefore σ is decomposable.

Conversely, suppose that σ is decomposable. Hence σ admits nontrivial modules. Consider a module M of σ that is maximal under inclusion among the nontrivial modules of σ. Since σ is primitive, M is not a strong module of σ. Thus there exists a module N of σ such that M ∩ N ≠ ∅, M ∖ N ≠ ∅, and

N ∖ M ≠ ∅. By Assertion (M5) of Proposition 2.5, M ∪ N is a module of σ because M ∩ N ≠ ∅. Since N ∖ M ≠ ∅, we have M ⊊ M ∪ N . By the maximality of M , we obtain M ∪ N = V (σ). Thus N ∖ M = V (σ) ∖ M . By Assertion (M6) of Proposition 2.5, N ∖ M = V (σ) ∖ M is a module of σ because M ∖ N ≠ ∅. Consequently, M is a nontrivial modular cut of σ. Therefore σ is cuttable.
Proposition 2.20. Given a 2-structure σ, σ is primitive and cuttable if and only if σ is constant or linear, with v(σ) ≥ 2.

Proof. Suppose that σ is constant or linear, with v(σ) ≥ 2. By Lemma 2.14, σ is primitive. Since σ is constant or linear, it follows from Corollary 2.11 that there exists v ∈ V (σ) such that {v} is a modular cut of σ. Since v(σ) ≥ 2, {v} is a nontrivial modular cut of σ, so σ is cuttable.

Conversely, suppose that σ is primitive and cuttable. By Proposition 2.12, there exist e, f ∈ E(σ) such that σ is not {e, f }-connected. Furthermore, by Lemma 2.13, each {e, f }-component of σ is a strong module of σ. Since σ is primitive and not {e, f }-connected, each {e, f }-component of σ is reduced to a singleton. Consequently, the function V (σ) → C {e,f } (σ), defined by v ↦ {v} for every v ∈ V (σ), realizes an isomorphism from σ onto σ C {e,f } (σ). It follows from Proposition 2.8 that σ is constant or linear. Note that v(σ) ≥ 2 because σ is cuttable.

There is another approach to establish the forward direction of Proposition 2.20. It reveals the importance of the notion of a modular cut in the study of non connected 2-structures.

Second proof of the forward direction of Proposition 2.20. Let σ be a cuttable and primitive 2-structure. We consider a maximal set S under inclusion among the sets of modular cuts of σ that are linearly ordered by inclusion. By the maximality of S, we have ∅, V (σ) ∈ S, and S ∖ {∅, V (σ)} ≠ ∅ because σ is cuttable. We denote the elements of S by X 0 , . . . , X n , where n ≥ 2, in such a way that

∅ = X 0 ⊊ ⋯ ⊊ X n = V (σ).
Let m ∈ {0, . . . , n -1}. We show that

X m+1 ∖ X m is a strong module of σ. Since V (σ) ∖ X m is a module of σ, X m+1 ∩ (V (σ) ∖ X m ) = X m+1 ∖ X m is a module of σ by Assertion (M4) of Proposition 2.5. Now, consider a module M of σ such that M ∩ (X m+1 ∖ X m ) ≠ ∅, and M ∖ (X m+1 ∖ X m ) ≠ ∅. We have to verify that (X m+1 ∖X m ) ⊆ M . Since M ∖(X m+1 ∖X m ) ≠ ∅, we have M ∩X m ≠ ∅ or M ∩ (V (σ) ∖ X m+1 ) ≠ ∅. The set {V (σ) ∖ X p ∶ 0 ≤ p ≤ n}
(σ) ∖ X p ∶ 0 ≤ p ≤ n} if necessary, we can assume that M ∩ X m ≠ ∅.
We verify that

X m ∪ (M ∩ X m+1 ) is a modular cut of σ. Since X m ∩ (M ∩ X m+1 ) = M ∩ X m , X m ∪ (M ∩ X m+1 ) is a module of σ by Assertion (M5) of Proposition 2.5. Clearly, V (σ)∖(X m ∪(M ∩X m+1 )) = (V (σ)∖ X m ) ∖ (M ∩ X m+1 ). Since (M ∩ X m+1 ) ∖ (V (σ) ∖ X m ) = M ∩ X m , it follows from Assertion (M6) of Proposition 2.5 that (V (σ) ∖ X m ) ∖ (M ∩ X m+1 ) = V (σ) ∖ (X m ∪ (M ∩ X m+1 )) is a module of σ. Therefore, X m ∪ (M ∩ X m+1 ) is a modular cut of σ. Since X m ∪ (M ∩ X m+1 ) = X m ∪ (M ∩ (X m+1 ∖ X m )), we have X m ⊊ X m ∪ (M ∩ X m+1 ) ⊆ X m+1 .
It follows from the maximality of S that

X m ∪ (M ∩ X m+1 ) = X m+1 or, equiva- lently, (X m+1 ∖ X m ) ⊆ M . Consequently, X m+1 ∖X m is a strong module of σ for every m ∈ {0, . . . , n-1}. Since σ is primitive, X m+1 ∖X m = 1 for every 0 ≤ m ≤ n-1. Denote by x m+1 the unique element of X m+1 ∖ X m . We have X m = {x 1 , . . . , x m } for each 0 < m ≤ n. In particular, V (σ) = X n = {x 1 , . . . , x n }. Consider p, q ∈ {1, . . . , n} such that p < q. Since X p = {x 1 , . . . , x p } is a module of σ, we obtain [x p , x q ] σ = [x 1 , x q ] σ . Since V (σ) ∖ X 1 = {x 2 , . . . , x n } is a module of σ, [x 1 , x q ] σ = [x 1 , x 2 ] σ . Thus [x p , x q ] σ = [x 1 , x 2 ] σ for any p, q ∈ {1, . . . , n} such that p < q. It follows that σ is constant if (x 1 , x 2 ) σ = (x 2 , x 1 ) σ , and σ is linear if (x 1 , x 2 ) σ ≠ (x 2 , x 1 ) σ .
The characterization of primitive 2-structures follows.

Theorem 2.21 (Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF] 8 ). Given a 2-structure σ, σ is primitive if and only if σ is prime, constant or linear.

Proof. By Lemma 2.14, if σ is prime, constant or linear, then σ is primitive. Conversely, we verify that if σ is primitive and decomposable, then σ is constant or linear. Hence, suppose that σ is primitive and decomposable. Obviously, v(σ) ≥ 3 because σ is decomposable. It follows from Proposition 2.19 that σ is cuttable, and it suffices to apply Proposition 2.20.

The next result, called Gallai's decomposition theorem, is a direct consequence of Proposition 2.18 and Theorem 2.21.

Theorem 2.22 (Gallai [18, 28] 9 ). Given a 2-structure σ, with v(σ) ≥ 2, the quotient σ Π(σ) is prime, constant or linear.

Remark 2.23. Chein, Habib and Maurer [START_REF] Chein | Partitive hypergraphs[END_REF] adopted a different approach to establish Theorem 2.22 for partitive hypergraphs, which constitutes a nice generalization of Theorem 2.22. We transcribe it in terms of symmetric 2structures. (The set of the modules of a symmetric 2-structure is a partitive hypergraph.) Given a symmetric 2-structure σ, define a partial order O on the set of the modular partitions of σ as follows. Given distinct modular partitions P and Q of σ, (P,

Q) ∈ A(O) if for every X ∈ P , there exists Y ∈ Q such that X ⊆ Y . Clearly, {{v} ∶ v ∈ V (σ)} is the least vertex of O,
and {V (σ)} is the greatest one. Furthermore, with modular partitions P and Q of σ associate their join join 8 Ehrenfeucht, Harju and Rozenberg [START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF]Theorem 5.3] established a more general result. They associate with a decomposable and primitive 2-structure σ a graph Γ defined on V (Γ) = V (σ) as follows. Given distinct v, w ∈ V (Γ), vw ∈ E(Γ) if {v, w} is a module of σ. Then, they proved that either Γ is complete or Γ is a path. In the first instance, σ is constant whereas σ is linear in the second one. 9 Gallai [START_REF] Gallai | Transitiv orientierbare Graphen[END_REF] demonstrated this theorem for graphs; Boussaïri, Ille, Lopez, Thomassé [8, Theorem 5] for digraphs; Ehrenfeucht, Harju and Rozenberg [14, Theorem 5.5] for 2-structures; Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF]Theorem 2] for binary structures. P ∨ Q, and their meet P ∧ Q defined as follows. First, given distinct v, w ∈ V (σ), meet v and w belong to the same block of P ∨ Q if there exist X 0 , . . . , X n ∈ P ∪ Q satisfying v ∈ X 0 , w ∈ X n , and for 0

≤ i ≤ n -1 (when n ≥ 1), X i ∩ X i+1 ≠ ∅.
Second, given distinct v, w ∈ V (σ), v and w belong to the same block of P ∧ Q if there exist X ∈ P and Y ∈ Q such that v, w ∈ X ∩ Y . Clearly, P ∨ Q and P ∧ Q are modular partitions of σ. Therefore, O is a lattice, that is, for any lattice modular partitions P , Q and R of σ, we have: if

(P, R), (Q, R) ∈ A(O), then ((P ∨ Q), R) ∈ A(O), and if (R, P ), (R, Q) ∈ A(O), then (R, (P ∧ Q)) ∈ A(O).
The maximal vertices of O -{V (σ)} are called the coatoms of O. Lastly, Chein, coatoms Habib and Maurer observed that Π(σ) is the meet of all the coatoms of O.

We specify Theorem 2.22 as follows.

Theorem 2.24. Given a 2-structure σ, with v(σ) ≥ 2, the assertions below hold.

1. There exists e ∈ E(σ) such that σ is not {e}-connected if and only if

Π(σ) = C {e} (σ) and σ Π(σ) is constant. 2. There exist distinct e, f ∈ E(σ) such that σ is not {e, f }-connected if and only if Π(σ) = C {e,f } (σ) and σ Π(σ) is linear.
(σ) such that [X, V (σ) ∖ X] σ = (e, f ). Therefore, σ is not {e, f }-connected.
Lastly, suppose that σ is {e, f }-connected for any e, f ∈ E(σ). By Proposition 2.12, σ is uncuttable. It follows that Π(σ) ≥ 3. For every modular cut Q of σ Π(σ), ∪Q is a modular cut of σ by Lemma 2.10. Thus σ Π(σ) is uncuttable as well. Moreover, by Proposition 2.18, σ Π(σ) is primitive. Since Π(σ) ≥ 3, it follows from Proposition 2.19 that σ Π(σ) is prime. Conversely, suppose that σ Π(σ) is prime. Since Π(σ) ≥ 3, σ Π(σ) is neither constant nor linear. It follows from the first two assertions that σ is connected. Notation 2.25. Let σ be a non connected 2-structure. It follows from Theorem 2.24 that there exists a unique subset ν(σ) of E(σ) such that ν(σ) = 1 or 2, and σ is not ν(σ)-connected. The ν(σ)-components of σ are simply called the components of σ, and C ν(σ) (σ) is denoted by C(σ). component Finally, the last assertion of Theorem 2.24 is developed as follows.

Theorem 2.26. Given a 2-structure σ, with v(σ) ≥ 2, the following assertions are equivalent

1. σ is connected; 2. σ is uncuttable; 3. σ Π(σ) is prime;
4. There exists a modular partition P of σ such that σ P is prime; 5. Π(σ) ≥ 3 and Π(σ) is the set of the maximal modules of σ under inclusion among the proper modules of σ.

Proof. We denote by M the set of the maximal modules of σ under inclusion among the proper modules of σ. Hence, the fifth assertion is restated as follows

Π(σ) ≥ 3 and Π(σ) = M.
By Proposition 2.12, the first two assertions are equivalent. Hence, it follows from the last assertion of Theorem 2.24 that the first three assertions are equivalent.

Clearly, the third assertion implies the fourth one. Now, we show that the fourth assertion implies the fifth one. Suppose that there exists a modular partition P of σ such that σ P is prime. First, we prove that for every module

M of σ, if M P ≥ 2, then M P = P . (2.6)
Let M be a module of σ such that M P ≥ 2. By Lemma 2.10, M P is a module of σ P . Thus M P = P because σ P is prime. Therefore (2.6) holds. Second, we prove that for every module M of σ,

if M P ≥ 2, then M = V (σ). (2.7)
Let M be a module of σ such that M P ≥ 2. By (2.6), M P = P . For a contradiction, suppose that M ≠ V (σ), and consider X ∈ P such that X ∖M ≠ ∅. By Assertion (M6) of Proposition 2.5, M ∖ X is a module of σ. We have (M ∖X) P = P ∖{X}. Since P ≥ 3, we obtain (M ∖X) P ≥ 2 and (M ∖X) P ≠ P , which contradicts (2.6). It follows that X ⊆ M for every X ∈ P , so M = V (σ). Therefore (2.7) holds. Third, we prove that P = M.

Given X ∈ P , consider a module M of σ such that X ⊊ M . Since X ⊊ M , M P ≥ 2. By (2.7), M = V (σ). Thus P ⊆ M. Conversely, consider Y ∈ M. Since Y ≠ V (σ)
, it follows from (2.7) that there exists X ∈ P such that Y ⊆ X. By the maximality of Y , Y = X, so Y ∈ P . Therefore M ⊆ P . It follows that P = M. Fourth, we verify that the blocks of P are strong modules of σ. Given X ∈ P , consider a module M of σ such that X ∩ M ≠ ∅ and M ∖ X ≠ ∅. We have M P ≥ 2. By (2.7), M = V (σ).

Hence X ⊆ M . It follows that X is a strong module of σ. Since P = M, X ∈ M. It follows that X ∈ Π(σ). Therefore P ⊆ Π(σ). Since P and Π(σ) are partitions of V (σ), we obtain P = Π(σ). Consequently

P = Π(σ) = M.
Note that Π(σ) ≥ 3 because Π(σ) = P and σ P is prime. It follows that the fourth assertion implies the fifth one. Lastly, we show that the fifth assertion implies the third one. Hence suppose that Π(σ) ≥ 3 and Π(σ) = M. Since Π(σ) ≥ 3, we have to show that σ Π(σ) is indecomposable. Let Q be a module of σ Π(σ) such that Q ≥ 2. We have to verify that Q = Π(σ). By Lemma 2.10, ∪Q is a module of σ.

Consider X ∈ Q. Since Π(σ) = M, we have X ∈ M. Since Q ≥ 2, we obtain X ⊊ ∪Q. It follows from the maximality of X that ∪Q = V (σ). Hence Q = Π(σ).

Chapter 3

Prime 2-substructures of a prime 2-structure: the first results Notation 3.1. Let σ be a 2-structure. For n ∈ {3, . . . , v(σ) -1}, we denote by P n (σ) the set of X ⊆ V (σ) such that σ[X] is prime and X = n. Furthermore, we denote by R n (σ) the union of the elements of P n (σ). Question 3.2. Let σ be a prime 2-structure. A natural question is for which n ∈ {3, . . . , v(σ) -1}, is P n (σ) ≠ ∅?

Obviously, we can refine the question as follows. Given v ∈ V (σ), for which n ∈ {3, . . . , v(σ) -1}, do we have v ∈ R n (σ)?

For instance, given n ≥ 2, consider the graph B 2n+1 defined on V (B 2n+1 ) = {0, . . . , 2n} by

E(B 2n+1 ) ={{i, j} ∶ i, j ∈ {0, . . . , n -1}, i ≠ j} ∪ {{i, i + n} ∶ i ∈ {0, . . . , n -1}} ∪ {{i, 2n} ∶ i ∈ {0, . . . , n -1}} (see Figure 3.1).
The graph B 5 is called the bull. the bull Proof. Consider a module M of B 2n+1 -2n such that M ≥ 2. We have to show that M = {0, . . . , 2n-1}. For a contradiction, suppose that M ∩{0, . . . , n-1} = ∅. Hence M ⊆ {n, . . . , 2n -1}. Since M ≥ 2, there exist distinct i, j ∈ {0, . . . , n -1} such that i + n, j

' & $ % N B7 (6) • 1 d d d d • 0 • 2 • 6 ' & $ % • 4 • 3 • 5 N B7 (6)
+ n ∈ M , which is impossible because i ∈ M with {i, i + n} ∈ E(B 2n+1 ) and {i, j + n} ∈ E(B 2n+1 ). It follows that M ∩ {0, . . . , n -1} ≠ ∅. Similarly, M ∩ {n, . . . , 2n -1} ≠ ∅.
We prove that for every i ∈ {0, . . . , n -1},

i ∈ M ⇒ i + n ∈ M. (3.1) Indeed, consider i ∈ M ∩{0, . . . , n-1}. Since M ∩{n, . . . , 2n-1} ≠ ∅, there exists j ∈ {0, . . . , n -1} such that j + n ∈ M ∩ {n, . . . , 2n -1}. If i = j, then i + n ∈ M . Suppose that i ≠ j. We have {i, i + n} ∈ E(B 2n+1 ) and {j + n, i + n} ∈ E(B 2n+1 ).
Since M is a module of B 2n+1 -2n such that i, j + n ∈ M , we obtain i + n ∈ M . Hence (3.1) holds. Now, we prove that for every i ∈ {0, . . . , n -1},

i ∈ M ⇐⇒ i + n ∈ M. (3.2) Since (3.1) holds, consider i ∈ {0, . . . , n -1} such that i + n ∈ M . Since M ∩ {0, . . . , n -1} ≠ ∅, there exists j ∈ {0, . . . , n -1} ∩ M . If j = i, then i ∈ M .
Suppose that i ≠ j. Since (3.1) holds, j+n ∈ M . We have {i, i+n} ∈ E(B 2n+1 ) and {i, j +n} ∈ E(B 2n+1 ). Since M is a module of B 2n+1 -2n such that i+n, j +n ∈ M , we obtain i ∈ M . Hence (3.2) holds.

Lastly, since (3.2) holds, there exists i ∈ {0, . . . , n -1} such that i, i + n ∈ M . For each j ∈ {0, . . . , n -1} ∖ {i}, we have {i, j} ∈ E(B 2n+1 ) and {i + n, j} ∈ E(B 2n+1 ). Since M is a module of B 2n+1 -2n such that i, i + n ∈ M , we obtain j ∈ M . Therefore {0, . . . , n -1} ⊆ M . It follows from (3.2) that M = {0, . . . , 2n -1}.

Claim 3.4. B 2n+1 is prime.

Proof. Consider a module M of B 2n+1 such that M ≥ 2. We have to show that M = {0, . . . , 2n}. For a contradiction, suppose that 2n

∈ M . By Assertion (M2) of Proposition 2.5, M is a module of B 2n+1 -2n. Since M ≥ 2, it follows from Claim 3.3 that M = {0, . . . , 2n -1}, which is impossible because {0, 2n} ∈ E(B 2n+1 ) and {n, 2n} ∈ E(B 2n+1 ). Thus 2n ∈ M .
We prove that M ∖ {2n} ≥ 2. We have

M ∖ {2n} ≠ ∅ because M ≥ 2. First, suppose that there exists i ∈ M ∩ {0, . . . , n -1}. Since {i, i + n} ∈ E(B 2n+1 ) and {i + n, 2n} ∈ E(B 2n+1 ), we have i + n ∈ M . Second, suppose that there exists i ∈ {0, . . . , n -1} such that i + n ∈ M . Consider j ∈ {0, . . . , n -1} ∖ {i}. Since {j, 2n} ∈ E(B 2n+1 ) and {j, i + n} ∈ E(B 2n+1 ), we have j ∈ M . It follows that M ∖ {2n} ≥ 2. By Assertion (M2) of Proposition 2.5, M ∖ {2n} is a module of B 2n+1 -2n. Moreover, B 2n+1 -2n is prime by Claim 3.3. Since M ∖ {2n} ≥ 2, we obtain M ∖ {2n} = {0, . . . , 2n -1}. Thus M = {0, . . . , 2n} because 2n ∈ M . Claim 3.5. We have 2n ∈ R 5 (B 2n+1 ) ∖ (R 3 (B 2n+1 ) ∪ R 4 (B 2n+1 )). Proof. It follows from Claim 3.4 that the 2-substructure B 2n+1 [{0, 1, n, n+1, 2n}] is prime because it is isomorphic to B 5 . Thus 2n ∈ R 5 (B 2n+1 ). Now, consider X ⊆ V (B 2n+1
) such that X = 3 or 4, and 2n ∈ X. We have to prove that B 2n+1 [X] is decomposable.

First, suppose that for all i ∈ X ∩ {0, . . . , n -1} and all j ∈ X ∩ {n, . . . , 2n -1}, we have ji ≠ n. We have (X ∩ {0, . . . , n -1}) ∪ {2n} and X ∩ {n, . . . , 2n -1} are modules of

B 2n+1 [X]. If (X ∩ {0, . . . , n -1}) ∪ {2n} is a trivial module of B 2n+1 [X], then X ⊆ {n, . . . , 2n}, and hence X ∩ {n, . . . , 2n -1} is a nontrivial module of B 2n+1 [X]. Therefore, (X ∩{0, . . . , n-1})∪{2n} or X ∩{n, . . . , 2n-1} are nontrivial modules of B 2n+1 [X].
Second, suppose that there exists i ∈ {0, . . . , n -1} such that i, i

+ n ∈ X. If X ∩ {0, . . . , n -1} = {i}, then {i + n, 2n} is a nontrivial module of B 2n+1 [X]. Otherwise, if there exists j ∈ {0, . . . , n -1} ∖ {i} such that X = {i, j, i + n, 2n}, then {j, 2n} is a nontrivial module of B 2n+1 [X].
In Corollary 3.9, we establish that such a vertex 2n is unique.

Sumner's theorem

Remark 3.6. Let σ be a prime 2-structure. Consider

v ∈ R 3 (σ) ∪ R 4 (σ). (3.3)
Let X ⊆ V (σ) such that v ∈ X and X = 3 or 4. We verify that σ[X] is not connected. Otherwise, σ[X] is connected, and it follows from Theorem 2.26 that σ

[X] Π(σ[X]) is prime. Consider a subset X ′ of X such that v ∈ X ′ and X ′ ∩ Y = 1 for each Y ∈ Π(σ[X]). The function f ∶ Π(σ[X]) → X ′ , satisfying X ′ ∩ Y = {f (Y )} for each Y ∈ Π(σ[X]), realizes an isomorphism from σ[X] Π(σ[X]) onto σ[X ′ ]. Thus, σ[X ′ ] is prime with v ∈ X ′ and X ′ = 3 or 4, which contradicts (3.3). Notation 3.7. Consider a 2-structure σ. Let v ∈ V (σ). For e, f ∈ E(σ), set N (e,f ) σ (v) = {w ∈ V (σ) ∖ {v} ∶ [v, w] σ = (e, f )}. Proposition 3.8 (Ille [23] 1 ). Given a prime 2-structure σ, consider v ∈ R 3 (σ)∪ R 4 (σ). 1. For each e ∈ E(σ), σ[N (e,e) σ (v)] is e-constant. 2. For distinct e, f ∈ E(σ), σ[N (e,f ) σ (v)
] is e-linear (and f -linear).

Proof. Consider e, f ∈ E(σ) such that N (e,f ) σ (v) ≠ ∅. We prove that each {e, f }- component C of σ[N (e,f ) σ (v)] is a module of σ. Consider x ∈ V (σ) ∖ C. We have to verify that x ←→ σ C. Since C ⊆ N (e,f ) σ (v), we have [v, C] σ = (e, f ). Hence, suppose that x ≠ v. Moreover, since C is a module of σ[N (e,f ) σ (v)] by Lemma 2.4, suppose that x ∈ N (e,f ) σ (v). Thus suppose that x ∈ N (e,f ) σ (v) ∪ {v}. Let γ ∈ C. Since γ ∈ N (e,f ) σ (v) and x ∈ N (e,f ) σ (v), {γ, x} is not a module of σ[{v, x, γ}]. Since σ[{v, x, γ}] is decomposable, {v, x} or {v, γ} are modules of σ[{v, x, γ}]. First, suppose that {v, γ} is a module of σ[{v, x, γ}]. We obtain x ←→ σ {v, γ}. Thus [x, γ] σ = [x, v] σ , so γ ∈ N [x,v]σ σ (x). Second, suppose that {v, x} is a module of σ[{v, x, γ}]. We obtain γ ←→ σ {v, x}. Hence [x, γ] σ = [v, γ] σ . Since C ⊆ N (e,f ) σ (v), we obtain γ ∈ N (e,f ) σ (x). Therefore C ⊆ N (e,f ) σ (x) ∪ N [x,v]σ σ (x). (3.4) Suppose that ≺ v, x ≻ σ = {e, f } (see Notation 1.1). We obtain e ≠ f and [v, x] σ = (f, e) because x ∈ N (e,f ) σ (v). It follows from (3.4) that [x, C] σ = (e, f ). Now, suppose that ≺v, x≻ σ ≠ {e, f }. We obtain (C ∩ N [x,v]σ σ (x)) ∩ (C ∩ N (e,f ) σ (x)) = ∅. Consider γ ∈ C ∩N [x,v]σ σ (x) and δ ∈ C ∩N (e,f ) σ (x). By Remark 3.6, σ[{v, x, γ, δ}] is not connected. Since ≺v, γ ≻ σ =≺v, δ ≻ σ =≺x, δ ≻ σ = {e, f }, we obtain ν(σ[{v, x, γ, δ}]) = {e, f } (see Notation 2.25). Furthermore, since ≺v, x≻ σ ≠ {e, f } and ≺x, γ ≻ σ =≺v, x≻ σ , we have {v, x, γ} ∈ C(σ[{v, x, γ, δ}]).
By Lemma 2.4, δ ←→ σ {v, x, γ}, and hence

[γ, δ] σ = [v, δ] σ = (e, f ). Conse- quently, if C ∩ N [x,v]σ σ (x) ≠ ∅ and C ∩ N (e,f ) σ (x) ≠ ∅, then [C ∩ N [x,v]σ σ (x), C ∩ N (e,f ) σ (x)] σ = (e, f ). Since σ[C] is {e, f }-connected and C ⊆ N (e,f ) σ (x) ∪ N [x,v]σ σ (x) by (3.4), we have C ⊆ N [x,v]σ σ (x) or C ⊆ N (e,f ) σ (x). In both in- stances, we obtain x ←→ σ C. Consequently, the {e, f }-components of σ[N (e,f ) σ (v)
] are modules of σ. Since σ is prime, they are reduced to singletons. Thus, the function from

N (e,f ) σ (v) onto C {e,f } (σ[N (e,f ) σ (v)), defined by u ↦ {u} for every u ∈ N (e,f ) σ (v), is an iso- morphism from σ[N (e,f ) σ (v)] onto σ[N (e,f ) σ (v)] C {e,f } (σ[N (e,f ) σ (v)]). It follows from Lemma 2.4 and Proposition 2.8 that σ[N (e,f ) σ (v)] is e-constant if e = f , and σ[N (e,f ) σ (v)] is e-linear if e ≠ f .
Corollary 3.9 (Ille [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF] 2 ). Given a prime 2-structure σ, we have

V (σ) ∖ (R 3 (σ) ∪ R 4 (σ)) ≤ 1.
Proof. For a contradiction, suppose that there exist distinct x, y ∈ V (σ) ∖ (R 3 (σ) ∪ R 4 (σ)). We prove that σ is decomposable. This is the case if {x, y} is a module of σ because v(σ) ≥ 3. Hence, suppose that {x, y} is not a module of σ, and consider v ∈ V (σ) ∖ {x, y} such that v ←→ σ {x, y}. Since σ[{x, y, v}] is decomposable, {x, v} or {y, v} are modules of σ [{x, y, v}]. Suppose that {x, v} is a module of σ [{x, y, v}]

. Thus x, v ∈ N [y,x]σ σ (y). By Proposition 3.8, ≺x, v ≻ σ =≺x, y ≻ σ . It follows that ≺x, v ≻ σ =≺y, v ≻ σ =≺x, y ≻ σ . (3.5) Since v ←→ σ {x, y} and ≺ x, v ≻ σ =≺ y, v ≻ σ , we obtain ≺ x, v ≻ σ = 2 and [x, v] σ = [v, y] σ . Since {x, v} is a module of σ[{x, y, v}], we have [x, v] σ = [v, y] σ = [x, y] σ . (3.6)
We obtain also that (3.6) is satisfied when {y, v} is a module of σ [{x, y, v}]. Consequently, by setting In both cases, we obtain that (3.7) holds. It follows that W ∪ {x, y} is a module of σ. By Assertion (M3) of Proposition 2.5,

W = {v ∈ V (σ) ∖ {x, y} ∶ v ←→ σ {x, y}}, we obtain that W ∪ {x} is a module of σ[W ∪ {x, y}]. We show that W ∪
σ = [x, y] σ or [y, x] σ . Suppose that [w, {x, y}] σ = [x, y] σ . Since ≺ x, y ≻ σ = 2 and [v, x] σ = [y, x] σ by (3.6), {v, w} is not a module of σ[{x, v, w}]. Since σ[{x, u, w}] is decomposable, we have {x, v} is a module of σ[{x, v, w}] and [w, v] σ = [w, x] σ = [x, y] σ or {x, w} is a module of σ[{x, v, w}] and [w, v] σ = [x, v] σ . Since [x, v] σ = [x,
W ∪ {x} is a module of σ. Hence σ is decomposable. Consequently V (σ) ∖ (R 3 (σ) ∪ R 4 (σ)) ≤ 1.
Sumner's theorem is an immediate consequence of Corollary 3.9.

Theorem 3.10 (Sumner [START_REF] Sumner | Graphs indecomposable with respect to the X-join[END_REF] 3 ). Given a prime 2-structure σ, we have

P 3 (σ) ∪ P 4 (σ) ≠ ∅.
Sumner's theorem is improved as follows.

Theorem 3.11 (Cournier, Ille [START_REF] Cournier | Minimal indecomposable graph[END_REF] 4 ). Given a prime 2-structure σ, we have

V (σ) = R 3 (σ) ∪ R 4 (σ) ∪ R 5 (σ).
Proof. We prove that

V (σ) ∖ (R 3 (σ) ∪ R 4 (σ)) ⊆ R 5 (σ). Hence, consider v ∈ V (σ) ∖ (R 3 (σ) ∪ R 4 (σ))
. By Corollary 3.9, V (σ) ∖ (R 3 (σ) ∪ R 4 (σ)) = {v}. Thus, by considering an element of V (σ) ∖ {v}, we obtain

X ∈ P 3 (σ) ∪ P 4 (σ) such that X ⊆ V (σ) ∖ {v}. We prove that σ[X ∪ {v}] is prime. Otherwise, σ[X ∪ {v}] admits a nontrivial module M . By Assertion (M2) of Proposition 2.5, M ∩ X is a module of σ[X]. Since M ≥ 2, M ∩ X ≠ ∅. Since σ[X] is prime, we obtain M ∩ X = 1 or M ∩ X = X. In the first instance, there is y ∈ X such that M = {y, v}. Since {y, v} is a module of σ[X ∪ {v}], the function X → (X ∖ {y}) ∪ {v}, defined by y ↦ v and z ↦ z for each z ∈ X ∖ {y}, is an isomorphism from σ[X] onto σ[(X ∖ {y}) ∪ {v}]. Thus σ[(X ∖ {u}) ∪ {v}] is prime, which contradicts v ∈ V (σ) ∖ (R 3 (σ) ∪ R 4 (σ)). In the second instance, v ←→ σ X. Hence there exist e, f ∈ E(σ) such that X ⊆ N (e,f ) σ (v). By Proposition 3.8, σ[N (e,f ) σ (v)
] is constant or linear. Therefore, σ[X] is constant or linear as well, which contradicts the fact that σ[X] is prime. Consequenlty σ[X ∪ {v}] is prime.

The Ehrenfeucht-Rozenberg theorem

We continue examining the existence of prime 2-substructures of cardinality > 5 in a prime 2-structure (see Question 3.2). Notation 3.12. Given a 2-structure σ, suppose that there exists X ⊊ V (σ) such that σ[X] is prime. By Theorem 3.10, such a subset X exists if σ is prime with v(σ) ≥ 5. The discussion on M ∩ X in the proof of Theorem 3.11, where M is a module of σ[X ∪ {v}], leads us to consider the following subsets of V (σ) ∖ X

• Ext σ (X) denotes the set of v ∈ V (σ) ∖ X such that σ[X ∪ {v}] is prime; • ⟨X⟩ σ denotes the set of v ∈ V (σ)∖X such that X is a module of σ[X ∪{v}]; • For each y ∈ X, X σ (y) denotes the set of v ∈ V (σ) ∖ X such that {y, v} is a module of σ[X ∪ {v}]. Furthermore, p (σ,X) denotes the set {Ext σ (X), ⟨X⟩ σ } ∪ {X σ (y) ∶ y ∈ X}. Lemma 3.13. Given a 2-structure σ, consider X ⊆ V (σ) such that σ[X] is prime. The set p (σ,X) constitutes a partition of V (σ) ∖ X.
Proof. To begin, we verify that the union of the elements of

p (σ,X) equals V (σ)∖ X. Let v ∈ V (σ) ∖ X. If σ[X ∪ {v}] is prime, then v ∈ Ext σ (X). Suppose that σ[X ∪ {v}] is decomposable. Hence, σ[X ∪ {v}] admits a nontrivial module M . By Assertion (M2) of Proposition 2.5, M ∩ X is a module of σ[X]. Since M is a nontrivial module of σ[X ∪ {v}], we have M ≥ 2, so M ∩ X ≠ ∅. Since σ[X]
is prime, we obtain M = 1 or M ∩ X = X. In the first instance, there exists y ∈ X such that M ∩ X = {y}. Since M ≥ 2, we obtain M = {y, v}, and hence v ∈ X σ (y). In the second instance, we obtain M = X because M ≠ X ∪ {v}. It follows that v ∈ ⟨X⟩ σ . Now, we show that the elements of p (σ,X) are pairwise disjoint. By definition of the elements of p (σ,X) , we have Ext σ (X)∩⟨X⟩ σ = ∅, and Ext σ (X)∩X σ (y) = ∅ for every y ∈ X.

Let y ∈ X. Suppose for a contradiction that there exists v ∈ X σ (y) ∩ ⟨X⟩ σ . We obtain that {y, v} and X are modules of σ

[X ∪ {v}]. By Assertion (M6) of Proposition 2.5, X ∖ {y, v} = X ∖ {y} is a module of σ[X ∪ {v}] because v ∈ {y, v}∖X. By Assertion (M2) of Proposition 2.5, X∖{y} is a module of σ[X], which contradicts the fact that σ[X] is prime. It follows that X σ (y)∩⟨X⟩ σ = ∅.
Lastly, consider distinct y, z ∈ X. Suppose for a contradiction that there is v ∈ X σ (y) ∩ X σ (z). We obtain that {y, v} and {z, v} are modules of σ[X ∪ {v}]. By Assertion (M5) of Proposition 2.5, {y, v} ∪ {z, v} = {y, z, v} is a module of σ[X ∪ {x}] because v ∈ {y, v} ∩ {z, v}. By Assertion (M2) of Proposition 2.5, X ∩ {y, z, v} = {y, z} is a module of σ[X], which contradicts the fact that σ[X] is prime. Lemma 3.13 justifies the following definition.

Definition 3.14. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. By Lemma 3.13, p (σ,X) is a partition of V (σ) ∖ X.
It is called the outside partition induced by σ and X. outside partition Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. We study the modules of σ[X ∪ {v, w}], where v, w ∈ V (σ) ∖ X. We begin with two remarks.

Remark 3.15. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime.

• For every v ∈ ⟨X⟩ σ , X is a module of σ[X ∪ {v}], that is, v ←→ σ X. Thus, X is a module of σ[X ∪ ⟨X⟩ σ ], and X is a module of σ[X ∪ {v, w}] for v, w ∈ ⟨X⟩ σ . • Let y ∈ X. For z ∈ X ∖ {y} and v ∈ X σ (y), we have z ←→ σ {y, v} be- cause {y, v} is a module of σ[X ∪ {v}]. Therefore z ←→ σ {y} ∪ X σ (y). Consequently {y} ∪ X σ (y) is a module of σ[X ∪ X σ (y)], and {y, v, w} is a module of σ[X ∪ {v, w}] for v, w ∈ X σ (y). Remark 3.16. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that σ admits a nontrivial module M . By Assertion (M2) of Proposition 2.5, X ∩ M is a module of σ[X]. Since σ[X] is prime, we obtain X ∩ M = ∅, X ∩ M = 1 or X ∩ M = X.
We consider the three cases below.

• Suppose that X ∩ M = ∅. We prove that there exists B ∈ p (σ,X) such that M ⊆ B. For distinct v, w ∈ M , we have (X ∪ {v, w}) ∩ M = {v, w} is a module of σ[X ∪ {v, w}] by Assertion (M2) of Proposition 2.5. Therefore, the function f ∶ X ∪{v} → X ∪{w}, defined by v ↦ w and y ↦ y for every

y ∈ X, is an isomorphism from σ[X ∪ {v}] onto σ[X ∪ {w}]. Consequently, if v ∈ Ext σ (X), that is, σ[X ∪ {v}] is prime, then σ[X ∪ {w}] is prime too, so w ∈ Ext σ (X). Furthermore, if v ∈ ⟨X⟩ σ , that is, if X is a module of σ[X ∪{v}], then f (X) = X is a module of σ[X ∪{w}], so w ∈ ⟨X⟩ σ . Lastly, given y ∈ X, if v ∈ X(y), that is, {y, v} is a module of σ[X ∪ {v}], then f ({y, v}) = {y, w} is a module of σ[X ∪ {w}]
, so w ∈ X σ (y). Therefore, v and w belong to the same block of p (σ,X) .

• Suppose that there is y ∈ X such that X ∩ M = {y}. We verify that M ∖ {y} ≠ ∅ and M ∖ {y} ⊆ X σ (y). We have

M ∖ {y} ≠ ∅ because M ≥ 2. For each v ∈ M ∖ {y}, it follows from Assertion (M2) of Proposition 2.5 that (X ∪ {v}) ∩ M = {y, v} is a module of σ[X ∪ {v}] or, equivalently, v ∈ X σ (y).
• Suppose that X ⊆ M . Since M is a nontrivial module of σ, we have

M ⊊ V (σ). Moreover, (V (σ) ∖ M ) ⊆ ⟨X⟩ σ . Indeed, for each v ∈ V (σ) ∖ M , it follows from Assertion (M2) of Proposition 2.5 that (X ∪ {v}) ∩ M = X is a module of σ[X ∪ {v}] or, equivalently, v ∈ ⟨X⟩ σ . Lemma 3.17. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime.
The following statements hold.

(P1) For v ∈ ⟨X⟩ σ and w ∈ X σ (y), where y ∈ X, if σ[X∪{v, w}] is decomposable, then X ∪ {w} and {y, w} are the only nontrivial modules of σ[X ∪ {v, w}].

(P2) For v ∈ ⟨X⟩ σ and w ∈ Ext σ (X), if σ[X ∪ {v, w}] is decomposable, then X ∪ {w} is the unique nontrivial module of σ[X ∪ {v, w}]. (P3) Given distinct y, z ∈ X, for v ∈ X σ (y) and w ∈ X σ (z), if σ[X ∪ {v, w}]
is decomposable, then {y, v} and {z, w} are the only nontrivial modules of σ[X ∪ {v, w}].

(P4) Given y ∈ X, for v ∈ X σ (y) and w ∈ Ext σ (X), if σ[X ∪ {v, w}] is decom- posable, then {y, v} is the unique nontrivial module of σ[X ∪ {v, w}]. (P5) Given distinct v, w ∈ Ext σ (X), if σ[X ∪ {v, w}] is decomposable, then {v, w} is the unique nontrivial module of σ[X ∪ {x, y}]. Proof. For Statements (P1),...,(P5) above, consider distinct v, w ∈ V (σ) ∖ X. Suppose that σ[X ∪ {v, w}] admits a nontrivial module M . By Assertion (M2) of Proposition 2.5, X ∩ M is a module of σ[X]. Since σ[X] is prime, we obtain X ∩M ≤ 1 or X ∩M = X.
Observe that in Statements (P1), (P2), (P3), and (P4) above, v and w do not belong to the same block of p (σ,X) . By Remark 3.16, we have X ∩ M ≠ ∅. Hence, we have

X ∩ M = 1 or X ∩ M = X in Statements (P1), ( P2
), (P3), and (P4) above.

For Statement (P1), suppose that v ∈ ⟨X⟩ σ and w ∈ X σ (y), where y ∈ X. As above observed, X ⊆ M or there is z ∈ X such that X ∩ M = {z}. First, suppose that X ⊆ M . Since w ∈ X σ (y), w ∈ ⟨X⟩ σ by Lemma 3.13. It follows from Remark 3.16 that w ∈ M . Since M ≠ X ∪ {v, w}, we obtain M = X ∪ {w}. Thus v ←→ σ X ∪{w}, so v ←→ σ {y, w}. Since w ∈ X σ (y), that is, {y, w} is a module of σ[X ∪{w}], we obtain that {y, w} is a module of σ[X ∪{v, w}]. Second, suppose that X ∩ M = {z}. By Lemma 3.13, v ∈ X σ (z). It follows from Remark 3.16 that v ∈ M . Therefore M = {z, w} because M ≥ 2. We obtain w ∈ X σ (z). By Lemma 3.13, y = z. Since {y, w} is a module of σ[X ∪ {v, w}], we have v ←→ σ {y, w}. We have also v ←→ σ X because v ∈ ⟨X⟩ σ . Thus v ←→ σ X ∪ {w}, and hence X ∪ {w} is a module of σ[X ∪ {v, w}].

For Statement (P2), suppose that v ∈ ⟨X⟩ σ and w ∈ Ext σ (X). We verify that X ∩ M ≥ 2. Otherwise, there exits y ∈ X such that X ∩ M = {y}. By Remark 3.16, M ∖{y} ≠ ∅ and M ∖{y} ⊆ X σ (y), which contradicts Lemma 3.13. Therefore, X ∩ M ≥ 2, and hence X ⊆ M . Since w ∈ ⟨X⟩ σ , we obtain w ∈ M by Remark 3.16. Hence M = X ∪ {w} because M ⊊ X{v, w}.

For Statement (P3), suppose that v ∈ X σ (y) and w ∈ X σ (z), where y, z ∈ X and y ≠ z. Suppose for a contradiction that X ⊆ M . By Remark 3.16, {v, w} ∖ M ≠ ∅ and {v, w} ∖ M ⊆ ⟨X⟩ σ , which contradicts Lemma 3.13. Consequently, X ∖M ≠ ∅, and hence there exists t ∈ X such that X ∩M = {t}. By Remark 3.16, M ∖{t} ≠ ∅ and M ∖{t} ⊆ X σ (t). It follows from Lemma 3.13 that t ∈ {y, z}. By interchanging y and z, and hence v and w if necessary, we can assume that y = t. As previously, M ∖ {y} ≠ ∅ and M ∖ {y} ⊆ X σ (y). By Lemma 3.13, w ∈ X σ (y), and hence w ∈ M . Since M ≥ 2, we obtain M = {y, v}. It remains to show that {z, w} is a module of σ[X ∪ {x, y}] as well. Since {z, w} is a module of σ[X ∪ {w}], it suffices to verify that v ←→ σ {z, w}. We have

[z, y] σ = [z, v] σ and [w, y] σ = [w, v] σ because {y, v} is a module of σ[X ∪ {v, w}]. Furthermore, we have [z, y] σ = [w, y] σ because {z, w} is a module of σ[X ∪ {w}]. Therefore [z, v] σ = [w, v] σ .
For Statement (P4), suppose that v ∈ X σ (y), where y ∈ X, and w ∈ Ext σ (X). Suppose for a contradiction that X ⊆ M . By Remark 3.16, {v, w} ∖ M ≠ ∅ and {v, w} ∖ M ⊆ ⟨X⟩ σ , which contradicts Lemma 3.13. Consequently, X ∖ M ≠ ∅, and hence there exists z ∈ X such that X ∩M = {z}. By Remark 3.16, M ∖{z} ≠ ∅ and M ∖ {z} ⊆ X σ (z). By Lemma 3.13, w ∈ X σ (v), so w ∈ M . Since M ≥ 2, we obtain M = {z, v}. By Lemma 3.13, we have z = y.

For Statement (P5), suppose that v, w ∈ Ext σ (X). First, suppose that X ⊆ M . By Remark 3.16, {v, w} ∖ M ≠ ∅ and {v, w} ∖ M ⊆ ⟨X⟩ σ . It follows from Lemma 3.13 that X ∖ M ≠ ∅. Second, suppose that there exists y ∈ X such that X ∩ M = {y}. By Remark 3.16, M ∖ {y} ≠ ∅ and M ∖ {y} ⊆ X σ (y). It follows from Lemma 3.13 that X ∩ M ≠ 1. Consequently, we have X ∖ M ≠ ∅, and X ∩ M ≠ 1. By Remark 3.16, X ∩ M = ∅, and hence M = {v, w}.

The following result is a direct consequence of Lemma 3.17.

Corollary 3.18. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. The following two assertions hold.

(Q1) For v ∈ ⟨X⟩ σ and w ∈ V (σ) ∖ (X ∪ ⟨X⟩ σ ), if σ[X ∪ {v, w}] is decomposable, then X ∪ {w} is a nontrivial module of σ[X ∪ {v, w}]. (Q2) Given y ∈ X, for v ∈ X σ (y) and w ∈ V (σ) ∖ (X ∪ X σ (y)), if σ[X ∪ {v, w}] is decomposable, then {y, v} is a nontrivial module of σ[X ∪ {v, w}].
At present, we are ready to establish the Ehrenfeucht-Rozenberg theorem.

Theorem 3.19 (Ehrenfeucht and Rozenberg [START_REF] Ehrenfeucht | Primitivity is hereditary for 2-structures[END_REF]). Given a 2-structure σ, consider X ⊊ V (σ) such that V (σ) ∖ X ≥ 2 and σ[X] is primitive. If σ is prime, then there exist distinct v, w ∈ V (σ) ∖ X such that σ[X ∪ {v, w}] is primitive. More precisely, if σ is prime, then the following two statements hold.

1. If ⟨X⟩ σ ≠ ∅, then there exist v ∈ ⟨X⟩ σ and w ∈ V (σ) ∖ (X ∪ ⟨X⟩ σ ) such that σ[X ∪ {v, w}] is prime. 2. For each y ∈ X, if X σ (y) ≠ ∅, then there exist v ∈ X σ (y) and w ∈ V (σ) ∖ (X ∪ X σ (y)) such that σ[X ∪ {v, w}] is prime. Proof. First, suppose that ⟨X⟩ σ ≠ ∅. Since σ is prime, V (σ) ∖ ⟨X⟩ σ is not a module of σ. Thus, there exists v ∈ ⟨X⟩ σ such that v ←→ σ V (σ) ∖ ⟨X⟩ σ . But v ←→ σ X because v ∈ ⟨X⟩ σ . It follows that there exists w ∈ V (σ) ∖ (X ∪ ⟨X⟩ σ ) such that v ←→ σ X ∪ {w}. It follows from Assertion (Q1) of Corollary 3.18 that σ[X ∪ {v, w}] is prime. Second, consider y ∈ X such that X σ (y) ≠ ∅. Since σ is prime, {y} ∪ X σ (y) is not a module of σ. By Remark 3.15, {y}∪X σ (y) is a module of σ[X ∪X σ (y)].
Consequently, there exists w ∈ V (σ)∖(X ∪X σ (y)) such that w ←→ σ {y}∪X σ (y). Observe that for u ∈ V (σ) ∖ ({y} ∪ X σ (y)), we have u ←→ σ {y} ∪ X σ (y) if and only if u ←→ σ {y, v} for every v ∈ X σ (y). It follows that there is v ∈ X σ (y) such that w ←→ σ {y, v}. Therefore, {y, v} is not a module of σ[X ∪{v, w}]. It follows from Assertion (Q2) of Corollary 3.18 that σ[X ∪ {v, w}] is prime.

Finally, suppose that ⟨X⟩ σ = ∅, and X σ (y) = ∅ for each y ∈ X. By Lemma 3.13, we have V (σ) ∖ X = Ext σ (X). Since σ is prime, V (σ) ∖ X is not a module of σ. Therefore, there exist y ∈ X and distinct v, w ∈ V (σ) ∖ X such that y ←→ σ {v, w}. We obtain that {v, w} is not a module of σ[X ∪ {v, w}]. Since v, w ∈ Ext σ (X), it follows from Statement (P5) of Lemma 3.17 that σ[X∪{v, w}] is prime.

The next result, called the parity property, follows by applying Theorem 3.19 several times. It also provides an upward hereditary property of primality.

Corollary 3.20 (Ehrenfeucht and Rozenberg [START_REF] Ehrenfeucht | Primitivity is hereditary for 2-structures[END_REF]). Given a prime 2-structure σ, consider

X ⊊ V (σ) such that σ[X] is prime. For each n ≥ 0 such that V (σ) ∖ X ≥ 2n, there exists Y ⊆ V (σ) ∖ X such that Y = 2n and σ[X ∪ Y ] is prime.
The next result is a simple consequence of Corollary 3.20.

Corollary 3.21 (Ehrenfeucht and Rozenberg [START_REF] Ehrenfeucht | Primitivity is hereditary for 2-structures[END_REF]). Given a prime 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. There exist v, w ∈ V (σ) ∖ X such that σ -{v, w} is prime.

Proof. It suffices to apply Corollary 3.20 with n

= ⌈ v(σ)-X 2 ⌉ -1.
The first downward hereditary property of primality ends the section. It is an immediate consequence of Theorem 3.10 and Corollary 3.21. The second downward hereditary property of primality is the Schmerl-Trotter theorem (see Theorem 5.3). Proposition 3.22 (Ehrenfeucht and Rozenberg [START_REF] Ehrenfeucht | Primitivity is hereditary for 2-structures[END_REF]). Given a prime 2-structure σ, with v(σ) ≥ 5, there exist v, w ∈ V (σ) such that σ -{v, w} is prime.

Proof. By Theorem 3.10, there exists X ⊆ V (σ) such that σ[X] is prime, and X = 3 or 4. To conclude, it suffices to apply Corollary 3.21.

Chapter 4 Critical 2-structures

Given n ≥ 2, the tournament T 2n+1 (see Figure 1.2) is prime by Fact 2.7. Moreover, we have • T 2n+1 -2n = L 2n and, for instance, {0, 1} is a nontrivial module of T 2n+1 -2n;

• {2, . . . , 2n} is a nontrivial module of T 2n+1 -0;

• {0, . . . , 2n -3} is a nontrivial module of T 2n+1 -(2n -1);

• for 1 ≤ p ≤ 2n -2, {p -1, p + 1} is a nontrivial module of T 2n+1 -p.
This leads us to the following definition.

Definition 4.1. Given a prime 2-structure σ, a vertex v of σ is critical (in terms critical of primality) if σv is decomposable. The set of the noncritical vertices of σ is called the support of σ, it is denoted by S (σ). Generally, a proper subset W of support V (σ) is critical if σ -W is decomposable. A primitive 2-structure is critical if all its vertices are critical.

From the example above, given n ≥ 2, the tournament T 2n+1 is critical. Since critical 2-structures exist, the only attempt to improve Proposition 3.22 is to answer the following question positively.

Question 4.2. Let σ be a prime 2-structure. If v(σ) is large enough, then does there exist Z ⊊ V (σ) such that σ[Z] is prime and V (σ) ∖ Z = 2?
The second downward hereditary property of primality, that is, the Schmerl-Trotter theorem (see Theorem 5.3) answered Question 4.2 positively. Before providing such an answer, Schmerl and Trotter [START_REF] Schmerl | Critically indecomposable partially ordered sets, graphs, tournaments and other binary relational structures[END_REF] characterized the critical partial orders, graphs, tournaments, etc... Bonizzoni [START_REF] Bonizonni | Primitive 2-structures with the (n -2)-property[END_REF] independently characterized the critical 2-structures. To describe the structure of the critical digraphs, Boubabbous and Ille [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF] study the components of the primality graph 1 associated with every prime 2-structure. The primality graph was introduced by Ille [START_REF] Ille | Recognition problem in reconstruction for decomposable relations[END_REF] as below. It plays a decisive role in the structural study of the prime 2-structures.

The primality graph

Definition 4.3. Given a prime 2-structure σ, the primality graph P(σ) of σ is primality graph the graph defined on V (P(σ)) = V (σ), the edges of which are the noncritical unordered pairs. Therefore, given v, w ∈ V (σ), with v ≠ w, we have {v, w} ∈ E(P(σ)) if and only if σ -{v, w} is prime.

To begin, given a prime 2-structure σ, we examine the neighbourhood N P(σ) (v) of a critical vertex v of σ. Lemma 4.4 (Ille [START_REF] Ille | Recognition problem in reconstruction for decomposable relations[END_REF]). Let σ be a prime 2-structure with v(σ) ≥ 5. For every v ∈ V (σ) ∖ S (σ), we have d P(σ) (v) ≤ 2. Moreover, we have

1. if d P(σ) (v) = 1, then V (σ)∖(N P(σ) (v)∪{v}) is the unique nontrivial module of σ -v; 2. if d P(σ) (v) = 2, then N P(σ) (v) is the unique nontrivial module of σ -v.
Proof. To begin, we prove that

d P(σ) (v) ≤ 2 for each v ∈ V (σ) ∖ S (σ). Consider v ∈ V (σ) ∖ S (σ) such that N P(σ) (v) ≠ ∅. Let w ∈ N P(σ) (v). Set X = V (σ) ∖ {v, w}.
Hence, σ[X] is prime. Since v ∈ S (σ), σv is decomposable. Thus, w ∈ Ext σ (X) (see Notation 3.12). (

By Lemma 3.13, w ∈ ⟨X⟩ σ or w ∈ X σ (y), where y ∈ X. Therefore, we distinguish the following two cases.

• Suppose that w ∈ ⟨X⟩ σ . For every y ∈ X, X ∖ {y} is a nontrivial module of σ -{v, y}. Therefore y ∈ N P(σ) (v). Consequently, if there exists w

∈ N P(σ) (v) ∩ ⟨X⟩ σ , then N P(σ) (v) = {w}. (4.2)
• Suppose that there exists y ∈ X such that w ∈ X σ (y). For every z ∈ X ∖{y}, {y, w} is a nontrivial module of σ -{v, z}. Consequently, z ∈ N P(σ) (v), and hence N P(σ) (v) ⊆ {y, w}. Since {y, w} is a module of σ[X ∪ {w}], the function X → (X ∖ {y}) ∪ {w}, defined by y ↦ w and z ↦ z for every z ∈ X ∖ {y}, is an isomorphism from σ -{v, w} onto σ -{v, y}. It follows that y ∈ N P(σ) (v). Consequently, given y ∈ X, if there exists w ∈ N P(σ) (v) ∩ X σ (y), then N P(σ) (v) = {y, w}.

It follows from both cases above that d P(σ) (v) ≤ 2. Now, consider v ∈ V (σ) ∖ S (σ) such that d P(σ) (v) = 1. Denote by w the unique neighbour of v in P(σ). Set X = V (σ) ∖ {v, w}. It follows from (4.3) that w ∈ X σ (y) for every y ∈ X. Moreover, w ∈ Ext σ (X) by (4.1). By Lemma 3.13, w ∈ ⟨X⟩ σ , and V (σ) ∖ {v, w} is the only nontrivial module of σv.

Lastly, consider v ∈ V (σ) ∖ S (σ) such that d P(σ) (v) = 2. Denote by w and w ′ the neighbours of v in P(σ). Set X = V (σ) ∖ {v, w}. It follows from (4.2) that w ∈ ⟨X⟩ σ . Moreover, w ∈ Ext σ (X) by (4.1). By Lemma 3.13, there exists y ∈ X such that w ∈ X σ (y). By (4.3), N P(σ) (v) = {y, w}. Therefore, w ′ = y, so w ∈ X σ (w ′ ). It follows from Lemma 3.13 that {w, w ′ } is the only nontrivial module of σv.

Given a prime 2-structure σ, consider a component C of P(σ) such that v(C) ≥ 2 and V (C) ⊆ V (σ) ∖ S (σ). It follows from Lemma 4.4 that C is a cycle or a path. Proposition 4.5 (Boudabbous and Ille [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF]). Let σ be a prime 2-structure with v(σ) ≥ 5. For every component C of P(σ) such that v(C) ≥ 2 and V (C) ⊆ V (σ) ∖ S (σ), the following statements hold.

1. If C is a cycle, then its length is odd and

V (C) = V (σ); 2. If C is a path of odd length, then V (σ) ∖ V (C) ≤ 1; 3. If C is a path of even length, then V (C) = V (σ).
Proof. We denote the vertices of C by 0, . . . , v(C) -1 in such a way that C = C v(C) or P v(C) .

First, suppose that v(C) ≥ 3 and C = C v(C) . For a contradiction, suppose that v(C) is even. Hence, v(C) = 2n, where n ≥ 2. We show that {1, 2n -1} is a nontrivial module of σ. Since N P(σ) (0) = {1, 2n -1}, {1, 2n -1} is a module of σ -0 by Lemma 4.4. To show that {1, 2n -1} is a nontrivial module of σ, it suffices to verify that [0, 1] σ = [0, 2n -1] σ . For m ∈ {1, . . . , n -1}, we have

N P(σ) (2m) = {2m -1, 2m + 1}. By Lemma 4.4, {2m -1, 2m + 1} is a module of σ -2m. In particular, we obtain [0, 2m -1] σ = [0, 2m + 1] σ . Therefore, we have [0, 1] σ = [0, 3] σ = ⋯ = [0, 2n -1] σ . Consequently, {1, 2n -1} is a nontrivial module of σ, which contradicts the fact that σ is prime. It follows that v(C) is odd. Hence v(C) = 2n + 1, where n ≥ 1. For a contradiction, suppose that V (C) ⊊ V (σ). We show that V (C) is a module of σ. Consider v ∈ V (σ)∖V (C). For m ∈ {0, . . . , n-1}, we have N P(σ) (2m+1) = {2m, 2m+2}. By Lemma 4.4, {2m, 2m+2} is a module of σ-(2m+1). We obtain [v, 0] σ = [v, 2] σ = ⋯ = [v, 2n] σ , so v ←→ σ {0, 2, . . . , 2n}. Similarly, since for m ∈ {1, . . . , n -1}, N P(σ) (2m) = {2m-1, 2m+1}, we have v ←→ σ {1, 3, . . . , 2n-1}. Since N P(σ) (0) = {1, 2n}, [v, 2n] σ = [v, 1] σ . It follows that v ←→ σ V (C). Consequently, V (C) is a nontrivial module of σ, which contradicts the fact that σ is prime. Therefore V (C) = V (σ).
Second, suppose that v(C) ≥ 2, v(C) is even, and C = P v(C) . Hence v(C) = 2n, where n ≥ 1. For a contradiction, suppose that n = 1. We obtain N P(σ) (0) = {1} and N P(σ) (1) = {0}. By Lemma 4.4, V (σ) ∖ {0, 1} is a module of σ -0 and σ -1. Thus V (σ) ∖ {0, 1} is a nontrivial module of σ. Therefore n ≥ 2. We show that V (σ)∖V (C) is a module of σ. Consider v ∈ V (σ)∖V (C). Since N P(σ) (2m+ 1) = {2m, 2m + 2} for m ∈ {0, . . . , n -2}, we have v ←→ σ {0, 2, . . . , 2n -2}. Moreover, since

N P(σ) (2n-1) = {2n-2}, [v, 2n-2] σ = [1, 2n-2] σ . It follows that for any v, w ∈ V (σ)∖V (C) and m ∈ {0, . . . , n-1}, [v, 2m] σ = [w, 2m] σ . Similarly, for v ∈ V (σ) ∖ V (C), we have v ←→ σ {1, 3, . . . , 2n -1} because N P(σ) (2m) = {2m -1, 2m + 1} for m ∈ {1, . . . , n -1}. Now, since N P(σ) (0) = {1}, [v, 1] σ = [2n -2, 1] σ . Consequently, for any v, w ∈ V (σ) ∖ V (C) and m ∈ {0, . . . , n -1}, [v, 2m + 1] σ = [w, 2m + 1] σ . Consequently, V (σ) ∖ V (C) is a module of σ. Since σ is prime, we obtain V (σ) ∖ V (C) ≤ 1.
Lastly, suppose that v(C) is odd. Hence v(C) = 2n + 1, where n ≥ 1. For a contradiction, suppose that V (C) ⊊ V (σ). We show that V (σ) ∖ {1} is a nontrivial module of σ.

Since N P(σ) (0) = {1}, V (σ) ∖ {0, 1} is a module of σ -0 by Lemma 4.4. Let v ∈ V (σ) ∖ V (C). It suffices to verify that [1, v] σ = [1, 0] σ .
We distinguish the following two cases.

• Suppose that n = 1. We have

N P(σ) (2) = {1}. By Lemma 4.4, V (σ)∖{1, 2}
is a module of σ -2. In particular, we obtain [1, v]

σ = [1, 0] σ . • Suppose that n ≥ 2. Since for m ∈ {1, . . . , n -1}, N P(σ) (2m) = {2m - 1, 2m + 1}, we obtain v ←→ σ {1, 3, . . . , 2n -1} and 0 ←→ σ {1, 3, . . . , 2n -1}. Therefore, [1, v] σ = [2n -1, v] σ and [1, 0] σ = [2n -1, 0] σ . Furthermore, since N P(σ) (2n) = {2n -1}, [2n -1, v] σ = [2n -1, 0] σ . It follows that [1, v] σ = [1, 0] σ .
In both cases above, V (σ) ∖ {1} is a nontrivial module of σ, which contradicts the fact that σ is prime. Consequently, V (C) = V (σ).

V (σ) ∖ V (C) ≤ 1. Obviously, if V (C) = V (σ), then n ≥ 3 and P(σ) ≃ P 2n . Suppose that V (σ) ∖ C = 1. The single element of V (σ) ∖ V (C)
is an isolated vertex of P(σ) because C is a component of P(σ). Therefore, P(σ) ≃ P 2n ⊕ K {2n} .

We end the section with some specific results on critical 2-structures. The first one follows from Corollary 3.20.

Corollary 4.7. Let σ be a critical 2-structure σ such that v(σ) ≥ 5. Let X ⊊ V (σ).

1. If σ[X] is prime, then v(σ) -X is even. 2. Moreover, if σ[X] is prime and X ≥ 4, then σ[X] is critical. Proof. Let X ⊊ V (σ) such that σ[X] is prime. For a contradiction, suppose that v(σ) -X = 2n + 1, where n ≥ 0. By Corollary 3.20, there exists Y ⊆ V (σ) ∖ X such that Y = 2n and σ[X ∪ Y ] is prime. We have V (σ) ∖ (X ∪ Y ) = 1.
By denoting by v the unique element of V (σ) ∖ (X ∪ Y ), we obtain σv is prime, which contradicts the fact that σ is critical. Consequently, v(σ) -X is even. Now, suppose that X ≥ 4. For each x ∈ X, we have v(σ)-X ∖{x} is odd. It follows from the first assertion that σ[X ∖ {x}] is decomposable. Consequently, σ is critical.

The second result follows from Lemma 4.4 and Corollary 4.6.

Corollary 4.8. Let σ be a critical 2-structure σ such that v(σ) ≥ 5. For e, f ∈ E(P(σ)), we have σe ≃ σf . Proof. Consider distinct e, f ∈ E(P(σ)). It follows from Corollary 4.6 that e and f are contained in the same component of P(σ). Consequently, there exist distinct vertices v 0 , . . . , v p of σ satisfying

• {v 0 , v 1 } = e;
• p ≥ 2, and {v p-1 , v p } = f ;

• for i ∈ {0, . . . , p -1}, {v i , v i+1 } ∈ E(P(σ)).

Let i ∈ {0, . . . , p -1}. We have

v i-1 , v i+1 ∈ N P(σ) (v i ). Since v i ∈ S (σ), it follows from Lemma 4.4 that N P(σ) (v i ) = {v i-1 , v i+1 }, and {v i-1 , v i+1 } is a module of σ -v i . Thus, σ -{v i-1 , v i } ≃ σ -{v i , v i+1 }. It follows that σ -{v 0 , v 1 } ≃ σ - {v p-1 , v p }, that is, σ -e ≃ σ -f .
The third result is an easy consequence of Corollaries 4.7 and 4.8. Corollary 4.9. Let σ be a critical 2-structure σ such that v(σ) ≥

6. Let X, Y ⊊ V (σ) such that X = Y and X ≥ 4. If σ[X] and σ[Y ] are prime, then σ[X] ≃ σ[Y ].
Proof. By Corollary 3.21, there exist x, x ′ ∈ V (σ) ∖ X such that σ -{x, x ′ } is prime. Since σ is critical, we have x ≠ x ′ . Thus, v(σ) -X ≥ 2. We proceed by induction on v(σ) -X ≥ 2. If v(σ) -X = 2, then it suffices to apply Corollary 4.8. Hence, suppose that v(σ) -X ≥ 3. Similarly, there exist distinct y, y ′ ∈ V (σ) ∖ Y such that σ -{y, y ′ } is prime. By Corollary 4.8, σ -{x, x ′ } ≃ σ -{y, y ′ }. Therefore, there exits

Y ′ ⊊ V (σ) ∖ {x, x ′ } such that σ[Y ′ ] ≃ σ[Y ]. Since σ -{x, x ′ } is prime and v(σ) ≥ 6, σ -{x, x ′ }
v(σ) ≥ 6. Consider X, Y ⊊ V (σ) such that σ[X] and σ[Y ] are prime. If 4 ≤ X ≤ Y , then σ[X] embeds into σ[Y ]. Proof. By Corollary 4.9, σ[X] ≃ σ[Y ] if X = Y . Hence, suppose that X < Y . By Corollary 4.7, there exist m > n ≥ 0 such that v(σ) -X = 2m and v(σ) -Y = 2n. By applying (m -n) times Theorem 3.19 from σ[X], we obtain X ⊊ X ′ ⊆ V (σ) such that σ[X ′ ] is prime, and X ′ = Y . By Corollary 4.9, we have σ[X ′ ] ≃ σ[Y ]. It follows that σ[X] embeds into σ[Y ].

The characterization of critical 2-structures

Given a critical 2-structure σ, it follows from Corollary 4.6 that σ has four possible types according to whether P(σ) is isomorphic to P 2n , P 2n ⊕ K {2n} , P 2n+1 or C 2n+1 . The following remark is very useful in the characterization of critical 2-structures of a given type.

Remark 4.11. Consider a set S. We denote by ∆(S) the set of all 2-structures defined on S. We consider the partial order < S defined on ∆(S) as follows. Given σ, τ ∈ ∆(S), σ < S τ if (σ ≠ τ and) for every e ∈ E(σ), there exists f ∈ E(τ ) such that e ⊆ f . Consider σ, τ ∈ ∆(S). As in Remark 2.23, we define their meet σ ∧ τ and their join σ ∨ τ as follows. Given x, y, v, w ∈ V (σ), with x ≠ y and v ≠ w, (x, y) ≡ σ∧τ (v, w) if (x, y) ≡ σ (v, w) and (x, y) ≡ τ (v, w). Hence

E(σ ∧ τ ) = {e ∩ f ∶ e ∈ E(σ), f ∈ E(τ ), e ∩ f ≠ ∅}.
Given x, y, v, w ∈ V (σ), with x ≠ y and v ≠ w, (x, y) ≡ σ∨τ (v, w) if there exists a sequence (e 0 , . . . , e n ) of elements of E(σ)∪E(τ ) such that (x, y) ∈ e 0 , (v, w) ∈ e n , and (when n ≥ 1,) e i ∩ e i+1 ≠ ∅ for 0 ≤ i ≤ n -1. Hence, ∆(S) ordered by < S is a lattice.

Since it is easy to verify that the next fact holds, we omit its proof.

Fact 4.12. Given σ, τ ∈ ∆(S), the following statements hold.

1. If σ < S τ , then all the modules of σ are modules of τ .

2. The modules of σ ∧ τ are exactly the modules of both σ and τ .

We obtain the following consequences.

Fact 4.13. Given σ, τ ∈ ∆(S) such that σ < S τ , the next statements hold.

1. If τ is prime, then σ is prime too.

2. For each n ∈ {3, . . . , S -1}, we have

P n (τ ) ⊆ P n (σ) (see Notation 3.1).
In particular, when τ is prime, we have S (τ ) ⊆ S (σ) and P(τ ) ⊆ P(σ).

It follows from the first statement of Fact 4.13 that the set of the prime 2structures defined on S is an ideal of the lattice (∆(S), < S ). We end the remark with the following consequence of Lemma 4.4, Fact 4.12 and Fact 4.13. Fact 4.14. Consider σ, τ ∈ ∆(S). Suppose that σ and τ are critical. Suppose also that P(σ) = P(τ ). Lastly, suppose that P(σ) does not have isolated vertices. Under these assumptions, we obtain that σ ∧ τ is critical, and

P(σ ∧ τ ) = P(σ).
Proof. To begin, we verify that σ ∧ τ is prime. We have σ ∧ τ ≤ S σ and σ is prime. By the first statement of Fact 4.13, σ ∧ τ is prime. Now, we show that σ ∧ τ is critical, and P(σ ∧ τ ) = P(σ). Let v ∈ V (σ). Since P(σ) does not have isolated vertices, d P(σ) (v) ≠ 0. Since v is a critical vertex of σ, it follows from Lemma 4.4 that d P(σ) (v) = 1 or 2. We distinguish the following two cases.

• Suppose that d P(σ) (v) = 2. Since P(σ) = P(τ ), we have N P(σ) (v) = N P(τ ) (v). Furthermore, since v is a critical vertex of σ and τ , it follows from Lemma 4.4 that

N P(σ) (v) is a nontrivial module of σ -v and τ -v. Note that (σ -v) ∧ (τ -v) = (σ ∧ τ ) -v.
Therefore, it follows from the second statement of Fact 4.12 that N P(σ) (v) is a nontrivial module of (σ ∧ τ )v. Thus, v is a critical vertex of σ ∧ τ , and

N P(σ∧τ ) (v) ⊆ N P(σ) (v).
Lastly, it follows from the second statement of Fact 4.13 that N P(σ) (v) ⊆ N P(σ∧τ ) (v). Consequently, we obtain N P(σ∧τ ) (v) = N P(σ) (v).

• Suppose that d P(σ) (v) = 1. Since P(σ) = P(τ ), we have N P(σ) (v) = N P(τ ) (v). Furthermore, since v is a critical vertex of σ and τ , it follows from Lemma 4.4 that V (σ) ∖ (N P(σ) (v) ∪ {v}) is a nontrivial module of σv and τv. We conclude as in the preceding case.

It follows from both cases above that σ ∧ τ is critical, and P(σ ∧ τ ) = P(σ).

The type P 2n

Proposition 4.15. Given n ≥ 3, consider a 2-structure τ defined on V (τ ) = {0, . . . , 2n -1}. The following two statements are equivalent 1. τ is critical and P(τ ) = P 2n ;

2. ≺0, 1≻ τ ≠≺0, 2≻ τ (see Notation 1.1), and for p, q ∈ {0, . . . , 2n -1} such that p < q, we have

[p, q] τ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [0, 1] τ if p is even and q is odd, [0, 2] τ otherwise. (4.4)
Proof. To begin, suppose that τ is critical and P(τ ) = P 2n . First, we show that (4.4) holds. Consider p, q ∈ {0, . . . , 2n -1} such that p < q. We prove that there exist p ′ ∈ {0, 1} and q ′ ∈ {2n -2, 2n -1} such that p ′ ≡ p mod 2, q ′ ≡ q mod 2, and [p, q] τ = [p ′ , q ′ ] τ . (4.5)

For instance, suppose that p ≥ 2. Since P(τ ) = P 2n , we have N P(τ ) (p -1) = {p -2, p}. By Lemma 4.4, {p -2, p} is a module of τ -(p -1). In particular, we obtain [p, q] τ = [p -2, q] τ . By iteration, we obtain p ′ ∈ {0, 1} such that p ′ ≡ p mod 2 and [p, q] τ = [p ′ , q] τ .

Similarly, we obtain q ′ ∈ {2n -2, 2n -1} such that q ′ ≡ q mod 2 and [p ′ , q] τ = [p ′ , q ′ ] τ . Therefore, (4.5) holds. It follows from (4.5) that for any p ′ , q ′ ∈ {0, . . . , 2n -1}, if p ′ < q ′ , p ′ ≡ p mod 2 and q ′ ≡ q mod 2, then [p, q] τ = [p ′ , q ′ ] τ . (4.6)

We distinguish the following four cases, where p, q ∈ {0, . . . , 2n -1} such that p < q.

• Suppose that p and q are even. By (4.6), [p, q] τ = [0, 2] τ .

• Suppose that p and q are odd. By (4.6), [p, q] τ = [

τ = [0, 2] τ .
• Suppose that p is even and q is odd. By (4.6), [p, q] τ = [0, 1] τ .

• Suppose that p is odd and q is even. By (4.6), [p, q] τ = [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ . Since N P(τ ) (0) = {1}, we have {2, . . . , 2n -1} is a module of τ -0. In particular, we obtain [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ = [1, 2n -2] τ . Since N P(τ ) (2n -1) = {2n -2}, we have {0, . . . , 2n -3} is a module of τ -(2n -1). In particular, we obtain

[1, 2n - 2] τ = [0, 2n -2] τ . By (4.6), [0, 2n -2] τ = [0, 2] τ . Consequently, we obtain [p, q] τ = [0, 2] τ .
It follows from the four cases above that (4.4) holds. Second, we show that ≺ 0, 1

≻ τ ≠≺ 0, 2 ≻ τ . Since τ is prime, τ is neither constant nor linear. It follows from (4.4) that [0, 1] τ ≠ [0, 2] τ . Furthermore, since N P(τ ) (1) = {0, 2}, it follows from Lemma 4.4 that {0, 2} is a module of τ -1. Since τ is prime, {0, 2} is not a module of τ . Therefore, we have [0, 1] τ ≠ [2, 1] τ . Since [2, 1] τ = [2, 0] τ by (4.4), we obtain [0, 1] τ ≠ [2, 0] τ . Consequently, we obtain [0, 1] τ ≠ [0, 2] τ and [0, 1] τ ≠ [2, 0] τ . It follows that ≺0, 1≻ τ ≠≺0, 2≻ τ .
Conversely, suppose that ≺ 0, 1 ≻ τ ≠≺ 0, 2 ≻ τ and (4.4) holds. Since ≺ 0, 1 ≻ τ ≠≺ 0, 2≻ τ , we have

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [0, 1] τ ≠ [0, 2] τ and [0, 1] τ ≠ [2, 0] τ . ( 4
• Let p ∈ {1, . . . , 2n -2}. Consider v ∈ V (τ ) ∖ {p -1, p, p + 1}. Since p -1 ≡ p + 1mod 2, it follows from (4.4) that [v, p -1] τ = [v, p + 1] τ . Therefore, {p -1, p + 1} is a module of τ -p.
It follows that τ is critical. Lastly, we have to prove that P(τ ) = P 2n . Let p ∈ {0, . . . , 2n-2}. The function {0, . . . , 2n -1} ∖ {p, p + 1} → {0, . . . , 2n -3}, defined by q ↦ q if q ≤ p -1 and q ↦ q -2 if q ≥ p + 2, is an isomorphism from τ -{p, p + 1} onto τ [{0, . . . , 2n -3}]. It follows from (4.8) that τ [{0, . . . , 2n-3}] is prime. Hence τ -{p, p+1} is prime too, so {p, p + 1} ∈ E(P(τ )). It follows that E(P 2n ) ⊆ E(P(τ )).

(4.9) By (4.9), 2n -2 ∈ N P(τ ) (2n -1). Furthermore, it follows from (4.10) that N P(τ ) (2n -1) ∩ {1, . . . , 2n -3} = ∅. Finally, by (4.11), 0 ∈ N P(τ ) (2n -1). Thus,

N P(τ ) (2n -1) = {2n -2}.
Consequently, we obtain that P(τ ) = P 2n .

By using Proposition 4.15, we construct critical graphs, digraphs or 2structures of type P 2n that allow us to characterize the critical 2-structures of type P 2n . We use the following notation. Notation 4.16. Let n ≥ 2. Recall that A(L n ) is the set of ordered pairs (p, q), where 0 ≤ p < q ≤ n -1. Given i, j ∈ {0, 1}, set

A(L n ) (i,j) = {(p, q) ∈ A(L n ) ∶ p ≡ i mod 2, q ≡ j mod 2}.
Let τ be a 2-structure defined on V (τ ) = {0, . . . , 2n-1}, where n ≥ 3. Suppose that ≺ 0, 1 ≻ τ ≠≺ 0, 2 ≻ τ . Suppose also that τ satisfies (4.4). By Proposition 4.15, τ is critical and P(τ ) = P 2n . We distinguish the following cases.

1. Suppose that (0, 2) τ = (2, 0) τ .

(a) Suppose that (0, 1) τ = (1, 0) τ . Since ≺ 0, 1 ≻ τ ≠≺ 0, 2 ≻ τ , (0, 1) τ ≠ (0, 2) τ . Therefore, we obtain

E(τ ) = {A(L 2n ) (0,1) ∪ (A(L 2n ) (0,1) ) ⋆ , (see Notation 1.2) (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) (4.12) ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }.
In fact, τ is the 2-structure associated with a graph (see the end of Section 1.2). Given m ≥ 1, we consider the half-graph H 2m defined half-graph on V (H 2n ) = {0, . . . , 2m -1} as follows (see Figure 4.1). For p, q ∈ {0, . . . , 2m -1}, with p ≠ q, {p, q} ∈ E(H 2m ) if there exist 0 ≤ i ≤ j ≤ m -1 such that {p, q} = {2i, 2j + 1}. It follows from (4.12) that τ = σ(H 2n ).

0 • . . 1 • . . 3 • 2 • . . . . . . 2m -2 • 2m -1 • ¨¨¨¨ ¨¨¨¨¨¨¨F igure 4.1: The half-graph H 2m
(b) Suppose that (0, 1) τ ≠ (1, 0) τ . We distinguish the following three subcases. i. Suppose that (1, 0) τ = (0, 2) τ . We obtain

E(τ ) = {A(L 2n ) (0,1) , (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ (4.13) ∪ (A(L 2n ) (0,1) ) ⋆ }.
In fact, τ is the 2-structure associated with a partial order (see Section 1.3). Given m ≥ 2, we consider the partial order Q 2m defined on V (Q 2m ) = {0, . . . , 2m -1} as follows (see Figure 4.2). For p, q ∈ {0, . . . , 2m -1}, with p ≠ q, (p, q) ∈ A(Q 2m ) if there exist 0 ≤ i ≤ j ≤ m -1 such that (p, q) = (2i, 2j + 1).

0 • . . 1 • . . 3 • 2 • . . . . . . 2m -2 • 2m -1 • T ¨¨¨B T ¨¨¨¨¨¨¨B T Figure 4.2: The partial order Q 2m Observe that Comp(Q 2m ) = H 2m 2 . It follows from (4.13) that τ = σ(Q 2n ).
ii. Suppose that (0, 1) τ = (0, 2) τ . We obtain

E(τ ) = {(A(L 2n ) (0,1) ) ⋆ , (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ (4.14) ∪ A(L 2n ) (0,1) }. It follows from (4.14) that τ = σ((Q 2n ) ⋆ ).
iii. Suppose that ≺0, 1≻ τ ∩ ≺0, 2≻ τ = ∅. We obtain

E(τ ) = {A(L 2n ) (0,1) , (A(L 2n ) (0,1) ) ⋆ , (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) (4.15) ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }.
It follows from (4.13), (4.14) and (4.15) that Notation 4.11). 2 The following theorem is due to Gallai [START_REF] Gallai | Transitiv orientierbare Graphen[END_REF][START_REF] Maffray | A translation of Tibor Gallai's Paper: Transitiv orientierbare Graphen[END_REF] . Theorem 4.17 (Gallai [18,[START_REF] Maffray | A translation of Tibor Gallai's Paper: Transitiv orientierbare Graphen[END_REF]). A partial order is prime if and only if its comparability graph is prime.

τ = σ(Q 2n ) ∧ σ((Q 2n ) ⋆ ) (see
Let m ≥ 3. As showed above, σ(H 2n ), and hence H 2n are critical. Moreover, P(H 2n ) = P 2n . It follows from Theorem 4.17 that Q 2n is critical, and P(Q 2n ) = P 2n .

Recall that by Proposition 4.15, σ(Q 2n ) ∧ σ((Q 2n ) ⋆ ) is critical and P(σ(Q 2n )∧σ((Q 2n ) ⋆ )) = P 2n . Note that we obtain the same by using Fact 4.14 because σ(Q 2n ) and σ((Q 2n ) ⋆ ) are critical, and

P(σ(Q 2n )) = P(σ((Q 2n ) ⋆ )) = P 2n . Note also that σ(Q 2n ) ∧ σ((Q 2n ) ⋆ ) = σ(H 2n ) ∧ σ(Q 2n ) = σ(H 2n ) ∧ σ((Q 2n ) ⋆ ).
2. Suppose that (0, 2) τ ≠ (2, 0) τ .

(a) Suppose that (0, 1) τ = (1, 0) τ . We distinguish the following three subcases.

i. Suppose that (0, 1) τ = (2, 0) τ . We obtain

E(τ ) = {(A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ), A(L 2n ) (0,1) ∪ (A(L 2n ) (0,1) ) ⋆ (4.16) ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }.
In fact, τ is the 2-structure associated with a partial order. Given m ≥ 2, we consider the partial order R 2m defined on V (R 2m ) = {0, . . . , 2m -1} as follows (see Figure 4.3). For p, q ∈ {0, . . . , 2m -1}, with p ≠ q, (p, q) ∈ A(R 2n ) if p < q and either p is odd or q is even. Equivalently, R 2m is obtained from the linear order L 2m by removing the arcs (2i, 2j + 1) for 0 ii. Suppose that (0, 1) τ = (0, 2) τ . We obtain

≤ i ≤ j ≤ m -1. 0 • T 2 • 1 • T 3 •
E(τ ) = {A(L 2n ) (0,1) ∪ (A(L 2n ) (0,1) ) ⋆ ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ), (4.17) (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }.
It follows from (4.17) that

τ = σ((R 2n ) ⋆ ).
iii. Suppose that ≺0, 1≻ τ ∩ ≺0, 2≻ τ = ∅. We obtain

E(τ ) = {A(L 2n ) (0,1) ∪ (A(L 2n ) (0,1) ) ⋆ , (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ), (4.18) (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }.
It follows from (4.16), (4.17) and (4.18) that

τ = σ(R 2n ) ∧ σ((R 2n ) ⋆ ). Note that σ(R 2n ) ∧ σ((R 2n ) ⋆ ) = σ(H 2n ) ∧ σ(R 2n ) = σ(H 2n ) ∧ σ((R 2n ) ⋆ ).
(b) Suppose that (0, 1) τ ≠ (1, 0) τ . We distinguish the following five subcases.

i. Suppose that (0, 1) τ = (0, 2) τ . Since ≺ 0, 1 ≻ τ ≠≺ 0, 2 ≻ τ , we have (1, 0) τ ≠ (2, 0) τ . We obtain

E(τ ) = {A(L 2n ) (0,1) ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ), (A(L 2n ) (0,1) ) ⋆ , (4.19) (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }. It follows from (4.19) that τ = σ((Q 2n ) ⋆ ) ∧ σ((R 2n ) ⋆ ).
ii. Suppose that (1, 0) τ = (0, 2) τ . Since ≺ 0, 1 ≻ τ ≠≺ 0, 2 ≻ τ , we have (0, 1) τ ≠ (2, 0) τ . We obtain

E(τ ) = {(A(L 2n ) (0,1) ) ⋆ ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ), A(L 2n ) (0,1) , (4.20) (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }. It follows from (4.20) that τ = σ(Q 2n ) ∧ σ((R 2n ) ⋆ ).
iii. Suppose that (0, 1) τ = (2, 0) τ . Since ≺ 0, 1 ≻ τ ≠≺ 0, 2 ≻ τ , we have

(1, 0) τ ≠ (0, 2) τ . We obtain

E(τ ) = {A(L 2n ) (0,1) ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ , (A(L 2n ) (0,1) ) ⋆ , (4.21) (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) )}. It follows from (4.21) that τ = σ((Q 2n ) ⋆ ) ∧ σ(R 2n ). iv. Suppose that (1, 0) τ = (2, 0) τ . Since ≺ 0, 1 ≻ τ ≠≺ 0, 2 ≻ τ , we have (0, 1) τ ≠ (0, 2) τ .
We obtain

E(τ ) = {(A(L 2n ) (0,1) ) ⋆ ∪ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ , A(L 2n ) (0,1) , (4.22)
(A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) )}. It follows from (4.22) that τ = σ(Q 2n ) ∧ σ(R 2n ).
v. Suppose that ≺0, 1≻ τ ∩ ≺0, 2≻ τ = ∅. We obtain

E(τ ) = {A(L 2n ) (0,1) , (A(L 2n ) (0,1) ) ⋆ (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ), (4.23) (A(L 2n ) (0,0) ∪ A(L 2n ) (1,0) ∪ A(L 2n ) (1,1) ) ⋆ }. It follows from (4.23) that τ = σ(Q 2n ) ∧ σ((Q 2n ) ⋆ ) ∧ σ(R 2n ).
Remark 4.18. We showed previously that σ(H 2n ), σ(Q 2n ), σ((Q 2n ) ⋆ ), σ(R 2n ) and σ((R 2n ) ⋆ ) are critical. Furthermore, their primality graph equals P 2n . We also obtained that some of their meets are also critical, and admit P 2n as primality graph. Observe that, by Fact (4.14), all their meets are also critical, and admit P 2n as primality graph.

We summarize the previous examination in the next theorem. [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF]). Consider a 2-structure τ defined on V (τ ) = {0, . . . , 2n -1}, where n ≥ 3. The following two statements are equivalent • τ is critical, and P(τ ) = P 2n ;

• τ = σ(H 2n ), σ(Q 2n ), σ((Q 2n ) ⋆ ), σ(R 2n ), σ((R 2n ) ⋆ ), σ(Q 2n )∧σ((Q 2n ) ⋆ ), σ(R 2n ) ∧ σ((R 2n ) ⋆ ), σ(Q 2n ) ∧ σ(R 2n ), σ(Q 2n ) ∧ σ((R 2n ) ⋆ ), σ((Q 2n ) ⋆ ) ∧ σ(R 2n ), σ((Q 2n ) ⋆ ) ∧ σ((R 2n ) ⋆ ) or σ(Q 2n ) ∧ σ((Q 2n ) ⋆ ) ∧ σ(R 2n ).
The following result is an immediate consequence of Theorem 4.19.

Corollary 4.20. Consider a reversible 2-structure τ defined on V (τ ) = {0, . . . , 2n-1}, where n ≥ 3. The following two statements are equivalent

• τ is critical, and P(τ ) = P 2n ;

• τ = σ(H 2n ), σ(Q 2n )∧σ((Q 2n ) ⋆ ), σ(R 2n )∧σ((R 2n ) ⋆ ) or σ(Q 2n )∧σ((Q 2n ) ⋆ )∧ σ(R 2n ).
The next remark completes Subsection 4.2.1. We use the following notation.

Notation 4.21. Given n ≥ 2, π n denotes the permutation of {0, . . . , n-1} which exchanges i and (n -1)i for i ∈ {0, . . . , n -1}.

Remark 4.22. Given n ≥ 3, consider a critical 2-structure τ defined on V (τ ) = {0, . . . , 2n -1}, and such that P(τ ) = P 2n . Set

E(τ ) = {A(L 2n ) (0,0) ∪ A(L 2n ) (1,1) ∪ A(L 2n ) (1,0) , (A(L 2n ) (0,0) ∪ A(L 2n ) (1,1) ∪ A(L 2n ) (1,0) ) ⋆ , A(L 2n ) (0,1) , A(L 2n ) ⋆ (0,1) }.
We obtain π 2n (e) = e ⋆ for each e ∈ E(τ ). Thus, we obtain

π 2n (e) = ⋃ f ∈Be π 2n (f ) = ⋃ f ∈Be f ⋆ (by (4.24)) = e ⋆ .
Consequently, π 2n is an isomorphism from τ onto τ ⋆ . If τ is reversible, then τ = τ ⋆ , and hence π 2n is an automorphism of τ .

The type P 2n ⊕ K {2n}

Proposition 4.23. Given n ≥ 2, consider a 2-structure τ defined on V (τ ) = {0, . . . , 2n}. The following two statements are equivalent 1. τ is critical, and P(τ

) = P 2n ⊕ K {2n} ;
2. (0, 1) τ ≠ (1, 0) τ , and for p, q ∈ {0, . . . , 2n -1}, we have

if p < q, then [p, q] τ = [0, 1] τ ; (4.25)
moreover, for i ∈ {0, . . . , n -1}, we have

[2i, 2n] τ = [1, 0] τ and [2i + 1, 2n] τ = [0, 1] τ . (4.26)
Proof. To begin, suppose that τ is critical, and P(τ ) = P 2n ⊕ K {2n} . In a similar way as in the proof of Proposition 4.15, we verify that (4.4) holds.

For a contradiction, suppose that ≺ 0, 1

≻ τ ≠≺ 0, 2 ≻ τ . If n = 2,
Since P(τ ) = P 2n ⊕K {2n} , we have

N P(τ ) (0) = {1}. By Lemma 4.4, {2, . . . , 2n} is a module of τ -0. Since τ is primitive, {0}∪{2, . . . , 2n} is not a module of τ , so [1, 0] τ ≠ [1, 2] τ . Since (4.4) holds, we have [1, 2] τ = [0, 2] τ . Therefore, we obtain [1, 0] τ ≠ [0, 2] τ . It follows from (4.27) that (0, 1) τ ≠ (1, 0) τ and [0, 1] τ = [0, 2] τ . Since (4.4) holds and [0, 1] τ = [0, 2] τ , (4.25) holds.
Lastly, we show that (4.26) holds. As previously seen, {2, . . . , 2n} is a module of τ -0. Hence we have [1, 2n]

τ = [1, 2] τ . Since (4.25) holds, we have [1, 2] τ = [0, 1] τ . We obtain [1, 2n] τ = [0, 1] τ . Let i ∈ {0, . . . , n -2}. Since P(τ ) = P 2n ⊕K {2n} , we have N P(τ ) (2i+2) = {2i+1, 2i+3}. By Lemma 4.4, {2i+1, 2i+3} is a module of τ -(2i). In particular, we have [2i + 1, 2n] τ = [2i + 3, 2n] τ . It follows that [0, 1] τ = [1, 2n] τ = [3, 2n] τ = ⋯ = [2n -1, 2n] τ . Since P(τ ) = P 2n ⊕ K {2n} , we have N P(τ ) (2n -1) = {2n -2}. By Lemma 4.4, {0, . . . , 2n -3} ∪ {2n} is a module of τ -(2n -1). In particular, we have [2n - 2, 2n] τ = [2n-2, 0] τ . Since (4.25) holds, we have [2n-2, 0] τ = [1, 0] τ . We obtain [2n -2, 2n] τ = [1, 0] τ . Let i ∈ {1, . . . , n -1}. Since P(τ ) = P 2n ⊕ K {2n} , we have N P(τ ) (2i -1) = {2i -2, 2i}. By Lemma 4.4, {2i -2, 2i} is a module of τ -(2i -1).
In particular, we have

[2i -2, 2n] τ = [2i, 2n] τ . It follows that [1, 0] τ = [2n -2, 2n] τ = [2n -4, 2n] τ = ⋯ = [0, 2n] τ .
Consequently, (4.26) holds. Conversely, suppose that (4.25) and (4.26) hold. Moreover, suppose that (0, 1) τ ≠ (1, 0) τ . It follows that τ = σ(T 2n+1 ) (see Figure 1.2). By Fact 2.7, τ is prime. We continue with the following observation for every p ∈ {0, . . . , 2n -2}, τ -{p, p + 1} is prime.

(4.28) Lastly, we prove that τ is critical, and P(τ Finally, it follows from (4.25) and (4.26) that {0, . . . , 2n-3}∪{2n} is a module of τ -(2n-1). Therefore, 2n-1 is a critical vertex of τ . By (4.30), 2n ∈ N P(σ) (2n-1). By (4.31), 2n -2 ∈ N P(σ) (2n -1) and N P(σ) (2n -1) ∩ {1, . . . , 2n -3} = ∅. By (4.32), 0 ∈ N P(σ) (2n -1). Consequently, N P(σ) (2n -1) = {2n -2}. It follows that τ is critical, and P(τ

Indeed, consider p ∈ {0, . . . , 2n-2}. Recall that τ -{p, p+1} = σ(T 2n+1 )-{p, p+1}. The bijection ϕ ∶ {0, . . . , 2n} ∖ {p, p + 1} → {0, . . . , 2n -2} q ≤ p -1 → q, q ≥ p + 2 → q -2 is an isomorphism from T 2n+1 -{p, p+1} onto T 2n-1 . Hence, ϕ is an isomorphism from σ(T 2n+1 ) -{p, p + 1} onto σ(T 2n-1
) = P 2n ⊕ K {2n} . As already observed, τ -(2n) = σ(L 2n ). Hence τ -(2n) is decomposable. Furthermore, since τ -(2n) = σ(L 2n ), τ -{p, 2n} is decomposable for each p ∈ {0, . . . , 2n -1}. Thus, N P(σ) (2n) = ∅. ( 4
) = P 2n ⊕ K {2n} .
The next characterization is a simple consequence of Proposition 4.23 and its proof.

) = P 2n ⊕ K {2n} ; • τ = σ(T 2n+1 ) (see Figure 1.2).
In Remark 4.26, we determine the automorphism group of σ(T 2n+1 ), where n ≥ 2. We use the following note. Note 4.25. Consider a tournament T . We denote by Iso(T, T ⋆ ) the set of the isomorphism from T onto its dual. We prove that Aut(σ(T )) = Aut(T ) ∪ Iso(T, T ⋆ ).

(4.33)

Given x, y, v, w ∈ V (σ(T )), with x ≠ y and v ≠ w,

(x, y) ≡ σ(T ) (v, w) if ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (x, y), (v, w) ∈ A(T ) or (x, y), (v, w) ∈ A(T ).
Therefore,

(x, y) ≡ σ(T ) (v, w) if ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (x, y), (v, w) ∈ A(T ) or (x, y), (v, w) ∈ (A(T )) ⋆ . It follows that E(σ(T )) = {A(T ), (A(T )) ⋆ }.
Given a permutation ϕ of V (T ), we have

ϕ ∈ Aut(σ(T )) if and only if ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ϕ(A(T )) = A(T ) and ϕ((A(T )) ⋆ ) = (A(T )) ⋆ or ϕ(A(T )) = (A(T )) ⋆ and ϕ((A(T )) ⋆ ) = A(T ). Clearly, if ϕ(A(T )) = A(T ), then ϕ((A(T )) ⋆ ) = (A(T )) ⋆ . Similarly, if ϕ(A(T )) = (A(T )) ⋆ , then ϕ((A(T )) ⋆ ) = A(T ). Therefore, ϕ ∈ Aut(σ(T )) if and only if ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ϕ(A(T )) = A(T ) or ϕ(A(T )) = (A(T )) ⋆ . We have ϕ(A(T )) = A(T ) if and only if ϕ ∈ Aut(T ). Analogously, ϕ(A(T )) = (A(T )) ⋆ if and only if ϕ ∈ Iso(T, T ⋆ ). It follows that (4.33) holds. Remark 4.26. Let n ≥ 2. We verify that T 2n+1 is rigid. Let ϕ ∈ Aut(T 2n+1 ). Since n ≥ 2, 2n is the only vertex of T 2n+1 such that T 2n+1 -(2n) is a linear order. Consequently, ϕ(2n) = 2n. It follows that ϕ ↾{0,...,2n-1} ∈ Aut(T 2n+1 -(2n)). Since T 2n+1 -(2n) is a linear order, T 2n+1 -(2n) is rigid. Therefore, ϕ ↾{0,...,2n-1} = Id {0,...,2n-1} . Since ϕ(2n) = 2n, we obtain ϕ = Id {0,...,2n} .
We denote by π 2n the extension of π 2n to {0, . . . , 2n} defined by π 2n (2n) = 2n (see Notation 4.21). Clearly, π 2n is an isomorphism from T 2n+1 onto (T 2n+1 ) ⋆ . Conversely, consider an isomorphism ϕ from T 2n+1 onto (T 2n+1 ) ⋆ . Recall that 2n is the only vertex of T 2n+1 such that T 2n+1 -(2n) is a linear order. Hence, 2n is the only vertex of (T 2n+1 ) ⋆ such that (T 2n+1 ) ⋆ -(2n) is a linear order. It follows that ϕ(2n) = 2n. Therefore, ϕ ↾{0,...,2n-1} is an isomorphism from

T 2n+1 -(2n) onto (T 2n+1 ) ⋆ -(2n). Since T 2n+1 -(2n) = L 2n ,
) = P 2n+1 ; 2. (0, 1) τ ≠ (1, 0) τ , [0, 1] τ ≠ [0, 2
] τ , and for p, q ∈ {0, . . . , 2n} such that p < q, we have

[p, q] τ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [0, 2] τ if p and q are even, [0, 1] τ otherwise. (4.35)
Proof. To begin, suppose that τ is critical, and P(τ ) = P 2n+1 . First, we show that (4.35) holds. Consider p, q ∈ {0, . . . , 2n} such that p < q. We prove that there exist p ′ ∈ {0, 1} and q ′ ∈ {2n -1, 2n} such that p ′ ≡ p mod 2, q ′ ≡ q mod 2, and

[p, q] τ = [p ′ , q ′ ] τ . (4.36)
For instance, suppose that p ≥ 2. Since P(τ ) = P 2n+1 , we have N P(τ ) (p -1) = {p -2, p}. By Lemma 4.4, {p -2, p} is a module of τ -(p -1). In particular, we obtain [p, q] τ = [p -2, q] τ . By iteration, we obtain p ′ ∈ {0, 1} such that

p ′ ≡ p mod 2 and [p, q] τ = [p ′ , q] τ .
Similarly, we obtain q ′ ∈ {2n -2, 2n -1} such that q ′ ≡ q mod 2 and [p ′ , q] τ = [p ′ , q ′ ] τ . Therefore, (4.36) holds. It follows from (4.36) that for any p ′ , q ′ ∈ {0, . . . , 2n} such that p ′ < q ′ , if p ′ ≡ p mod 2 and q ′ ≡ q mod 2, then

[p, q] τ = [p ′ , q ′ ] τ . (4.37)
We distinguish the following four cases, where p, q ∈ {0, . . . , 2n} such that p < q.

• Suppose that p and q are even. By (4.37), [p, q] τ = [0, 2] τ .

• Suppose that p and q are odd. By (4.37),

[p, q] τ = [1, 2n -1] τ . Since P(τ ) = P 2n+1 , we have N P(τ ) (2n) = {2n-1}. By Lemma 4.4, {0, . . . , 2n-2} is a module of τ -(2n). In particular, we obtain [1, 2n -1] τ = [0, 2n -1] τ . By (4.37), [0, 2n -1] τ = [0, 1] τ . Consequently, we obtain [p, q] τ = [0, 1] τ .
• Suppose that p is even and q is odd. By (4.37), [p, q] τ = [0, 1] τ .

• Suppose that p is odd and q is even. By (4.37), [p, q] τ = [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ . Since N P(τ ) (0) = {1}, we have {2, . . . , 2n} is a module of τ -0. In particular, we obtain

[1, 2] τ = [1, 2n -1] τ . Since N P(τ ) (2n) = {2n -1}, we have {0, . . . , 2n-2} is a module of τ -(2n). In particular, we obtain [1, 2n-1] τ = [0, 2n -1] τ . By (4.37), [0, 2n -1] τ = [0, 1] τ . Consequently, we obtain [p, q] τ = [0, 1] τ .
It follows from the four cases above that (4.35) holds. Second, we verify that (0,

1) τ ≠ (1, 0) τ and [0, 1] τ ≠ [0, 2] τ . If (0, 1) τ = (1, 0) τ , then {2i ∶ i ∈ {0, .
. . , n}} is a module of τ , which contradicts the fact that τ is critical, and hence prime. Hence (0,

1) τ ≠ (1, 0) τ . Furthermore, if [0, 1] τ = [0, 2] τ , then τ = σ(L 2n+1 ), which contradicts the fact that τ is prime. Thus [0, 1] τ ≠ [0, 2] τ .
Conversely, suppose that (0,

1) τ ≠ (1, 0) τ , [0, 1] τ ≠ [0, 2
] τ , and (4.35) holds. To begin, we prove that τ is prime. We show by induction that for each m ∈ {1, . . . , n} that τ [{0, . . . , 2m}] is prime.

(4.38)

Since [0, 1] τ ≠ [0, 2] τ , {0, 1} and {1, 2} are not modules of τ [{0, 1, 2}]. More- over, {0, 2} is not a module of τ [{0, 1, 2}] because (0, 1) τ ≠ (1, 0) τ . It follows that τ [{0, 1, 2}] is prime. Now, consider m ∈ {1, . . . ,
{0, . . . , 2n} ∖ {p, p + 1} → {0, . . . , 2n -2} q ≤ p -1 → q, (if p ≤ 2n -2) q ≥ p + 2 → q -2. is an isomorphism from τ -{p, p + 1} onto τ [{0, . . . , 2n -2}]. It follows from (4.38) that τ [{0, . . . , 2n -2}] is prime, so τ -{p, p + 1} is as well. E(P 2n+1 ) ⊆ E(P(τ )). (4.40)
Lastly, we prove that τ is critical, and P(τ ) = P 2n+1 . Let p ∈ {1, . . . , 2n -1}. Since (4.35) holds, {p -1, p + 1} is a module of τp. Thus, p is a critical vertex of τ . By (4.40), {p -1, p + 1} ⊆ N P(τ ) (p). Since p is a critical vertex of τ , it follows from Lemma 4.4 that N P(τ ) (p) = {p -1, p + 1}. Therefore, for each p ∈ {1, . . . , 2n -1}, we have

N P(τ ) (p) = {p -1, p + 1}. ( 4
) = P 2n+1 .
Let τ be a 2-structure defined on V (τ ) = {0, . . . , 2n}, where n ≥ 2. Suppose that (0, 1) τ ≠ (1, 0) τ and [0, 1] τ ≠ [0, 2] τ . Suppose also that τ satisfies (4.35). By Proposition 4.27, τ is critical, and P(τ ) = P 2n+1 . We distinguish the following cases.

1. Suppose that (0, 2) τ = (2, 0) τ . We distinguish the following three subcases.

(a) Suppose that (0, 2) τ = (1, 0) τ . We obtain

E(τ ) = {A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) , (A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) ) ⋆ (4.43) ∪ A(L 2n+1 ) (0,0) ∪ (A(L 2n+1 ) (0,0) ) ⋆ } (see Notation 4.16).
In fact, τ is the 2-structure associated with a digraph (see Section 1.3). Given m ≥ 2, we consider the digraph D 2m+1 obtained from the linear order L 2m+1 by removing all the arcs between the even integers (see Figure 4

.4). 0 • 2 • . . . 2i • 2i + 2 • . . . 2m -2 • 2m • 1 • 2i + 1 • 2m -1 • . . . E . . . . . . E . . . j j £ £ £ £ # t t t t s ¡ ¡ ¡ ¡ ! e e e e s
Q e e e e (b) Suppose that (0, 2) τ = (0, 1) τ . We obtain

E(τ ) = {A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) ∪ A(L 2n+1 ) (0,0) ∪ (A(L 2n+1 ) (0,0) ) ⋆ , (4.44) (A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) ) ⋆ }. It follows from (4.44) that τ = σ((D 2n+1 ) ⋆ ). (c) Suppose that ≺0, 1≻ τ ∩ ≺0, 2≻ τ = ∅. We obtain E(τ ) = {A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) , (A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) ) ⋆ , (4.45)
A(L 2n+1 ) (0,0) ∪ (A(L 2n+1 ) (0,0) ) ⋆ }.

Suppose that

(0, 2) τ ≠ (2, 0) τ . Since [0, 1] τ ≠ [0, 2] τ , we have (0, 1) τ = (2, 0) τ and (1, 0) τ = (0, 2) τ or ≺ 0, 1 ≻ τ ∩ ≺ 0, 2 ≻ τ = ∅.
We distinguish the following two subcases.

(a) Suppose that (0, 1) τ = (2, 0) τ and (1, 0) τ = (0, 2) τ . We obtain

E(τ ) = {A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) ∪ (A(L 2n+1 ) (0,0) ) ⋆ , (see Notation 4.16) (A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) ) ⋆ (4.46) ∪ A(L 2n+1 ) (2,0) )} (see Notation 1.2).
In fact, τ is the 2-structure associated with a graph (see Section 1.2). Given m ≥ 2, we consider the tournament U 2m+1 obtained from the linear order L 2m+1 by reversing all the arcs between the even integers (see Figure 4.5).

0 • 2 • . . . 2i • 2i + 2 • . . . 2m -2 • 2m • i ' ' ' 1 • 2i + 1 • 2m -1 • . . . E . . . . . . E . . . j j £ £ £ £ # t t t t s ¡ ¡ ¡ ¡ ! e e e e s
Q e e e e

τ = σ(U 2n+1 ). (b) Suppose that ≺0, 1≻ τ ∩ ≺0, 2≻ τ = ∅.
We obtain We summarize the previous examination in the next theorem.

E(τ ) = {A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) , (A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ (A(L 2n+1 ) (1,1) ) ⋆ , A(L 2n+1 ) (0,0) , (4.47) (A(L 2n+1 ) (0,0) ) ⋆ }.
Theorem 4.28 (Boudabbous and Ille5 [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF]). Consider a 2-structure τ defined such that V (τ ) = {0, . . . , 2n}, where n ≥ 2. The following two statements are equivalent

• τ is critical, and P(τ ) = P 2n+1 ;

• τ = σ(D 2n+1 ), σ((D 2n+1 ) ⋆ ), σ(D 2n+1 )∧σ((D 2n+1 ) ⋆ ), σ(U 2n+1 ), or σ(D 2n+1 )∧ σ((D 2n+1 ) ⋆ ) ∧ σ(U 2n+1 ).
The following result is an immediate consequence of Theorem 4.28.

Corollary 4.29. Consider a reversible 2-structure τ defined such that V (τ ) = {0, . . . , 2n}, where n ≥ 2. The following two statements are equivalent

• τ is critical, and P(τ ) = P 2n+1 ;

• τ = σ(U 2n+1 ), σ(D 2n+1 ) ∧ σ((D 2n+1 ) ⋆ ), or σ(D 2n+1 ) ∧ σ((D 2n+1 ) ⋆ ) ∧ σ(U 2n+1 ).
The next remark completes the subsection.

Remark 4.30. Given n ≥ 2, consider a critical 2-structure τ defined on V (τ ) = {0, . . . , 2n}, and such that P(τ ) = P 2n+1 . Set Thus, we obtain

E(τ ) = {A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) , (A(L 2n+1 ) (0,1) ∪ A(L 2n+1 ) (1,0) ∪ A(L 2n+1 ) (1,1) ) ⋆ , A(L 2n+1 ) (0,0) , A(L 2n+1 ) ⋆ (0,0) }.
π 2n+1 (e) = ⋃ f ∈Be π 2n+1 (f ) = ⋃ f ∈Be f ⋆ (by (4.48)) = e ⋆ .
Consequently, π 2n+1 is an isomorphism from τ onto τ ⋆ . If τ is reversible, then τ = τ ⋆ , and hence π 2n+1 is an automorphism of τ .

The type C 2n+1

Given m ≥ 1, we consider the tournament W 2m+1 obtained from the tournament U 2m+1 by reversing all the arcs between the odd integers (see Figure 4.6). The next remark is useful to establish Proposition 4.36 below.

0 • 2 • . . . 2i • 2i + 2 • . . . 2m -2 • 2m • i ' ' ' 1 • 2i + 1 • 2m -1 • . . . ' . . . . . . ' . . . % % £ £ £ £ # t t t t s ¡ ¡ ¡ ¡ ! e e e e s
Q e e e e Remark 4.31. Let Γ be a group of odd order with identity element e. Consider Ω ⊆ Γ ∖ {e} such that Ω ∩ {x, x -1 } = 1 for each x ∈ Γ ∖ {e}. We associate with Γ and Ω the Cayley tournament Cay(Γ, Ω) defined on V (Cay(Γ, Ω)) = Γ as follows. Given x, y ∈ Γ, (x, y) ∈ A(Cay(Γ, Ω)) if yx -1 ∈ Ω. For each a ∈ Γ, the permutation of Γ, defined by x ↦ xa for every x ∈ Γ, is an automorphism of Cay(Γ, Ω). Consequently, Cay(Γ, Ω) is vertex-transitive. Let m ≥ 1. We consider the cyclic group (Z 2m+1 , +). We consider also the permutation

ψ 2m+1 ∶ {0, . . . , 2m} → {0, . . . , 2m} p → (m + 1) × p mod 2m + 1 of Z 2m+1 . We denote by ψ 2m+1 (W 2m+1 ) the unique tournament defined on Z 2m+1 such that ψ 2m+1 is an isomorphism from W 2m+1 onto ψ 2m+1 (W 2m+1 ). Fact 4.32. For m ≥ 1, we have (ψ 2m+1 (W 2m+1 )) ⋆ = Cay(Z 2m+1 , {1, . . . , m}).
For convenience, set Cay 2m+1 = Cay(Z 2m+1 , {1, . . . , m}).

P(Cay 2m+1 ) = ψ 2m+1 (C 2m+1 ),
where ψ 2m+1 (C 2m+1 ) denotes the unique graph defined on Z 2m+1 such that ψ 2m+1 is an isomorphism from C 2m+1 onto ψ 2m+1 (C 2m+1 ).

Proof. We have

W 2m+1 ∖ {2m -1, 2m} = W 2m-1 . By Fact 4.33, Cay 2m-1 is prime. Since (ψ 2m-1 (W 2m-1 )) ⋆ = Cay 2m-1 by Fact 4.32, W 2m-1 is prime. Hence W 2m+1 ∖ {2m -1,
∖ ψ 2m+1 ({0, -m}) = Cay 2m+1 ∖ ψ 2m+1 ({0, m + 1}) is prime. It follows that m, m + 1 ∈ N P(Cay 2m+1 ) (0).
Clearly, {m, m + 1} is a module of Cay 2m+1 -0. Thus 0 is a critical vertex of Cay 2m+1 . By Lemma 4.4, d P(Cay 2m+1 ) (0) ≤ 2. We obtain

N P(Cay 2m+1 ) (0) = {m, m + 1}.
Let q ∈ Z 2m+1 . Since the permutation of Z 2m+1 , defined by p ↦ (p+q) mod (2m+ 1) for each p ∈ Z 2m+1 , is an automorphism of Cay 2m+1 , we obtain that q is a critical vertex of Cay 2m+1 , and N P(Cay 2m+1 ) (q) = {q+m, q+m+1}. Consequently, Cay 2m+1 is critical, and

P(Cay 2m+1 ) = ψ 2m+1 (C 2m+1 ).
The next fact is an immediate consequence of Facts 4.32 and 4.34. 2. (0, 1) τ ≠ (1, 0) τ , and for p, q ∈ {0, . . . , 2n} such that p < q, we have

[p, q] τ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [1, 0] τ if p and q have the same parity, [0, 1] τ otherwise. ( 4

.51)

Proof. To begin, suppose that τ is critical, and P(τ ) = C 2n+1 . We verify that (4.51) holds in the following manner. Since E(P 2n+1 ) ⊆ E(C 2n+1 ), (4.36) holds. It follows that (4.37) holds. We distinguish the following cases, where p, q ∈ {0, . . . , 2n} such that p < q.

• Suppose that p and q are even. By (4.37),

[p, q] τ = [0, 2] τ . Since P(τ ) = C 2n+1 , we have N P(τ ) (2n) = {0, 2n -1}. By Lemma 4.4, {0, 2n -1} is a module of τ -(2n). In particular, we obtain [0, 2] τ = [2n -1, 2] τ . By (4.37), [2n -1, 2] τ = [1, 0] τ . Thus [p, q] τ = [1, 0] τ . (4.52)
• Suppose that p and q are odd. By (4.37),

[p, q] τ = [1, 2n -1] τ . Since {0, 2n -1} is a module of τ -(2n), we have [1, 2n -1] τ = [1, 0] τ . Hence [p, q] τ = [1, 0] τ .
• Suppose that p is even and q is odd. By (4.37), [p, q] τ = [0, 1] τ .

• Suppose that p is odd and q is even. By (4.37),

[p, q] τ = [1, 2] τ . Since P(τ ) = C 2n+1 , we have N P(τ ) (0) = {1, 2n}. By Lemma 4.4, {1, 2n} is a module of τ -0. In particular, we obtain [1, 2] τ = [2n, 2] τ . By (4.37), [2n, 2] τ = [2, 0] τ . By (4.52), [2, 0] τ = [0, 1] τ . Therefore [p, q] τ = [0, 1] τ .
It follows that (4.51) holds. Since τ is prime, τ is not constant. It follows from (4.51) that (0, 1) τ ≠ (1, 0) τ . Conversely, suppose that (4.51) holds, and (0, 1) τ ≠ (1, 0) τ . We obtain

τ = σ(W 2m+1 ).
By Fact 4.35, τ is critical, and

P(τ ) = C 2m+1 .
The next characterization is a simple consequence of Proposition 4.36 and its proof. In the next remark, we determine the automorphism group of σ(W 2n+1 ) when n ≥ 2.

Remark 4.38. Let n ≥ 2. By Fact 4.32, ψ 2n+1 is an isomorphism from W 2n+1 onto (Cay 2n+1 ) ⋆ . To determine Aut(Cay 2n+1 ), we consider the permutation

θ 2n+1 ∶ {0, . . . , 2n} → {0, . . . , 2n} p → p + 1 mod 2n + 1 of Z 2n+1 . We prove that Aut(σ(W 2n+1 )) =< θ 2n+1 , π 2n+1 > . (4.53)
To begin, we show that

Aut(Cay 2n+1 ) = < θ 2n+1 > . (4.54)
Clearly, θ 2n+1 ∈ Aut(Cay 2n+1 ), and hence

< θ 2n+1 > ⊆ Aut(Cay 2n+1 ). Con- versely, consider ϕ ∈ Aut(Cay 2n+1 ). Since < θ 2n+1 > ⊆ Aut(Cay 2n+1 ), (θ 2n+1 ) -ϕ(0) ○ ϕ ∈ Aut(Cay 2n+1 ). We have ((θ 2n+1 ) -ϕ(0) ○ϕ)(0) = 0. Since Cay 2n+1 [N - Cay 2n+1 (0)] and Cay 2n+1 [N +
Cay 2n+1 (0)] are linear orders, we obtain (θ 2n+1 ) -ϕ(0) ○ϕ = Id {0,...,2n} . Therefore, ϕ ∈ < θ 2n+1 >. It follows that (4.54) holds. Moreover, since ψ 2n+1 is an isomorphism from W 2n+1 onto (Cay 2n+1 ) ⋆ , we obtain

Aut(W 2n+1 ) = (ψ 2n+1 ) -1 ○ < θ 2n+1 > ○ ψ 2n+1 . We have (ψ 2n+1 ) -1 ○ θ 2n+1 ○ ψ 2n+1 = (θ 2n+1 ) 2 . Furthermore, we have ((θ 2n+1 ) 2 ) n+1 = θ 2n+1 . It follows that Aut(W 2n+1 ) = (ψ 2n+1 ) -1 ○ < θ 2n+1 > ○ ψ 2n+1 =< θ 2n+1 > . (4.55)
Now, we show that

Iso(Cay 2n+1 , (Cay 2n+1 ) ⋆ ) = < θ 2n+1 > ○ π 2n+1 . (4.56) Clearly, π 2n+1 ∈ Iso(Cay 2n+1 , (Cay 2n+1 ) ⋆ ). It follows from (4.54) that < θ 2n+1 > ○ π 2n+1 ⊆ Iso(Cay 2n+1 , (Cay 2n+1 ) ⋆ ). Conversely, let ϕ be an isomorphism from Cay 2n+1 onto (Cay 2n+1 ) ⋆ . Since θ 2n+1 ∈ Aut(Cay 2n+1 ) by (4.54), (θ 2n+1 ) n-ϕ(n) ○ ϕ ∈ Iso(Cay 2n+1 , (Cay 2n+1 ) ⋆ ). Set ϕ ′ = (θ 2n+1 ) n-ϕ(n) ○ ϕ. We have ϕ ′ (n) = n. Thus, ϕ ′ [N - Cay 2n+1 (n)] = N + Cay 2n+1 (n) and ϕ ′ [N + Cay 2n+1 (n)] = N - Cay 2n+1 (n). Recall that Cay 2n+1 [N - Cay 2n+1 (n)] = L {0,...,n-1} and Cay 2n+1 [N + Cay 2n+1 (n)] = L {n+1,...,2n} .
It follows that ϕ ′ = π 2n+1 , so ϕ = (θ 2n+1 ) ϕ(n)-n ○ π 2n+1 . Consequently, (4.57) holds. Lastly, since ψ 2n+1 is an isomorphism from W 2n+1 onto (Cay 2n+1 ) ⋆ , we obtain

Iso(W 2n+1 , (W 2n+1 ) ⋆ ) = (ψ 2n+1 ) -1 ○ < θ 2n+1 > ○ π 2n+1 ○ ψ 2n+1 . Since π 2n+1 ○ ψ 2n+1 = (θ 2n+1 ) n ○ ψ 2n+1 ○ π 2n+1 , Iso(W 2n+1 , (W 2n+1 ) ⋆ ) = (ψ 2n+1 ) -1 ○ < θ 2n+1 > ○ (θ 2n+1 ) n ○ ψ 2n+1 ○ π 2n+1 = (ψ 2n+1 ) -1 ○ < θ 2n+1 > ○ ψ 2n+1 ○ π 2n+1 =< θ 2n+1 > ○ π 2n+1 (by (4.55)). (4.57) By Note 4.25, Aut(σ(W 2n+1 )) = Aut(W 2n+1 ) ∪ Iso(W 2n+1 , (W 2n+1 ) ⋆ ). It follows from (4.55) and (4.57) that Aut(σ(W 2n+1 )) =< θ 2n+1 > ∪(< θ 2n+1 > ○ π 2n+1 ). Since π 2n+1 ○ (θ 2n+1 ) k = (θ 2n+1 ) -k ○ π 2n+1 , (4.53) holds.

Properties of critical 2-structures

Lemma 4.39. Let τ be a critical 2-structure, with v(τ ) ≥ 7. If u, v, x, y are distinct vertices of τ such that {u, x}, {x, y}, {y, v} ∈ E(P(τ )), then τ -{x, y} is critical, τ and τ -{x, y} share the same type, and

E(P(τ -{x, y})) = (E(P(τ )) ∖ {{u, x}, {x, y}, {y, v}}) ∪ {{u, v}}.
Proof. By Corollary 4.6, there exist n ≥ 3 and a bijection f defined on V (τ )

such that f (P(τ )) = P 2n ⊕ K {2n} , P 2n+1 , C 2n+1 or f (P(τ )) = P 2n , with n ≥ 4.
To begin, suppose that

f (P(τ )) = C 2n+1 .
We can assume that

f (u) = 2n -2, f (x) = 2n -1, f (y) = 2n and f (v) = 0. Since P(f (τ )) = f (P(τ )), we have P(f (τ )) = C 2n+1 .
By Proposition 4.36, (0, 1) f (τ ) ≠ (1, 0) f (τ ) , and f (τ ) satisfies (4.51). Clearly,

(0, 1) f (τ )-{2n-1,2n} ≠ (1, 0) f (τ )-{2n-1,2n}
, and f (τ ) -{2n -1, 2n} satisfies (4.51).

Since n ≥ 3, we can apply Proposition 4.36 to f (τ ) -{2n -1, 2n}. We obtain that f (τ ) -{2n -1, 2n} is critical, and

P(f (τ ) -{2n -1, 2n}) = C 2n-1 . Since f is an isomorphism from P(τ ) onto C 2n+1 , we obtain E(P(τ )) ={{f -1 (p), f -1 (p + 1)} ∶ 0 ≤ p ≤ 2n -1} ∪ {{f -1 (2n), f -1 (0)}}. (4.58)
Clearly, f ↾V (τ )∖{x,y} is an isomorphism from τ -{x, y} onto f (τ ) -{2n -1, 2n}. Hence, f ↾V (τ )∖{x,y} is an isomorphism from P(τ -{x, y}) onto P(f (τ ) -{2n -1, 2n}). Since P(f (τ ) -{2n -1, 2n}) = C 2n-1 , we obtain

E(P(τ -{x, y})) ={{f -1 (p), f -1 (p + 1)} ∶ 0 ≤ p ≤ 2n -3} ∪ {{f -1 (2n -2), f -1 (0)}}. (4.59)
It follows from (4.58) and (4.59) that

E(P(τ -{x, y}) =(E(P(τ )) ∖ {{f -1 (2n -2), f -1 (2n -1)}, {f -1 (2n -1), f -1 (2n)}, {f -1 (2n), f -1 (0)}}) ∪ {{f -1 (2n -2), f -1 (0)}} =(E(P(τ )) ∖ {{u, x}, {x, y}, {y, v}}) ∪ {{u, v}}.
Now, suppose that f (P(τ )) = P 2n , P 2n ⊕ K {2n} or P 2n+1 . We proceed in the same way for the three cases. For instance, assume that

f (P(τ )) = P 2n+1 . There exists p ∈ {0, . . . , 2n -3} such that f (u) = p, f (x) = p + 1, f (y) = p + 2 and f (v) = p + 3. Since P(f (τ )) = f (P(τ )), we have P(f (τ )) = P 2n+1 . By Proposition 4.27, (0, 1) f (τ ) ≠ (1, 0) f (τ ) , [0, 1] f (τ ) ≠ [0, 2] f (τ ) , and f (τ ) sat- isfies (4.35). Consider the bijection g ∶ {0, . . . , 2n} ∖ {p + 1, p + 2} → {0, . . . , 2n -2} q ≤ p → q, q ≥ p + 3 → q -2.
For each q ∈ {0, . . . , 2n -2}, we have q ≡ g -1 (q) mod 2. Moreover, for any q, r ∈ {0, . . . , 2n -2}, q < r if and only if g -1 (q) < g -1 (r). Since (0, 1)

f (τ ) ≠ (1, 0) f (τ ) , [0, 1] f (τ ) ≠ [0, 2] f (τ )
, and f (τ ) satisfies (4.35), we obtain

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (0, 1) g(f (τ )-{p+1,p+2}) ≠ (1, 0) g(f (τ )-{p+1,p+2}) , [0, 1] g(f (τ )-{p+1,p+2}) ≠ [0, 2] g(f (τ )-{p+1,p+2}) , and
g(f (τ ) -{p + 1, p + 2}) satisfies (4.35).
By Proposition 4.27 applied to g(f

(τ ) -{p + 1, p + 2}), g(f (τ ) -{p + 1, p + 2}) is critical, and P(g(f (τ ) -{p + 1, p + 2})) = P 2n-1 .
Since f is an isomorphism from P(τ ) onto P 2n+1 , we have

E(P(τ )) = {{f -1 (q), f -1 (q + 1)} ∶ 0 ≤ q ≤ 2n -1}. (4.60) Since P(g(f (τ ) -{p + 1, p + 2})) = P 2n-1 , g ○ (f ↾V (τ )∖{x,y}
) is an isomorphism from P(τ -{x, y}) onto P 2n-1 . It follows that

E(P(τ -{x, y})) = {{(g ○ (f ↾V (τ )∖{x,y} )) -1 (q), (g ○ (f ↾V (τ )∖{x,y} )) -1 (q + 1)} ∶ 0 ≤ q ≤ 2n -3}.
We obtain

E(P(τ -{x, y})) = {{f -1 (q), f -1 (q + 1)} ∶ 0 ≤ q ≤ p -1} (when p ≥ 1) ∪ {{f -1 (p), f -1 (p + 3)}} (4.61) ∪ {{f -1 (q), f -1 (q + 1)} ∶ p + 3 ≤ q ≤ 2n -1} (when p ≤ 2n -4).
It follows from (4.60) and (4.61) that Proof. By Corollary 4.6, there exist n ≥ 3 and a bijection f defined on V (τ ) such that f (P(τ )) = P 2n ⊕ K {2n} , P 2n+1 , C 2n+1 or f (P(τ )) = P 2n , with n ≥ 4. Since d P(τ ) (y) = 1, f (P(τ )) = P 2n , P 2n ⊕ K {2n} or P 2n+1 . As in the proof of Lemma 4.39, we treat only the case f (P(τ )) = P 2n+1 . We can assume that f (u) = 2n -2, f (x) = 2n -1 and f (y) = 2n. Since P(f (τ )) = f (P(τ )), we have

E(P(τ -{x, y}) =(E(P(τ )) ∖ {{f -1 (p), f -1 (p + 1)}, {f -1 (p + 1), f -1 (p + 2)}, {f -1 (p + 2), f -1 (p + 3)}}) ∪ {{f -1 (p), f -1 (p + 3)}} =(E(P(τ )) ∖ {{u, x}, {x, y}, {y, v}}) ∪ {{u, v}}.
P(f (τ )) = P 2n+1 .
By Proposition 4.27, (0, 1)

f (τ ) ≠ (1, 0) f (τ ) , [0, 1] f (τ ) ≠ [0, 2] f (τ )
, and f (τ ) satisfies (4.35). Therefore, we have

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (0, 1) f (τ )-{2n-1,2n} ≠ (1, 0) f (τ )-{2n-1,2n} , [0, 1] f (τ )-{2n-1,2n} ≠ [0, 2] f (τ )-{2n-1,2n} , and
f (τ ) -{2n -1, 2n} satisfies (4.35).
By Proposition 4.27 applied to f (τ ) -{2n -1, 2n}, f (τ ) -{2n -1, 2n} is critical, and

P(f (τ ) -{2n -1, 2n}) = P 2n-1 .
Since f is an isomorphism from P(τ ) onto P 2n+1 , we have

E(P(τ )) = {{f -1 (q), f -1 (q + 1)} ∶ 0 ≤ q ≤ 2n -1}. (4.62) Since P(f (τ ) -{2n -1, 2n}) = P 2n-1 , f ↾V (τ )∖{x,y} is an isomorphism from P(τ - {x, y}) onto P 2n-1 .
we obtain

E(P(τ -{x, y})) = {{f -1 (q), f -1 (q + 1)} ∶ 0 ≤ q ≤ 2n -3}. (4.63)
It follows from (4.62) and (4.63) that

E(P(τ -{x, y}) =E(P(τ )) ∖ {{f -1 (2n -2), f -1 (2n -1)}, {f -1 (2n -1), f -1 (2n)}} =E(P(τ )) ∖ {{u, x}, {x, y}}.
The following corollary is an immediate consequence of Lemmas 4.39 and 4.40. It is useful in the next chapter when disjoint edges of the primality graph of a critical 2-structure are considered.

Corollary 4.41. Given a critical 2-structure τ , with v(τ ) ≥ 7, consider distinct vertices x and y of τ such that {x, y} ∈ E(P(τ )). The following two statements hold.

1. τ -{x, y} is critical; 2. τ and τ -{x, y} share the same type; 3. For every e ∈ E(P(τ )), if e ∩ {x, y} = ∅, then e ∈ E(P(τ -{x, y}));

4. For every e ∈ E(P(τ -{x, y})), if e ∖ (N P(τ ) (x) ∪ N P(τ ) (y)) ≠ ∅, then e ∈ E(P(τ )).
Chapter 5

Noncritical unordered pair theorems

Given a prime 2-structure, a noncritical unordered pair theorem provides distinct vertices v and w of σ such that σ -{v, w} is prime as well.

We refine the notion of a support as follows. Given a 2-structure σ, the critical support of σ is the set of the vertices v of σ such that σv is critical. It critical support is denoted by S c (σ).

Remark 5.1. Let σ be a prime 2-structure. Suppose that

S (σ) ∖ S c (σ) ≠ ∅. Let v ∈ S (σ) ∖ S c (σ). Since v ∈ S (σ), σ -v is prime. Since v ∈ S c (σ), σ -v is not critical. Hence, there exists w ∈ V (σ -v) such that (σ -v) -w is prime. Therefore, {v, w} is a noncritical unordered pair of σ. Remark 5.2. Let σ be a prime 2-structure, with v(σ) ≥ 6. Suppose that S c (σ) ≤ 1.
It follows from Theorem 3.11 that there exists X ⊆ V (σ) such that 3 ≤ X ≤ 5 and S c (σ) ⊆ X.

(5.1) By Corollary 3.21, there exist v, w ∈ V (σ) ∖ X such that σ -{v, w} is prime.

Clearly, if v ≠ w, then {v, w} is a noncritical unordered pair of σ. Hence, suppose that v = w. We obtain v ∈ S (σ). Since S c (σ) ⊆ X, we have v ∈ S (σ) ∖ S c (σ), and we conclude as in Remark 5.1.

The Schmerl-Trotter theorem

Theorem 5.3 (Schmerl and Trotter [33] 1 ). Given a prime 2-structure σ such that v(σ) ≥ 7, there exist v, w ∈ V (σ) such that v ≠ w and σ -{v, w} is prime.

Theorem 5.3 is the second downward hereditary property of primality. We use the properties of critical 2-structures presented in Section 4.3 to prove it. Our approach is based on Remark 5.1.

In this section, we provide a proof of Theorem 5.3 when v(σ) ≥ 9. In Chapter 6, we provide a proof of Theorem 5.3, when v(σ) ≥ 7, by using Theorem 5.23. We begin with the following lemma.

Lemma 5.4. Let σ be a prime 2-structure, with v(σ) ≥ 8, such that S c (σ) ≠ ∅. Consider x ∈ S c (σ). Let E ⊆ E(P(σ -x)) such that e ∩ f = ∅ for distinct e, f ∈ E. If E ≥ 4, then E ∩ E(P(σ)) ≠ ∅.
Proof. Suppose that E ∩ E(P(σ)) = ∅. We have to show that E ≤ 3. Hence, suppose that E ≥ 3. We have to show that

E = 3. Given e ∈ E, set X e = V (σ) ∖ ({x} ∪ e). Since e ∈ E(P(σ -x)), σ[X e ] is prime. Since e ∈ E(P(σ)), σ[X e ∪ {x}]
is decomposable. By Lemma 3.13, x ∈ ⟨X e ⟩ σ or there exists u e ∈ X e such that x ∈ (X e ) σ (u e ).

Given distinct e, f ∈ E, set

X {e,f } = X e ∖ f.
Since e ∩ f = ∅, it follows from Corollary 4.41 applied to σ -

x that f ∈ E(P(σ[X e ])), that is, σ[X {e,f } ] is prime. (5.2)
For a contradiction, suppose that there is e ∈ E such that x ∈ ⟨X e ⟩ σ . For each f ∈ E ∖ {e}, we have

x ∈ ⟨X {e,f } ⟩ σ . (5.3) If there is f ∈ E ∖ {e} such that x ∈ ⟨X f ⟩ σ , then V (σ) ∖ {x} is a module of σ,
which contradicts the fact that σ is prime. Thus, suppose that for every

f ∈ E ∖ {e}, there is u f ∈ X f such that x ∈ (X f ) σ (u f ). Let f ∈ E ∖ {e}. If u f ∈ e, then x ∈ (X {e,f } ) σ (u f ). By (5.3), x ∈ (X {e,f } ) σ (u f ) ∩ ⟨X {e,f } ⟩ σ , which contradicts Lemma 3.13. Therefore, for every f ∈ E ∖ {e}, u f ∈ e.
(5.4)

Since E ≥ 3, consider distinct f, g ∈ E ∖ {e}. By (5.4), u f , u g ∈ e. If u f = u g , then {x, u f } is a module of σ, which contradicts the fact that σ is prime. Hence u f ≠ u g . Recall that σ[X {f,g}
] is prime by Corollary 4.41. We obtain that x ∈ (X {f,g} ) σ (u f ) ∩ (X {f,g} ) σ (u g ), which contradicts Lemma 3.13. It follows that for each e ∈ E, there is u e ∈ X e such that x ∈ (X e ) σ (u e ). Given distinct e, f ∈ E, if u e = u f , then {x, u e } is a module of σ, which contradicts the fact that σ is prime. Hence, for distinct e, f ∈ E, we have u e ≠ u f .

(5.5) The next result follows from Corollary 4.6 and Lemma 5.4.

Given distinct e, f ∈ E, if u e ∈ f and u f ∈ e, then x ∈ (X {e,f } ) σ (u e )∩(X {e,f } ) σ (u f )
Corollary 5.5. Given a prime 2-structure σ, consider x ∈ S c (σ). If v(σ) ≥ 9, then E(P(σ -x)) ∩ E(P(σ)) ≠ ∅.
A first proof of Theorem 5.3 when v(σ) ≥ 9. If σ is critical, then E(P(σ)) ≠ ∅ by Corollary 4.6. Suppose that σ is not critical, so S (σ) ≠ ∅. If S c (σ) ≠ ∅, then we conclude by using Corollary 5.5. Lastly, if S c (σ) = ∅, then we conclude as in Remark 5.1.

Remark 5.6. By using Corollary 4.6, we can directly verify that Corollary 5.5 holds when v(σ) = 7 or 8.

The next result improves the Schmerl-Trotter theorem when the critical support is nonempty.

Proposition 5.7. Let σ be a prime 2-structure such that v(σ) ≥ 9. If S c (σ) ≠ ∅, then E(P(σ)) ≥ ⌈ v(σ) 2 ⌉ -4. Proof. Consider x ∈ S c (σ). Set n = ⌈ v(σ) 2 ⌉.
We have n ≥ 5. We verify that P 2n-2 embeds into P(σx).

• Suppose that v(σ) is even. We obtain v(σ) = 2n, so v(σx) = 2n -1. It follows from Corollary 4.6 that P(σx) is isomorphic to P 2n-2 ⊕ K {2n-2} , P 2n-1 , or C 2n-1 . Thus, P 2n-2 embeds into P(σx).

• Suppose that v(σ) is odd. We obtain v(σ) = 2n -1, so v(σ -x) = 2n -2.
It follows from Corollary 4.6 that P(σx) is isomorphic to P 2n-2 .

Since P 2n-2 embeds into P(σx), there exists a function f ∶ {0, . . . , 2n

-3} → V (σ -x) such that f is an isomorphism from P 2n-2 onto P(σ -x)[{f (p) ∶ 0 ≤ p ≤ 2n -3}]. Set F = {{f (2m), f (2m + 1)} ∶ 0 ≤ m ≤ n -2}.
Clearly, F ⊆ E(P(σx)). It follows from Lemma 5.4 that F ∖ E(P(σ)) ≤ 3.

We obtain

E(P(σ)) ≥ F ∩ E(P(σ)) = F -F ∖ E(P(σ)) = (n -1) -F ∖ E(P(σ))
≥ n -4.

Ille's theorem

Ille [START_REF] Ille | Indecomposable graphs[END_REF] succeeded in providing conditions that ensure the existence of a noncritical unordered pair outside a prime substructure of a prime 2-structure.

Theorem 5.8 (Ille [START_REF] Ille | Indecomposable graphs[END_REF]). Given a prime 2-structure σ, consider

X ⊆ V (σ) such that σ[X] is prime. If V (σ) ∖ X ≥ 6, then there exist v, w ∈ V (σ) ∖ X such that v ≠ w and σ -{v, w} is prime.
The first proof of Theorem 5.8 is technical and unclear. A new clearer and shorter proof is provided in Section 9.6 at the end of Chapter 9. Belkhechine et al. [3] improved Theorem 5.8 in particular cases as follows.

Theorem 5.9 (Belkhechine et al. [3]). Given a prime 2-structure σ, consider X ⊆ V (σ) such that σ[X] is prime. Suppose that at least one of the following statements holds

(S 1 ) there exists v ∈ ⟨X⟩ σ such that (v, X) σ ≠ (X, v) σ (see Notation 2.1); (S 2 ) there exist y ∈ X and v ∈ X σ (y) such that (v, y) σ ≠ (y, v) σ .
Under these assumptions, if V (σ) ∖ X ≥ 4, then there exist v, w ∈ V (σ) ∖ X such that v ≠ w and σ -{v, w} is prime.

Sayar [32] proved Theorem 5.9 for tournaments. Obviously, Statements (S 1 ) and (S 2 ) above are satisfied by tournaments. We provide a proof of Theorem 5.9 in Section 9.6 as well.

The Boudabbous-Ille theorem

Boubabbous and Ille [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF] succeeded in finding a noncritical unordered pair which intersects the support. Note that the proof of the next result uses Theorem 5.3.

V (C) ∩ S (σ) ≠ ∅. Since C is connected, there exist distinct v, w ∈ V (C) such that {v, w} ∈ E(P(σ)) and v ∈ S (σ). Thus, v ∈ S c (σ).
As shown by the next result, Theorem 5.10 does not hold when S (σ) = 1. For convenience, we use the following notation. Notation 5.11. Given n ≥ 3, set Remark 5.12. Given n ≥ 3, it follows from Theorem 4.19 that the elements of R 2n are the critical 2-structures σ defined on {0, . . . , 2n-1} such that P(σ) = P 2n and (0, 2) σ ≠ (2, 0) σ . Theorem 5.13 (Boubabbous and Ille [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF] 3 ). Consider a prime 2-structure σ such that v(σ) ≥ 6, and S (σ) = 1. The primality graph P(σ) admits a unique component

R 2n = {σ(R 2n ), σ((R 2n ) ⋆ ), σ(R 2n ) ∧ σ((R 2n ) ⋆ ), σ(Q 2n ) ∧ σ(R 2n ), σ(Q 2n ∧ σ((R 2n ) ⋆ ), σ((Q 2n ) ⋆ ) ∧ σ(R 2n ), σ((Q 2n ) ⋆ ) ∧ σ((R 2n ) ⋆ ), σ(Q 2n ) ∧ σ((Q 2n ) ⋆ ) ∧ σ(R 2n )} (see Figures 4.2
C such that v(C) ≥ 2. Moreover, if V (C) ∩ S (σ) = ∅, then v(σ) = 2n + 1,
where n ≥ 3, and there exists an isomorphism ϕ from σ -S (σ) onto an element of R 2n satisfying 4

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [S (σ), ϕ -1 ({2i ∶ i ∈ {0, . . . , n -1}})] σ = [ϕ -1 (0), ϕ -1 (2)] σ , and [S (σ), ϕ -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ = [ϕ -1 (2), ϕ -1 (0)] σ .
(5.7)

2 Boubabbous and Ille [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF] proved this theorem for digraphs. 3 Boubabbous and Ille [START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF] proved this theorem for digraphs. 4 The digraph R 2n+1 (see Figure 5.1) is the extension of R 2n (sse Figure 4.3) to {0, . . . , 2n} defined by Proof. Denote by x the unique element of S (σ). By Theorem 3.11, there exists X ⊆ V (σ) such that x ∈ X, 3 ≤ X ≤ 5, and σ[X] is prime. It follows from Corollary 3.21 that there exist v, w ∈ V (σ) ∖ X such that σ -{v, w} is prime. Since S (σ) ⊆ X, we have v ≠ w. Denote by C the component of P(σ) containing v and w. For a contradiction, suppose that P(σ) admits a component D such that v(D) ≥ 2 and

A(R 2n+1 ) = A(R 2n ) ∪ {(2n, 2i) ∶ 0 ≤ i ≤ n -1} ∪ {(2i + 1, 2n) ∶ 0 ≤ i ≤ n -1}. By using the fact that σ(R 2n+1 ) -(2n) is prime, it is not difficult to verify that σ(R 2n+1 ) is prime. We have S (σ(R 2n+1 )) = {2n} and C = P 2n , so V (C) ∩ S (σ) = ∅. Furthermore, σ(R 2n+1 ) satisfies (5.7) with ϕ = Id {0,...,2n-1} . 0 • T 2 • 1 • T 3 • r r
D ≠ C. Since V (C) ∩ V (D) = ∅ and S (σ) = 1, we have V (C) ∩ S (σ) = ∅ or V (D) ∩ S (σ) = ∅. For instance, assume that V (C) ∩ S (σ) = ∅. Since V (C) ∩ V (D) = ∅ and v(D) ≥ 2, we obtain V (σ) ∖ V (C) ≥ 2, which contradicts Proposition 4.5. Consequently, C is the unique component of P(σ) such that v(C) ≥ 2. Now, suppose that V (C) ∩ S (σ) = ∅, so x ∈ V (C). Since x ∈ V (C), we have d P(σ) (x) = 0. Therefore, σ -x is critical. It follows from Proposition 4.5 that V (C) = V (σ)
∖{x}, and C is isomorphic to P 2n , where n = (v(σ)-1) 2. Consider an isomorphism ϕ from C onto P 2n . As in the proof of Proposition 4.15, we verify that

≺ϕ -1 (0), ϕ -1 (1)≻ σ ≠≺ϕ -1 (0), ϕ -1 (2)≻ σ (see Notation 1.1),
and for any p, q ∈ {0, . . . , 2n -1} such that p < q, we have

[ϕ -1 (p), ϕ -1 (q)] σ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [ϕ -1 (0), ϕ -1 (1)] σ if p is even and q is odd, [ϕ -1 (0), ϕ -1 (2)] σ otherwise. (5.8)
Let i ∈ {0, . . . , n -2}. Since ϕ -1 is an isomorphism from P 2n onto C, we have

N P(σ) (ϕ -1 (2i+1)) = {ϕ -1 (2i), ϕ -1 (2i+2)}. By Lemma 4.4, {ϕ -1 (2i), ϕ -1 (2i+2)} is a module of σ-ϕ -1 (2i+1). In particular, we obtain [x, ϕ -1 (2i)] σ = [x, ϕ -1 (2i+ 2)] σ . It follows that [x, ϕ -1 ({2i ∶ i ∈ {0, . . . , n -1}}] σ = [x, ϕ -1 (2n -2)] σ .
Since ϕ -1 is an isomorphism from P 2n onto C, we have

N P(σ) (ϕ -1 (2n -1)) = {ϕ -1 (2n -2)}. By Lemma 4.4, V (σ) ∖ {ϕ -1 (2n -2), ϕ -1 (2n -1)} is a module of σ-ϕ -1 (2n-1). In particular, we obtain [x, ϕ -1 (2n-2)] σ = [ϕ -1 (0), ϕ -1 (2n-2)] σ . Moreover, we have [ϕ -1 (0), ϕ -1 (2n-2)] σ = [ϕ -1 (0), ϕ -1 (2)] σ by (5.8). It follows that [x, ϕ -1 ({2i ∶ i ∈ {0, . . . , n -1}}] σ = [ϕ -1 (0), ϕ -1 (2)] σ .
Similarly, we show that

[x, ϕ -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}}] σ = [ϕ -1 (2), ϕ -1 (0)] σ .
Consequently, (5.7) holds. Since σ is prime,

V (σ) ∖ {x} is not a module of σ. It follows that [ϕ -1 (0), ϕ -1 (2)] σ ≠ [ϕ -1 (2), ϕ -1 (0)] σ . Therefore, [0, 2] τ ≠ [2, 0] τ ,
where τ is the unique 2-structure defined on {0, . . . , 2n -1} such that ϕ is an isomorphism from σx onto τ . Hence τ is critical and P(τ ) = P 2n . As observed in Remark 5.12, we have

τ ∈ R 2n because [0, 2] τ ≠ [2, 0] τ .
Since the elements of R 2n are not symmetric, the next result follows from Theorems 5.10 and 5.13.

Corollary 5.14. Consider a symmetric 2-structure σ such that v(σ) ≥ 7. If σ is prime and noncritical, then there exists v ∈ V (σ) such that σv is prime and noncritical, as well.

Proof. Suppose that σ is prime and noncritical. Hence, S (σ) ≠ ∅. If S (σ) ≥ 2, then we conclude by using Theorems 5.10. Therefore, suppose that S (σ) contains a unique element denoted by x. By Theorem 5.13, P(σ) admits a unique component C such that v(C) ≥ 2. Since σ is symmetric, σx is not isomorphic to an element of R 2n by Remark 5.12. It follows from Theorem 5.13 that x ∈ V (C). Since v(C) ≥ 2, we have d P(σ) (x) ≠ 0. Thus, σx is prime and noncritical. Theorem 5.13 leads us to introduce the following definition. It is useful to generalize the Chudnovsky-Seymour theorem (see Theorem 5.21). Definition 5.15. Consider a prime 2-structure σ such that v(σ) ≥ 5. Suppose that S (σ) admits a unique element, denoted by x. We say that σ is almost

critical if σ -x is critical (that is, S (σ) = S c (σ) = {x}).
almost critical Remark 5.16. Consider a prime 2-structure σ such that v(σ) ≥ 6. Suppose that S (σ) admits a unique element denoted by x. By Theorem 5.13, P(σ) admits a unique component C such that v(C) ≥ 2. Suppose also that σ is almost critical, that is, σx is critical. Since σx is critical, we have

N P(σ) (x) = ∅.
As seen in the proof of Theorem 5.13, it follows from Proposition 4.5 that

V (C) = V (σ) ∖ {x},
and there exists an isomorphism ϕ from C onto P 2n . Furthermore, ϕ is an isomorphism from σx onto a critical 2-structure τ such that P(τ ) = P 2n . Thus, ϕ is an isomorphism from P(σx) onto P 2n . It follows that

P(σ -x) = P(σ) -x = C.
The next result is an easy consequence of Theorem 5.13.

Corollary 5.17. Given a 2-structure σ such that v(σ) ≥ 7, the following two statements are equivalent

1. σ is almost critical; 2. v(σ) = 2n + 1,
where n ≥ 3, and there exist x ∈ V (σ) and an isomorphism ϕ from σx onto an element of R 2n (see Notation 5.11) such that (5.7) holds.

Proof. To begin, suppose that σ is almost critical. Hence, there exists x ∈ V (σ) such that S (σ) = S c (σ) = {x}.

As seen in Remark 5.16, V (σ)∖{x} is the unique component of P(σ) containing at least two elements. Clearly, (V (σ) ∖ {x}) ∩ S (σ) = ∅, and it suffices to apply Theorem 5.13 to obtain the second statement above. Conversely, suppose that v(σ) = 2n + 1, where n ≥ 3, and suppose that there exist x ∈ V (σ) and an isomorphism ϕ from σx onto an element τ of R 2n such that (5.7) holds. As observed in Remark 5.12, τ is critical and P(τ ) = P 2n . Hence, σx is critical.

(5.9)

Set X = V (σ) ∖ {x}.
We prove that σ is prime. As observed in Remark 5.12, we have

(0, 2) τ ≠ (2, 0) τ . It follows that (ϕ -1 (0), ϕ -1 (2)) σ ≠ (ϕ -1 (2), ϕ -1 (0)) σ . Since (5.7) holds, we have [x, ϕ -1 ({2i ∶ i ∈ {0, . . . , n -1}})] σ = [ϕ -1 (0), ϕ -1 (2)] σ and [x, ϕ -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ = [ϕ -1 (2), ϕ -1 (0)] σ . It follows that x ∈ ⟨X⟩ σ . Now, consider y ∈ X. Set p = ϕ(y).
There exists q ∈ {0, . . . , 2n -1} such that ≺p, q ≻ τ =≺0, 1≻ τ . It follows that

≺ϕ -1 (p), ϕ -1 (q)≻ σ =≺ϕ -1 (0), ϕ -1 (1)≻ σ .
Moreover, by Proposition 4.15, ≺0, 1≻ τ ≠≺0, 2≻ τ . It follows that

≺ϕ -1 (0), ϕ -1 (1)≻ σ ≠≺ϕ -1 (0), ϕ -1 (2)≻ σ .
Since (5.7) holds, we have

≺x, ϕ -1 (q)≻ σ =≺ϕ -1 (0), ϕ -1 (2)≻ σ .
Therefore, we obtain

≺x, ϕ -1 (q)≻ σ ≠≺ϕ -1 (p), ϕ -1 (q)≻ σ . It follows that x ∈ X σ (y).
By Lemma 3.13, x ∈ Ext σ (X), so σ is prime. Since σx is critical by (5.9), we obtain x ∈ S c (σ).

(5.10)

Lastly, we show that (V (σ) ∖ {x}) ∩ S (σ) = ∅. Consider y ∈ V (σ) ∖ {x}. We have to verify that σy is decomposable. Set p = ϕ(y).

Suppose that p ∈ {1, . . . , 2n -2}. Since P(τ ) = P 2n , we have

N P(τ ) (p) = {p - 1, p + 1}. By Lemma 4.4, {p -1, p + 1} is a module of τ -p. It follows that {ϕ -1 (p-1), ϕ -1 (p+1)} is a module of σ -{x, ϕ -1 (p)}. Since (5.7) holds, we have x ←→ σ {ϕ -1 (p-1), ϕ -1 (p+1)}. It follows that {ϕ -1 (p-1), ϕ -1 (p+1)} is a module of σ-ϕ -1 (p).
Hence, σ-ϕ -1 (p) is decomposable. Now, suppose that p = 0. Since P(τ ) = P 2n , we have N P(τ ) (0) = {1}. By Lemma 4.4, τ -{0, 1} is a module of τ -0. Precisely, since τ is critical and P(τ

) = P 2n , we have [1, {2, . . . , 2n -1}] τ = [0, 2] τ . It follows that [ϕ -1 (1), ϕ -1 ({2, . . . , 2n-1})] σ = [ϕ -1 (0), ϕ -1 (2)] σ . Since (5.7) holds, we have [ϕ -1 (1), x] σ = [ϕ -1 (0), ϕ -1 (2)] σ . It follows that V (σ) ∖ {ϕ -1 (0), ϕ -1 (1)} is a module of σ -ϕ -1 (0). Hence, σ -ϕ -1 (0) is decomposable. Similarly, σ -ϕ -1 (2n -1) is decomposable. Consequently, we obtain (V (σ) ∖ {x}) ∩ S (σ) = ∅. It follows from (5.10) that S c (σ) = S (σ) = {x}.
Thus, σ is almost critical.

We complete the section with the following properties of almost critical 2structures.

Fact 5.18. Consider an almost critical 2-structure σ such that v(σ) ≥ 7. The following two statements hold, where x denotes the unique element of S (σ),

1. given X ⊆ V (σ -x), if σ[X] is prime, then σ[X ∪ {x}] is prime; 2. for e, f ∈ E(P(σ -x)) (or for e, f ∈ E(P(σ))), we have σ -e ≃ σ -f .
Proof. Consider an isomorphism ϕ from σx onto an element ρ of R 2n , where n ≥ 3, satisfying (5.7).

For the first statement, consider X ⊆ V (σx) such that σ[X] is prime. Let y ∈ X. We verify that x ∈ X σ (y). Since (5.7) holds, we have

≺x, z ≻ σ =≺ϕ -1 (0), ϕ -1 (2)≻ σ for every z ∈ X ∖ {y}. Since ρ ∈ R 2n , it follows from Proposition 4.15 that there exists z ∈ X ∖ {y} such that ≺y, z ≻ σ =≺ϕ -1 (0), ϕ -1 (1)≻ σ .
Moreover, we have

≺ϕ -1 (0), ϕ -1 (2)≻ σ ≠≺ϕ -1 (0), ϕ -1 (1)≻ σ by Proposition 4.15. It follows that x ∈ X σ (y).
Now, we verify that x ∈ ⟨X⟩ σ . Consider i, j ∈ {0, . . . , n -1} such that i < j.

It follows from Proposition 4.15 that [ϕ -1 (2i), ϕ -1 (2j)] σ = [ϕ -1 (0), ϕ -1 (2)] σ . Therefore, σ[ϕ -1 ({2i ∶ i ∈ {0, . . . , n -1}}] is constant or linear. Since σ[X]
is prime, we obtain X ∖ ϕ -1 ({2i ∶ i ∈ {0, . . . , n -1}} ≠ ∅. Thus, there exits p ∈ {0, . . . , n-1} such that ϕ -1 (2p+1) ∈ X. Similarly, there exists q ∈ {0, . . . , n-1} such that ϕ -1 (2q) ∈ X. Since (5.7) holds, we have

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [x, ϕ -1 (2q)] σ = [ϕ -1 (0), ϕ -1 (2)] σ and [x, ϕ -1 (2p + 1)] σ = [ϕ -1 (2), ϕ -1 (0)] σ .
Since ρ ∈ R 2n , it follows from Remark 5.12 that

[ϕ -1 (0), ϕ -1 (2)] σ ≠ [ϕ -1 (2), ϕ -1 (0)].

Therefore, we have

x ∈ ⟨X⟩ σ .

It follows from Lemma 3.13 that

x ∈ Ext σ (X), that is, σ[X ∪ {x}] is prime.
For the second statement, consider e, f ∈ E(P(σx)). To begin, we make the following observation. By Remark 5.16, P(σ -x) = P(σ)-x, and x is isolated in P(σ). It follows that

E(P(σ -x)) = E(P(σ) -x).
Consequently, we can consider e, f ∈ E(P(σ)) as well.

By Remark 5.16, ϕ is an isomorphism from P(σ-x) onto P 2n . By exchanging e and f if necessary, we can suppose that e = {ϕ -1 (i), ϕ -1 (i + 1)} and f = {ϕ -1 (j), ϕ -1 (j + 1)}, where 0

≤ i < j ≤ 2n -2.
Consider the bijection f from {0, . . . , 2n -1} ∖ {i, i + 1} onto {0, . . . , 2n -1} ∖ {j, j + 1} defined as follows. Given m ∈ {0, . . . , 2n -1} ∖ {i, i + 1},

f (m) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ m if i ≥ 1 and 0 ≤ m ≤ i -1, m -2 if i + 2 ≤ m ≤ j + 1, m if j ≤ 2n -3 and j + 2 ≤ m ≤ 2n -1.
By Remark 5.12, P(ρ) = P 2n . It follows from Proposition 4.15 that for p, q ∈ {0, . . . , 2n -1}, with p < q, we have

[p, q] ρ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [0, 1] ρ if p is even and q is odd, [0, 2] ρ otherwise.
Since f is strictly increasing and preserves the parity, f is an isomorphism from ρ -{i, i + 1} onto ρ -{j, j + 1}. Now, consider the bijection ψ from V (σ) ∖ e onto V (σ) ∖ f defined by ψ(x) = x, and ψ(w) = (ϕ -1 ○ f ○ ϕ)(w) for every w ∈ V (σ) ∖ (e ∪ {x}). Since ϕ satisfies (5.7) and f preserves the parity, ψ is an isomorphism from σe onto σf . ). Let σ be a symmetric 2structure. If σ is prime and noncritical, then for every prime 2-structure τ such that τ embeds into σ, with 5 ≤ v(τ ) < v(σ), there exists X ⊊ V (σ) such that σ[X] ≃ τ and Ext σ (X) ≠ ∅ (see Notation 3.12).

The Chudnovsky-Seymour theorem

Proof. We consider a prime 2-structure τ , such that v(τ ) ≥ 5, and we proceed by induction on v(σ) ≥ v(τ ) + 1. The result is obvious when v(σ) = v(τ ) + 1. Hence, suppose that v(σ) ≥ v(τ ) + 2. We have v(σ) ≥ 7 because v(τ ) ≥ 5. By Corollary 5.14, we have S (σ) ∖ S c (σ) ≠ ∅.

To begin, we prove that there exists X ⊊ V (σ) such that

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ σ[X] ≃ τ and (V (σ) ∖ X) ∩ S (σ) ≠ ∅.
(5.11)

Consider Y ⊆ V (σ) such that σ[Y ] ≃ τ ,
V (C) ⊆ V (σ) ∖ S (σ). By Proposition 4.5, V (σ) ∖ V (C) ≤ 1, so S (σ) ≤ 1. Since σ is not critical, we have S (σ) = 1. By Theorem 5.13, C is the unique component of P(σ) such that v(C) ≥ 2. Since V (C) ∩ S (σ) = ∅, it
follows from Theorem 5.13 that σ is almost critical, which contradicts the fact that σ is symmetric (see Remark 5.12). Consequently, we have

V (C)∩S (σ) ≠ ∅. Since σ -u is decomposable for every u ∈ V (σ) ∖ Y , we have {v, w} ∩ S (σ) = ∅.
Since C is connected, there exist distinct vertices c 0 , . . . , c p of C satisfying

• {c 0 , c 1 } = {v, w}; • p ≥ 2, {c 0 , . . . , c p-1 } ⊆ V (σ) ∖ S (σ)
, and c p ∈ S (σ);

• for i ∈ {0, . . . , p -1}, {c i , c i+1 } ∈ E(P(σ)). Let i ∈ {1, . . . , p -1}. We have c i-1 , c i+1 ∈ N P(σ) (c i ). Since c i ∈ S (σ), it follows from Lemma 4.4 that N P(σ) (c i ) = {c i-1 , c i+1 }, and {c i-1 , c i+1 } is a module of σ-c i . Thus, σ-{c i-1 , c i } ≃ σ-{c i , c i+1 }. It follows that σ-{c 0 , c 1 } ≃ σ-{c p-1 , c p }, that is, σ -{v, w} ≃ σ -{c p-1 , c p }.
Since τ embeds into σ -{v, w}, τ embeds into σ -{c p-1 , c p } as well. Since c p ∈ S (σ), (5.11) holds. Now, we consider X ⊆ V (σ) such that (5.11) holds. There exists

v ∈ (V (σ) ∖ X) ∩ S (σ).
If there exists w ∈ (V (σ) ∖ X) ∩ (S (σ) ∖ S c (σ)), then it suffices to apply the induction hypothesis to σw. Hence, suppose that

(V (σ) ∖ X) ∩ (S (σ) ∖ S c (σ)) = ∅.
In particular, σv is critical. Since S (σ) ∖ S c (σ) ≠ ∅, there exists

x ∈ X ∩ (S (σ) ∖ S c (σ)). Since σ -v is a critical symmetric 2-
∈ (V (σ) -v) ∖ {x} such that {x, y} ∈ E(P(σ -v)). Since v(σ)-v(τ ) ≥ 2, we have X ⊊ V (σ -v). Since σ -v is critical, it follows from Corollary 3.21 that there exist distinct w, w ′ ∈ V (σ -v) ∖ X such that {w, w ′ } ∈ E(P(σ -v)). Thus, τ embeds into (σ -v) -{w, w ′ }. Since {x, y}, {w, w ′ } ∈ E(P(σ -v)), it follows from Corollary 4.8 that (σ -v) -{x, y} ≃ (σ -v) -{w, w ′ }.
Therefore, τ embeds into (σv) -{x, y} as well. To conclude, it suffices to apply the induction hypothesis to σx.

Remark 5.20. Theorem 5.19 does not hold for almost critical 2-structures. Indeed, given n ≥ 3, consider the 2-structure ρ 2n+1 defined on {0, . . . , 2n} by

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ρ 2n+1 -(2n) = σ(R 2n ) (see Figure 4.3) [2n, {2i ∶ i ∈ {0, . . . , n -1}}] ρ2n+1 = [0, 2] ρ2n+1 , and [2n, {2i + 1 ∶ i ∈ {0, . . . , n -1}}] ρ2n+1 = [2, 0] ρ2n+1 .
(5.12) By Corollary 5.17, ρ 2n+1 is almost critical. As observed in Remark 5.16, we have

P(ρ 2n+1 -(2n)) = P(ρ 2n+1 ) -(2n) = P 2n . Therefore, ρ 2n+1 -{2n -2, 2n -1} is prime. Set τ = ρ 2n+1 -{2n -2, 2n -1}. Consider X ⊆ {0, . . . , 2n} such that τ is isomorphic to ρ 2n+1 [X]. Since τ is prime, ρ 2n+1 [X] is prime. It follows that V (ρ 2n+1 ) ∖ X ∈ E(P(ρ 2n+1 )).
As observed in Remark 5.16, we have

P(ρ 2n+1 ) = P 2n ⊕ K {2n} .
It follows that there exists p ∈ {0, . . . , 2n -2} such that

X = V (ρ 2n+1 ) ∖ {p, p + 1}.
Finally, to establish that Theorem 5.19 does not hold for ρ 2n+1 , we verify that p, p + 1 ∈ Ext ρ2n+1 (X).

Since P(ρ 2n+1 ) = P 2n ⊕ K {2n} , we have

N P(ρ2n+1) (p + 1) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ {p, p + 2} if p ≤ 2n -3 or {p} if p = 2n -2.
It follows from Lemma 4.4 that p ∈ ⟨X⟩ ρ2n+1 ∪ X ρ2n+1 (p + 2).

In the same way, we verify that p + 1 ∈ X ρ2n+1 (p -1).

We generalize Theorem 5.19 as follows.

Theorem 5.21 (Liu [27]6 ). Let σ be a prime 2-structure σ. Suppose that σ is neither critical nor almost critical. For each prime 2-structure τ such that τ embeds into σ, with 5 ≤ v(τ ) < v(σ), there exists

X ⊆ V (σ) satisfying X ≠ V (σ), σ[X] ≃ τ , and Ext σ (X) ≠ ∅.
Theorem 5.21 is proved in Appendix A. The next result is obtained by applying Theorem 5.21 several times. Theorem 5.22. Let σ be a prime 2-structure. Suppose that σ is neither critical nor almost critical. Consider a prime 2-structure τ such that τ embeds into σ, with 5 ≤ v(τ ) < v(σ). Under these assumptions, there exists X ⊊ V (σ) such that σ[X] ≃ τ , and the elements of V (σ) ∖ X can be indexed as z 1 , . . . , z n in such a way that σ[X ∪ {z 1 , . . . , z i }] is prime for i ∈ {1, . . . , n}.

The critical support

The purpose of this section is to demonstrate the next theorem. We begin with the following lemma (compare with Corollary 4.10).

Lemma 5.24. Let σ be a prime 2-structure with v(σ) ≥ 6.

Consider X, Y ⊆ V (σ) such that σ[X] and σ[Y ] are critical. Suppose that X ≤ Y . If there exists Z ⊆ X ∩ Y such that σ[Z] is prime and Z ≥ 5, then σ[X] embeds into σ[Y ].
Proof. We can suppose that Z = 5 or 6. Indeed, suppose that Z ≥ 7. By Theorem 3.10, there exists Z ′ ⊆ Z such that σ[Z ′ ] is prime and Z ′ = 3 or 4. By Theorem 3.19, there exists

Z ′′ ⊆ Z such that Z ′ ⊆ Z ′′ , Z ′′ = Z ′ + 2,
(σ[X]) is isomorphic to C 2m+1 , P 2m ⊕ K {2m} or P 2m+1 . Suppose that P(σ[X]) ≃ C 2m+1 . Hence P(σ[Y ]) ≃ C 2n+1 . It follows from Theorem 4.37 that σ[X] embeds into σ[Y ]. Similarly, if P(σ[X]) ≃ P 2m ⊕ K {2m} , then it follows from Theorem 4.24 that σ[X] embeds into σ[Y ].
Therefore, suppose that P(σ[X]) ≃ P 2m+1 . Thus, P(σ[Y ]) ≃ P 2n+1 . Consider an isomorphism ϕ X from P(σ[X]) onto P 2m+1 . Denote by τ X the unique 2-structure defined on {0, . . . , 2m} such that ϕ X is an isomorphism from σ[X] onto τ X . We obtain that τ X is critical and P(τ X ) = P 2m+1 . It follows from Proposition 4.27 that (0, 1)

τ X ≠ (1, 0) τ X , [0, 1] τ X ≠ [0, 2] τ X ,
and for any p, q ∈ {0, . . . , 2m} such that p < q, we have

[p, q] τ X = [0, 2] τ X if p and q are even [0, 1] τ X otherwise.
(5.13)

There exist x 0 , . . . , x 4 ∈ {0, . . . , 2m} such that x 0 < ⋯ < x 4 and

ϕ X (Z) = {x 0 , . . . , x 4 }. Since σ[Z] is isomorphic to τ X [ϕ X (Z)], τ X [ϕ X (Z)] is prime too. By (5.13), if x 0 is odd, then ϕ X (Z)∖{x 0 } is a module of τ X [ϕ X (Z)]. Thus, x 0 is even. Given i ∈ {0, . . . , 3}, if x i ≡ x i+1 mod 2, then it follows from (5.13) that {x i , x i+1 } is a module of τ X [ϕ X (Z)]. Therefore, x i ≡ x i+1 mod 2. It follows that ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
x 0 , x 2 , x 4 are even and x 1 , x 3 are odd.

(

Let f X ∶ ϕ X (Z) ←→ {0, . . . , 4} defined by f X (x i ) = i for i ∈ {0, . . . , 4}. Clearly, f X is strictly increasing. By (5.14), f X preserves the parity. It follows that

f X is an isomorphism from τ X [ϕ X (Z)] onto τ X [{0, . . . , 4}]. Therefore, σ[Z] is isomorphic to τ X [{0, . . . , 4}]
. Since (5.13) holds, we have

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [0, 2] τ X = [0, 4] τ X = [2, 4] τ X and [0, 1] τ X = [0, 3] τ X = [1, 2] τ X = [1, 3] τ X = [1, 4] τ X = [2, 3] τ X = [3, 4] τ X . Since (0, 1) τ X ≠ (1, 0) τ X and [0, 1] τ X ≠ [0, 2] τ X ,
τ Y ≠ (1, 0) τ Y , [0, 1] τ Y ≠ [0, 2] τ Y ,
and for any p, q ∈ {0, . . . , 2n} such that p < q, we have

[p, q] τ Y = [0, 2] τ Y if p and q are even [0, 1] τ Y otherwise. ( 5
τ X = τ Y [{0, . . . , 2m}]. Since σ[X] ≃ τ X and σ[Y ] ≃ τ Y , σ[X] embeds into σ[Y ].
2. Suppose that ψ = π 5 . By Remark 4.30,

ψ is an isomorphism from τ X [{0, . . . , 4}] onto (τ X [{0, . . . , 4}]) ⋆ . Since ψ is also an isomorphism from τ X [{0, . . . , 4}] onto τ Y [{0, . . . , 4}], we obtain (τ X ) ⋆ [{0, . . . , 4}] = τ Y [{0, . . . , 4}].
Clearly, (τ X ) ⋆ is critical and P((τ X ) ⋆ ) = P 2m+1 . It follows from Proposition 4.27 that for p, q ∈ {0, . . . , 2m} such that p < q, we have

[p, q] (τ X ) ⋆ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [0, 2] (τ X ) ⋆ if p and q are even [0, 1] (τ X ) ⋆ otherwise. Since (τ X ) ⋆ [{0, . . . , 4}] = τ Y [{0, . . . , 4}
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ≺0, 1≻ µ ≠≺0, 2≻ µ [0, 1] µ = [0, 3] µ = [0, 5] µ = [2, 3] µ = [2, 5] µ = [4, 5] µ , and [0, 2] µ = [0, 4] µ = [1, 2] µ = [1, 3] µ = [1, 4] µ = [1, 5] µ = [2, 4] µ = [3, 4] µ = [3, 5] µ ,
It follows from Proposition 4.15 that P(µ) = P 6 . We obtain also a prime 2-

structure ν = τ Y [{0, . . . , 5}] such that ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ≺0, 1≻ ν ≠≺0, 2≻ ν [0, 1] ν = [0, 3] ν = [0, 5] ν = [2, 3] ν = [2, 5] ν = [4, 5] ν , and [0, 2] ν = [0, 4] ν = [1, 2] ν = [1, 3] ν = [1, 4] ν = [1, 5] ν = [2, 4] ν = [3, 4] ν = [3, 5] ν ,
It follows from Proposition 4.15 that P(ν) = P 6 . Furthermore, there exists an isomorphism ψ from µ onto ν. Thus, ψ is an automorphism of P 6 . We obtain ψ = Id {0,...,5} or π 6 . As previously, we deduce that τ

Y [{0, . . . , 2m -1}] = τ X or (τ X ) ⋆ . Since π 2m is an isomorphism from τ X onto (τ X ) ⋆ , τ X embeds into τ Y . Thus, σ[X] embeds into σ[Y ].
The next result follows from Lemma 5.24.

Corollary 5.25. Let σ be a prime 2-structure with v(σ) ≥ 7. Consider distinct s, t ∈ S c (σ). We have

N P(σ-s) (t) = N P(σ-t) (s), and N P(σ-s) (t) ≠ ∅. Moreover, if v(σ) ≥ 8, then σ -s ≃ σ -t. Proof. We have N P(σ-s) (t) = {x ∈ V (σ -s) ∖ {t} ∶ (σ -s) -{t, x} is prime}. Similarly, N P(σ-t) (s) = {x ∈ V (σ -t) ∖ {s} ∶ (σ -t) -{s, x} is prime}. Thus, N P(σ-s) (t) = N P(σ-t) (s).
For a contradiction, suppose that N P(σ-s) (t) = ∅. It follows from Corollary 4.6 that there exists an isomorphism ϕ s from P(σs) onto P 2n ⊕ K {2n} , where v(σ) = 2n + 2. Furthermore, since N P(σ-s) (t) = ∅, ϕ s (t) = 2n. Denote by τ s the unique 2-structure defined on {0, . . . , 2n} such that ϕ s is an isomorphism from σs onto τ s . We obtain that τ s is critical and P(τ s ) = P 2n ⊕ K {2n} . By Theorem 4.24, τ s = σ(T 2n+1 ). Similarly, there exists an isomorphism ϕ t from σt onto σ(T 2n+1 ) such that ϕ t (s) = 2n. Since ϕ s (t) = 2n and ϕ t (s) = 2n, (ϕ s ) ↾V (σ)∖{s,t} ○ ((ϕ t ) ↾V (σ)∖{s,t} ) -1 is an automorphism of σ(T 2n+1 ) -(2n). Furthermore, since T 2n+1 -(2n) = L 2n , σ(T 2n+1 ) -(2n) is linear, and hence σ(T 2n+1 ) -(2n) is rigid. Therefore, (ϕ s ) ↾V (σ)∖{s,t} = (ϕ t ) ↾V (σ)∖{s,t} . It follows that {s, t} is a module of σ, which contradicts the fact that σ is prime. Consequently, N P(σ-s) (t) ≠ ∅.

Lastly, suppose that v(σ) ≥ 8. Since N P(σ-s) (t) = N P(σ-t) (s) and N P(σ-s) (t) ≠ ∅, there exists v ∈ N P(σ-s) (t) ∩ N P(σ-t) (s). We have σ -{s, t, v} is prime. Since v(σ -{s, t, v}) ≥ 5, it follows from Lemma 5.24 that σs ≃ σt. Proposition 5.26. Let σ be a prime 2-structure with v(σ) ≥ 7. If there exists s ∈ S c (σ) such that P(σs) ≃ C 2n+1 , then S c (σ) = {s}. Proof. Let s ∈ S c (σ) be such that P(σs) ≃ C 2n+1 , where n ≥ 3. Up to isomorphism, we can assume that

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ V (σ) = {0, . . . , 2n + 1}, s = 2n + 1, and
P(σ -(2n + 1)) = C 2n+1 .
For a contradiction, suppose that

S c (σ) ≥ 2, ( 5
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ϕ(2n + 1) = 2n, ϕ(0) = 0, and ϕ(2n -1) = 2n -1.
(5.17)

Since σ -(2n + 1) is critical and P(σ

-(2n + 1)) = C 2n+1 , it follows from Propo- sition 4.36 that (0, 1) σ ≠ (1, 0) σ , (5.18)
and for p, q ∈ {0, . . . , 2n} such that p < q, we have

[p, q] σ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [0, 1] σ if p ≡ q mod 2 [1, 0] σ otherwise. (5.19)
Set E = {{2n -2, 2n -1}, {2n -1, 2n}, {2n, 0}}. Since E ⊆ E(P(σ -(2n + 1))), it follows from Lemma 4.39 that (σ -(2n + 1)) -{2n -1, 2n} is critical, and

E(P((σ -(2n + 1)) -{2n -1, 2n})) = (E(P(σ -(2n + 1)) ∖ E) ∪ {{2n -2, 0}}.
We obtain

P(σ -{2n -1, 2n, 2n + 1}) = C 2n-1 .
(5.20)

Similarly, we obtain (ϕ -1 (0), ϕ -1 (1)) σ ≠ (ϕ -1 (1), ϕ -1 (0)) σ , and for p, q ∈ {0, . . . , 2n} such that p < q, we have

[ϕ -1 (p), ϕ -1 (q)] σ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [ϕ -1 (0), ϕ -1 (1)] σ if p ≡ q mod 2 [ϕ -1 (1), ϕ -1 (0)] σ otherwise. (5.21) Furthermore, ϕ ↾(V (σ-(2n))∖{ϕ -1 (2n-1),ϕ -1 (2n)}) is an isomorphism from P((σ - (2n) -{ϕ -1 (2n -1), ϕ -1 (2n)}) onto C 2n-1
. By (5.17), ϕ ↾{0,...,2n-2} is an isomorphism from P(σ -{2n -1, 2n, 2n + 1}) onto C 2n-1 . It follows from (5.20) that ϕ ↾{0,...,2n-2} ∈ Aut(C 2n-1 ). Since ϕ(0) = 0, we obtain ϕ ↾{0,...,2n-2} = Id {0,...,2n-2} or π 2n-1 (see Notation 4.21).

We distinguish the following two cases. In each of them, we obtain a contradiction.

1. Suppose that ϕ ↾{0,...,2n-2} = Id {0,...,2n-2} . Hence, ϕ(i) = i for i ∈ {0, . . . , 2n-2}. Since ϕ(2n -1) = 2n -1 by (5.17), we obtain

ϕ(i) = i for i ∈ {0, . . . , 2n -1}. (5.22)
Consider k ∈ {0, . . . , 2n -1}. For instance, assume that k is even. We obtain

[k, 2n] σ = [1, 0] σ by (5.19) = [ϕ -1 (1), ϕ -1 (0)] σ by (5.22) = [ϕ -1 (k), ϕ -1 (2n)] σ by (5.21) = [k, ϕ -1 (2n)] σ by (5.22) = [k, 2n + 1] σ by (5.17).
The same holds when k is odd. It follows that {2n, 2n + 1} is a module of σ, which contradicts the fact that σ is prime.

2. Suppose that ϕ ↾{0,...,2n-2} = π 2n-1 . Therefore, for each i ∈ {0, . . . , 2n -2}, we have

ϕ -1 (i) = 2n -2 -i. (5.23)
It follows that Proposition 5.28. Let σ be a prime 2-structure with v(σ) ≥ 7. Suppose that there exists s ∈ S c (σ) such that P(σs) ≃ P 2n+1 . Also, suppose that there exists t ∈ S c (σ) ∖ {s}. Under these assumptions, the following statements hold

[0, 1] σ = [0, 2n -1] σ by (5.19) = [ϕ -1 (0), ϕ t ) -1 (2n -1)] σ by (5.17) = [ϕ -1 (0), ϕ -1 (
• d P(σ-s) (t) = 2;
• by denoting by x and y the elements of N P(σ-s) (t), the function

V (σ) ∖ {s} → V (σ) ∖ {t} t → s, x → y, y → x, v ∈ V (σ) ∖ {s, t, x, y} → v, (5.24)
is an isomorphism from σs onto σt;

• note that (x, y) σ = (y, x) σ ;

• S c (σ) = {s, t}.

Proposition 5.29. Let σ be a prime 2-structure with v(σ) ≥ 7. Suppose that there exists s ∈ S c (σ) such that P(σs) ≃ P 2n . Also, suppose that there exists t ∈ S c (σ) ∖ {s}. We can suppose that

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ V (σ) = {0, . . . , 2n}, s = 2n, t ∈ {n, . . . , 2n -1}, and
P(σ -(2n)) = P 2n .
(5.25)

Under these assumptions, one of the following two cases holds 1. d P(σ-(2n)) (t) = 1, and we have

1.1. t = 2n -1, 1.2. (0, 2) σ = (2, 0) σ , 1.3. the function {0, . . . , 2n -1} → {0, . . . , 2n -2} ∪ {2n} 0 → 2n, 1 → 2n -2, 2 ≤ k ≤ 2n -1 → k -2, ( 5.26)
is an isomorphism from σ -(2n) onto σt;

2. d P(σ-(2n)) (t) = 2, and we have

2.1. n ≤ t ≤ 2n -2, 2.2. the function {0, . . . , 2n -1} → {0, . . . , 2n} ∖ {t} t → 2n, t -1 → t + 1, t + 1 → t -1, v ∈ V (σ) ∖ {t -1, t, t + 1, 2n} → v, (5.27)
is an isomorphism from σ -(2n) onto σt; in particular, we have

(t -1, t + 1) σ = (t + 1, t -1) σ .
In both cases above, we have S c (σ) = {t, 2n}.

Chapter 6

Minimal prime 2-structures Definition 6.1. Let σ be a prime 2-structure. Consider a vertex subset W of σ. We say that σ is minimal for

W if for each W ′ ⊊ V (σ) such that W ⊆ W ′ minimal and W ′ ≥ 3, we have σ[W ′ ] is decomposable.
Cournier and Ille [START_REF] Cournier | Minimal indecomposable graph[END_REF] characterized the prime digraphs that are minimal for a vertex subset of size 1 or 2. The purpose of this chapter is to extend their characterization to prime 2-structures. The next question follows naturally. Question 6.2. Given k ≥ 3, characterize the prime 2-structures that are minimal for a vertex subset of size k1 .

Minimal and prime 2-structures for a singleton

Consider a prime 2-structure σ. Given v ∈ V (σ), suppose that σ is minimal for {v}. It follows from Theorem 3.11 that

v ∈ R 3 (σ) ∪ R 4 (σ) ∪ R 5 (σ) (see Notation 3.1).
Hence, there exists X ⊆ V (σ) such that 3 ≤ X ≤ 5, v ∈ X, and σ[X] is prime. Since σ is minimal for {v}, we obtain X = V (σ). Therefore, we have

3 ≤ v(σ) ≤ 5.
We examine only the minimal and prime 2-structures for one vertex that are defined on five vertices. For instance, it follows from Claims 3.4 and 3.5 that B 5 is prime and minimal for {4}. We use the following set of 2-structures. • σ is prime and minimal for {v};

• there exists an isomorphism f from σ onto an element of M 1 such that

f (v) = 4.
The proof of Theorem 6.5 is a long sequence of easy verifications. We omit it, but we provide the following hint.

Hint for a proof of Theorem 6.5. To begin, suppose that there exists an isomorphism from σ onto τ ∈ M 1 such that f (v) = 4. By Remark 6.4, τ is prime and minimal for {4}. Thus, σ is prime and minimal for {v}.

Conversely, suppose that σ is prime and minimal for {v}. Up to isomorphy, we can assume that V (σ) = {0, . . . , 4} and v = 4. We prove that σ ∈ M 1 .

Since σ is minimal for {4}, we have

4 ∈ R 3 (σ) ∪ R 4 (σ). (6.1)
We show that P 3 (σ) ∪ P 4 (σ) = {{0, . . . , 3}}. (6.2) By Theorem 3.10, there exists X ∈ P 3 (σ) ∪ P 4 (σ). By (6.1), v ∈ X. As in the proof of Theorem 3.11, we obtain that σ[X ∪ {4}] is prime. Since σ is minimal for {4}, we have V (σ) = X ∪ {4}, so X = {0, . . . , 3}. Consequently, (6.2) holds. It follows from (6.2) that σ[{0, . . . , 3}] is critical. Up to isomorphy, we can assume that

• {2, 3} is a module of σ[{0, . . . , 3}] -0; • {0, 2} is a module of σ[{0, . . . , 3}] -1; • {1, 3} is a module of σ[{0, . . . , 3}] -2; • {0, 1} is a module of σ[{0, . . . , 3}] -3.
2 By the first two assertions, σ -4 satisfies the second statement of Proposition 4.15. Proposition 4.15 does not hold for 2-structures of size 4 because the primality graph of a prime 2-structure of size 4 is empty. Nevertheless, we can directly verify that a 2-structure of size 4, which satisfies the second statement of Proposition 4.15, is critical. Therefore, we can deduce here that σ -4 is critical. 3 Cournier and Ille [START_REF] Cournier | Minimal indecomposable graph[END_REF] proved this theorem for digraphs.

It follows that

[0, 1] σ = [0, 3] σ = [2, 3] σ and [0, 2] σ = [1, 2] σ = [1, 3] σ . Therefore, we obtain ≺0, 1≻ σ ≠≺0, 2≻ σ .
We prove that ≺i, 4≻ σ = ≺0, 1≻ σ or ≺0, 2≻ σ (6.3) for i ∈ {0, . . . , 3}. By using Proposition 3.8 and (6.2), we show that (6.3) holds for i = 0 or 1. Since the permutation (03)( 12) is an isomorphism from σ -4 onto (σ -4) ⋆ , we obtain that (6.3) holds for i = 2 or 3. Finally, by using Proposition 3.8 and (6.2), we verify that [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF][START_REF] Bonizonni | Primitive 2-structures with the (n -2)-property[END_REF]

σ = [4, 3] σ = [0, 3] σ and [1, 4] σ = [4, 2] σ = [1, 2] σ . Therefore, σ ∈ M 1 .

Minimal and prime 2-structures for an unordered pair

Given n ≥ 4, it is easy to verify that σ(P n ) (see Figure 1.1) is prime and minimal for {0, n -1}. Furthermore, for n ≥ 3, M n is the tournament defined on V (M n ) = {0, . . . , n -1} as follows. Given i, j ∈ {0, . . . , n -1}, (i, j) ∈ A(M n ) if j = i + 1 or j < i -1 (see Figure 6.1). Given n ≥ 5, it is easy to verify that σ(M n ) is prime and minimal for {0, n -1}.

0 • 1 • . . . . n -2 • n -1 • E E E E s s © © Figure 6.1: The tournament M n .
We generalize σ(P n ) and σ(M n ) as follows. Notation 6.6. We denote by M 2 the set of the 2-structures σ defined on V (σ) = {0, . . . , n -1}, where n ≥ 3, and satisfying the following assertions

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ for i ∈ {0, . . . , n -3} and j ∈ {i + 2, . . . , n -1}, [i, j] σ = [0, n -1] σ and for i ∈ {0, . . . , n -2}, [i, i + 1] σ ≠ [0, n -1] σ . (6.4)
We use the next result to verify that the elements of M 2 are prime.

Lemma 6.7. Consider a 2-structure σ ∈ M 2 . If M is a nontrivial module of σ, then M = {0, n -1}. Proof. Consider i, j ∈ M such that i < j and {m ∈ M ∶ i ≤ m ≤ j} = {i, j}. Suppose that i ≥ 1. Since [i -1, i] σ ≠ [0, n -1] σ and [i -1, j] σ = [0, n -1] σ , we have i -1 ∈ M .
By proceeding by induction, we obtain {0, . . . , i} ⊆ M . Similarly, we have {j, . . . , n -1} ⊆ M . Therefore, we have M = {0, . . . , i} ∪ {j, . . . , n -1}.

Since M ≠ {0, . . . , n-1}, we have j ≥ i+2.

If i ≥ 1, then [i-1, i+1] σ = [0, n-1] σ and [i, i + 1] σ ≠ [0, n -1] σ , which contradicts the fact that M is a module of σ because i -1, i ∈ M and i + 1 ∈ M . It follows that i = 0.
Similarly, we have j = n -1. Consequently, we obtain M = {0, n -1}. Lemma 6.8. Given σ ∈ M 2 , if v(σ) ≥ 5, then σ is prime and minimal for {0, n -1}.

Proof. Let σ ∈ M 2 . We have V (σ) = {0, . . . , n -1}, where n ≥ 5. First, we verify that σ is prime. For a contradiction, suppose that σ admits a nontrivial module M . By Lemma 6.7, M = {0, n -1}. We have [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF][START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF]

σ = [0, n -1] σ and [n -1, 2] σ = [n -1, 0] σ . It follows that [0, n -1] σ = [n -1, 0] σ . We obtain [0, 1] σ ≠ [0, n -1] σ and [n -1, 1] σ = [n -1, 0] σ = [0, n -1] σ ,
which contradicts the fact that {0, n -1} is a module of σ. Consequently, σ is prime.

Second, we verify that σ is minimal for {0, n -1}. Let W ⊊ V (σ) such that 0, n -1 ∈ W and W ≥ 3. Since W ≠ V (σ), there exists i ∈ {1, . . . , n -2} such that i ∈ W . Set

W ′ = W ∩ {0, . . . , i -1}. For j ∈ W ′ and k ∈ W ∖ W ′ , we have k ≥ j + 2, and hence [j, k] σ = [0, n - 1] σ . It follows that W ′ and W ∖ W ′ are modules of σ[W ]. Thus, σ[W ] is decomposable.
Given n ≥ 6, the graph Q n is defined on V (Q n ) = {0, . . . , n -1} in the following way (see Figure 6.2)

1. Q n -{n -2, n -1} = P n-2 (see Figure 1.1); 2. for i ∈ {0, . . . , n -4}, {i, n -2} ∈ E(Q n ); 3. {n -2, n -1} ∈ E(Q n ). 0 • 1 • . . . n -4 • n -3 • ¡ ¡ ¡ ¡ ¡ ¡ ¡ ¡ d d d d d d d • n -2 • n -1 Figure 6.2: The graph Q n Furthermore, for n ≥ 6, O n is the tournament defined on V (O n ) = {0, .
. . , n -1} in the following way (see Figure 6.3)

1. O n -{n -2, n -1} = M n-2 ; 2. for i ∈ {0, . . . , n -4}, (i, n -2) ∈ A(O n ); 3. for i ∈ {0, . . . , n -3}, (i, n -1) ∈ A(O n ); 4. (n -2, n -3), (n -1, n -2) ∈ A(O n ). 0 • 1 • . . . . n -4 • n -3 • E E E E s s n -2 • ' $ d d d d d d d d d d d d d d T n -1 • ' $ ' Figure 6.3: The tournament O n .
Given n ≥ 6, it is not difficult to verify that σ(Q n ) and σ(O n ) are prime and minimal for {0, n -1}. We generalize σ(Q n ) and σ(O n ) as follows. Notation 6.9. We denote by N 2 the set of the 2-structures σ defined on V (σ) = {0, . . . , n -1}, where n ≥ 5, and satisfying the following assertions

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ σ -{n -2, n -1} satisfies (6.4), for i ∈ {0, . . . , n -4}, [n -2, i] σ = [n -2, n -1] σ , for i ∈ {0, . . . , n -3}, [n -1, i] σ = [0, n -3] σ , and [n -2, n -1] σ ≠ [n -2, n -3] σ and [n -2, n -1] σ ≠ [n -3, 0] σ .
(6.5) Lemma 6.10. The elements of N 2 are prime and minimal for {0, n -1}.

Proof. Let σ ∈ N 2 . We have V (σ) = {0, . . . , n -1}, where n ≥ 5. First, we verify that σ is prime. We distinguish the following two cases.

• Suppose that n ∈ {5, 6}. Using Assertion (M2) of Proposition 2.5 and Lemma 6.7 applied to σ[{0, . . . , n -3}] ∈ M 2 , it is not difficult to verify that σ is prime.

• Suppose that n ≥ 7. Since σ-{n-2, n-1} ∈ M 2 , it follows from Lemma 6.8 that σ -{n -2, n -1} is prime. Set X = {0, . . . , n -3}. It is not difficult to verify that n -2 ∈ Ext σ (X) and n -1 ∈ ⟨X⟩ σ . Since [n -2, n -1] σ ≠ [n -3, 0] σ and [n -1, n -3] σ = [0, n -3] σ , we have [n -1, n - 2] σ ≠ [n-1, n-3] σ . Thus, X ∪{n-2} is not a module of σ[X ∪{n-2, n-1}].
It follows from Assertion (P2) of Lemma 3.17 that σ[X ∪ {n -2, n -1}], that is σ, is prime.

Second, we verify that σ is minimal for

{0, n -1}. Consider W ⊊ V (σ) such that 0, n -1 ∈ W and W ≥ 3. If n -2 ∈ W , then W ∖ {n -1} is a nontrivial module of σ[W ]. Hence, suppose that {0, n -2, n -1} ⊆ W . If n -3 ∈ W , then W ∖ {n -2} is a nontrivial module of σ[W ]. Thus, suppose that {0, n-3, n-2, n-1} ⊆ W . If 1 ∈ W , then {0, n-1} is a nontrivial module of σ[W ]. Therefore, suppose that {0, 1, n -3, n -2, n -1} ⊆ W . Since W ≠ {0, . . . , n -1}, there exists i ∈ {2, . . . , n-4}∖W . We obtain that W ∩{0, . . . , i-1} is a nontrivial module of σ[W ].
Proposition 6.11. Consider a prime 2-structure σ such that v(σ) ≥ 6. Let v and w be distinct vertices of σ. Suppose that for every W ⊆ V (σ), we have

if 3 ≤ W ≤ 5 and v, w ∈ W , then σ[W ] is decomposable. (6.6)
Under these assumptions, there exists an isomorphism ϕ from an element of

M 2 ∪ N 2 onto σ[X]
, where X ⊆ V (σ), such that v, w ∈ X and X ≥ 6, satisfying ϕ({0, n -1}) = {v, w}.

Proof. Set e = (v, w) σ and f = (w, v) σ .

Consider

Z = {z ∈ V (σ) ∖ {v, w} ∶ z ←→ σ {v, w}} ∖ N (e,f ) σ (v)
(see Notation 2.1 and Notation 3.7).

Denote by

C(v) (respectively, C(w)) the {e, f }-component of σ -Z (see Defi- nition 2.
2) containing v (respectively, w).

To begin, suppose that

C(v) = C(w).
Let n be the least integer m ≥ 3 such that there exists a sequence v 0 , . . . , v m-1 of vertices of σ -Z satisfying

• v 0 = v and v m-1 = w; • for 0 ≤ i ≤ m -2, [v i , v i+1 ] σ ≠ (e, f ).
It follows from the minimality of n that for i ∈ {0, . . . , n-3} and j ∈ {i+2, . . . , n-1}, we have

[v i , v j ] σ = (e, f ).
We consider the bijection ϕ ∶ {0, . . . , n -1} → {v 0 , . . . , v n-1 } defined by ϕ(i) = v i for i ∈ {0, . . . , n -1}. Moreover, we denote by τ the unique 2-structure defined on V (τ ) = {0, . . . , n -1} such that ϕ is an isomorphism from τ onto

σ[{v 0 , . . . , v n-1 }]. For g ∈ E(σ[{v 0 , . . . , v n-1 }]), set ϕ -1 (g) = {(ϕ -1 (x), ϕ -1 (y)) ∶ (x, y) ∈ g}.
We obtain

E(τ ) = {ϕ -1 (g) ∶ g ∈ E(σ[{v 0 , . . . , v n-1 }])}.
In particular, we have

[0, n -1] τ = (ϕ -1 (e), ϕ -1 (f )). Let i ∈ {0, . . . , n -1}. Since [v i , v i+1 ] σ ≠ (e, f ), we have [i, i + 1] τ ≠ [0, n -1] τ . Furthermore, consider i ∈ {0, . . . , n -3} and j ∈ {i + 2, . . . , n -1}. Since [v i , v j ] σ = (e, f ), we have [i, j] τ = [0, n -1] τ .
It follows that τ satisfies (6.4). Hence, τ ∈ M 2 . Finally, we prove that τ is prime. For a contradiction, suppose that τ admits a nontrivial module M . By Lemma 6.7,

M = {0, n -1}. Hence, {v 0 , v n-1 }, that is {v, w}, is a module of σ[{v 0 , . . . , v n-1 }]. We obtain v 1 ←→ σ {v, w} and [v, v 1 ] σ ≠ (e, f ). Thus, v 1 ∈ Z, which contradicts the fact that C(v) ⊆ V (σ) ∖ Z. Consequently, τ is prime. Thus, σ[{v 0 , . . . , v n-1 }] is prime too. It follows from (6.6) that n ≥ 6. Now, suppose that C(v) ≠ C(w). It follows from Lemma 2.4 that C {e,f } (σ -Z) (see Definition 2.2) is a modular partition of σ -Z. In particular, for c ∈ C(v) and d ∈ C(w), we have [c, d] σ = [v, w] σ .
First, suppose that there exists z ∈ Z such that ≺v, z ≻ σ ≠≺v, w ≻ σ , we have

z ←→ σ C(v). (6.7)
We conclude in the following way. Since

C(v) is {e, f }-connected, there exist v 0 , . . . , v k-1 ∈ C(v),
where k ≥ 2, such that

• σ[{v 0 , . . . , v k-1 }] satisfies (6.4);

• v 0 = v; • if k = 2, then [v 0 , v k-1 ] σ ≠ (f, e); • if k ≥ 3, then [v 0 , v k-1 ] σ = (f, e); • [z, v 0 ] σ ≠ [z, v k-1 ] σ ; • for i ∈ {0, . . . , k -2}, [z, v 0 ] σ = [z, v i ] σ .
If k = 2, then it is not difficult to verify directly that σ[{v 0 , . . . , v k-1 } ∪ {z, w}] is prime, which contradicts (6.6). Therefore, we have k ≥ 3. Consider the bijection

ϕ ∶ {0, . . . , k + 1} → {v 0 , . . . , v k-1 } ∪ {z, w} 0 ≤ i ≤ k -1 → v i , k → z, k + 1 → w.
Denote by τ the unique 2-structure defined on V (τ ) = {0, . . . , k + 1} such that ϕ is an isomorphism from τ onto σ[{v 0 , . . . , v k-1 } ∪ {z, w}]. We have τ ∈ N 2 . By Lemma 6.10, τ is prime. Hence, σ[{v 0 , . . . , v k-1 } ∪ {z, w}] is prime too. It follows from (6.6) that k ≥ 4, so {v 0 , . . . , v k-1 } ∪ {z, w} ≥ 6. Second, suppose that for every for z ∈ Z such that ≺ v, z ≻ σ ≠≺ v, w ≻ σ , (6.7) does not hold, that is, z ←→ σ C(v). Similarly, for z ∈ Z such that ≺ v, z ≻ σ ≠≺ v, w ≻ σ , we can suppose that z ←→ σ C(w). Since z ←→ σ {v, w} for every z ∈ Z, we obtain To continue, suppose that there exists z ∈ Z such that z ←→ σ C(w). By (6.8), ≺v,

z ←→ σ C(v) ∪ C(w) for every z ∈ Z such that ≺v, z ≻ σ ≠≺v, w ≻ σ . ( 6
z ≻ σ =≺v, w ≻ σ . Since z ∈ N (e,f ) σ (v), we obtain [z, v] σ = (e, f ). Set C = {c ∈ C(w) ∖ {w} ∶ z ←→ σ {c, w}}.
If there exists c ∈ C such that ≺ c, w ≻ σ ≠≺ v, w ≻ σ , then σ[{c, v, w, z}] is prime, which contradicts (6.6). Furthermore, if there exists c, d ∈ C such that [c, w] σ = (e, f ) and [d, w] σ = (f, e), then σ[{c, d, v, w, z}] is prime, which contradicts (6.6). Hence, suppose that [w, C] σ = (e, f ) or (f, e). We distinguish the following two cases.

1. Suppose that [w, C] σ = (e, f ). Since C(w) is {e, f }-connected, there exist w 0 , . . . , w k-1 ∈ C(w), where k ≥ 3, such that

• σ[{w 0 , . . . , w k-1 }] satisfies (6.4);

• w 0 = w and [w 0 , w k-1 ] σ = (e, f );

• [z, w 0 ] σ ≠ [z, w k-1 ] σ ; • for i ∈ {0, . . . , k -2}, [z, w i ] σ = (e, f ).
Consider the bijection

ϕ ∶ {0, . . . , k + 1} → {w 0 , . . . , w k-1 } ∪ {z, v} 0 ≤ i ≤ k -1 → w i , k → z, k + 1 → v.
Denote by τ the unique 2-structure defined on V (τ ) = {0, . . . , k + 1} such that ϕ is an isomorphism from τ onto σ[{w 0 , . . . , w k-1 } ∪ {z, v}]. We have τ ∈ N 2 . By Lemma 6.10, τ is prime. Hence, σ[{w 0 , . . . , w k-1 } ∪ {z, v}] is prime too. It follows from (6.6) that k ≥ 4, so {w 0 , . . . , w k-1 } ∪ {z, v} ≥ 6.

2. Suppose that [w, C] σ = (f, e). Since C(w) is {e, f }-connected, there exist w 0 , . . . , w k-1 ∈ C(w), where k ≥ 3, such that

• σ[{w 0 , . . . , w k-1 }] satisfies (6.4);

• w 0 = w and [w 0 , w k-1 ] σ = (f, e); • [z, w 0 ] σ ≠ [z, w k-1 ] σ ; • for i ∈ {0, . . . , k -2}, [z, w i ] σ = (e, f ).
Consider the bijection

ϕ ∶ {0, . . . , k + 1} → {w 0 , . . . , w k-1 } ∪ {z, v} 0 ≤ i ≤ k -1 → w i , k → z, k + 1 → v.
Denote by τ the unique 2-structure defined on V (τ ) = {0, . . . , k + 1} such that ϕ is an isomorphism from τ onto σ[{w 0 , . . . , w k-1 } ∪ {z, v}]. We have τ ∈ M 2 . It follows from Lemma 6.8 that τ and hence σ[{w 0 , . . . , w k-1 } ∪ {z, v}] are prime. By (6.6), {w 0 , . . . , w k-1 } ∪ {z, v} ≥ 6.

Consequently, we can suppose that z ←→ σ C(w) for every z ∈ Z. Since C(w) is a module of σ -Z, C(w) is a module of σ as well. Since σ is prime, we obtain

C(w) = {w}. By Proposition 2.8, (σ -Z) C {e,f } (σ -Z) is linear. Set I {v,w} = {u ∈ V (σ) ∖ (Z ∪ C(v) ∪ {w}) ∶ [v, u] σ = [u, w] σ , [v, u] σ = (e, f )}. Given z ∈ Z, we verify that if ≺v, z ≻ σ ≠≺v, w ≻ σ , then z ←→ σ (C(v) ∪ I {v,w} ∪ {w}). ( 6
(C(v) ∪ I {v,w} ∪ {w}) ≠ V (σ). Hence, C(v) ∪ I {v,w} ∪ {w} is not a module of σ. Furthermore, since (σ-Z) C {e,f } (σ-Z) is linear, C(v)∪I {v,
w} ∪{w} is a module of σ-Z. Thus, there exists z ∈ Z such that z ←→ σ (C(v) ∪ I {v,w} ∪ {w}). By (6.9), ≺v, z ≻ σ =≺v, w ≻ σ , so [z, v] σ = (e, f ). We define by induction a sequence of pairwise disjoint subsets

(Z p ) p≥0 of {z ∈ Z ∶ [z, v] σ = (e, f )} as follows. Set Z 0 = {z ∈ Z ∶ [z, v] σ = (e, f ), z ←→ σ (C(v) ∪ I {v,w} ∪ {w})}.
Note that Z 0 ≠ ∅. Given Z 0 , . . . , Z i , where i ≥ 0, set

Z i+1 = {z ∈ Z ∖ (Z 0 ∪ ⋯ ∪ Z i ) ∶[z, v] σ = (e, f ), z ←→ σ (C(v) ∪ I {v,w} ∪ {w} ∪ (Z 0 ∪ ⋯ ∪ Z i ))}.
Denote by p the least integer i such that Z i = ∅. As previously noted,

Z 0 ≠ ∅, so p ≥ 1. We have [w, C(v) ∪ I {v,w} ∪ (Z 0 ∪ ⋯ ∪ Z p-1 )] σ = (f, e). Since σ is prime, we have C(v) ∪ I {v,w} ∪ {w} ∪ (Z 0 ∪ ⋯ ∪ Z p-1 ) ≠ V (σ). Therefore, there exists x ∈ V (σ) ∖ (C(v) ∪ I {v,w} ∪ {w} ∪ (Z 0 ∪ ⋯ ∪ Z p-1 )) such that x ←→ σ (C(v) ∪ I {v,w} ∪ {w} ∪ (Z 0 ∪ ⋯ ∪ Z p-1 )). Set Z ′ = {z ′ ∈ Z ∖ (Z 0 ∪ ⋯ ∪ Z p-1 ) ∶ [z ′ , v] σ = (e, f )}. For each z ′ ∈ Z ′ , we have z ′ ←→ σ (C(v) ∪ I {v,w} ∪ {w} ∪ (Z 0 ∪ ⋯ ∪ Z p-1 )). Thus, x ∈ Z ′ . It follows that either x ∈ N (e,f ) σ (v) ∩ N (e,f ) σ
(w) or x ∈ Z and ≺v, x≻ σ ≠≺v, w ≻ σ . In both cases, we obtain

[x, v] σ ≠ (e, f ). If x ∈ Z and ≺ v, x ≻ σ ≠≺ v, w ≻ σ , then x ←→ σ (C(v) ∪ I {v,
w} ∪ {w}) by (6.9). Furthermore, since (σ -Z) C {e,f } (σ -Z) is linear by Proposition 2.8, we have

[C(v) ∪ I {v,w} ∪ {w}, N (e,f ) σ (v) ∩ N (e,f ) σ (w)] σ = (e, f
). Therefore, in both cases, we have

x ←→ σ (C(v) ∪ I {v,w} ∪ {w}).

Consequently, there exists i ∈ {0, . . . , p -1} such that

x ←→ σ (C(v) ∪ I {v,w} ∪ {w}) ∪ (Z 0 ∪ ⋯ ∪ Z i ). Set j = min({i ∈ {0, . . . , p -1} ∶ x ←→ σ (C(v) ∪ I {v,w} ∪ {w}) ∪ (Z 0 ∪ ⋯ ∪ Z i )}).
We show that there exists a sequence (v 0 , . . . , v k-1 ) of elements of C(v) ∪ I {v,w} , where k ≥ 3, such that

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ σ[{v 0 , . . . , v k-1 }] satisfies (6.4), [v 0 , v k-1 ] σ = (f, e), v 0 = v, x ←→ σ {v 0 , . . . , v k-2 } and x ←→ σ {v 0 , v k-1 }. (6.10)
By minimality of j, we have

x ←→ σ (C(v) ∪ I {v,w} ∪ {w}) ∪ (Z 0 ∪ ⋯ ∪ Z j-1
)), when j ≥ 1, and x ←→ σ (C(v) ∪ I {v,w} ∪ {w}) ∪ Z j ). There exists a sequence (z 0 , . . . , z j ) satisfying

• for i ∈ {0, . . . , j}, z i ∈ Z i ;

• x ←→ σ (C(v) ∪ I {v,w} ∪ {w}) ∪ {z j });

• if j ≥ 1, then x ←→ σ (C(v) ∪ I {v,w} ∪ {w}) ∪ {z 0 , . . . , z j-1 });

• if j ≥ 1, then [z i , z i+1 ] σ ≠ (f, e) for i ∈ {0, . . . , j -1}.
Lastly, since z 0 ∈ Z 0 , we have z 0 ∈ Z, [z 0 , v] σ = (e, f ), and z 0 ←→ σ (C(v)∪I {v,w} ∪ {w}). Therefore, there exists u ∈ (C(v)∖{v})∪I {v,w} such that [z 0 , u] σ ≠ (e, f ).

To conclude, we distinguish the following two cases.

1. Suppose that u ∈ I {v,w} . Set k = j + 3 and

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ v 0 = v, v 1 = u, and for l ∈ {2, . . . , k -1}, v l = z l-2 . 2. Suppose that u ∈ C(v) ∖ {v}. Since C(v) is {e, f }-connected, there exist u 0 , . . . , u m-1 ∈ C(w)
, where m ≥ 2, such that

• u 0 = v; • for l ∈ {0, . . . , m -2}, [u l , u l+1 ] σ ≠ (f, e);
• if m ≥ 3, then for l ∈ {0, . . . , m -3} and l ′ ∈ {l + 2, . . . , m -1}, we have [u l , u l ′ ] σ = (f, e);

• for l ∈ {0, . . . , m -2}, [u l , z 0 ] σ = (f, e);

• [u m-1 , z 0 ] σ ≠ (f, e).
Set k = m + j + 1 and

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ for l ∈ {0, . . . , m -1}, v l = u l , and for l ∈ {m, . . . , k -1}, v l = z l-m .
In both cases, we obtain k ≥ 3 and (v 0 , . . . , v k-1 ) satisfies (6.10). Consider the bijection ϕ ∶ {0, . . . , k

+ 1} → {v 0 , . . . , v k-1 } ∪ {x, w} 0 ≤ i ≤ k -1 → v i , k → x, k + 1 → w.
Denote by τ the unique 2-structure defined on V (τ ) = {0, . . . , k + 1} such that ϕ is an isomorphism from τ onto σ[{v 0 , . . . , v k-1 } ∪ {z, w}]. We have τ ∈ N 2 . It follows from Lemma 6.10 that τ and hence σ[{v 0 , . . . , v k-1 } ∪ {z, w}] are prime. By (6.6), {v 0 , . . . , v k-1 } ∪ {z, w} ≥ 6.

The next characterization of prime 2-structures that are minimal for an unordered pair follows from Lemma 6.8, Lemma 6.10, and Proposition 6.11. Theorem 6.12 (Cournier and Ille4 [START_REF] Cournier | Minimal indecomposable graph[END_REF]). Consider a 2-structure σ such that v(σ) ≥ 6. Let v, w be distinct vertices of σ. The following two assertions are equivalent 1. σ is prime and minimal for {v, w}; It is not difficult to verify that σ is prime and minimal for {0, 4}. Nethertheless, σ is not isomorphic to an element of M 2 ∪ N 2 .

Proof of Theorem 5.3

A second proof of Theorem 5.3 when v(σ) ≥ 7. Consider a prime 2-structure σ such that v(σ) ≥ 7. By Theorem 5.23, S c (σ) ≤ 2. Therefore, there exist distinct v, w ∈ V (σ) such that S c (σ) ⊆ {v, w}.

First, suppose that σ is minimal for {v, w}. By Theorem 6.12, there exists an isomorphism ϕ from σ onto τ ∈ M 2 ∪ N 2 defined on V (τ ) = {0, . . . , n -1} such that ϕ({v, w}) = {0, n-1}. Clearly, we have τ -{0, 1} ∈ M 2 ∪N 2 . It follows from Lemmas 6.8 and 6.10 that τ -{0, 1} is prime. Hence, σ -{ϕ -1 (0), ϕ -1 (1)} is prime as well.

Second, suppose that σ is not minimal for {v, w}. There exists X ⊊ V (σ) such that σ[X] is prime and v, w ∈ X. We obtain

S c (σ) ⊆ X and X ⊊ V (σ).
We conclude as in Remark 5.2 from (5.1) by using Corollary 3.21.

Chapter 7

Infinite prime 2-structures

The purpose of this chapter is to prove the next theorem.

Theorem 7.1 (Ille 1 [START_REF] Ille | Graphes indécomposables infinis[END_REF][START_REF] Ille | A characterization of the indecomposable and infinite graphs[END_REF] ). Given an infinite 2-structure σ, the following two assertions are equivalent

• σ is prime; • for each finite F ⊆ V (σ), there exists F ′ ⊆ V (σ) such that ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ F ′ is finite, F ⊆ F ′ , and σ[F ′ ] is prime. (7.1)
We use the following definition. Definition 7.2. Let S be a set. A family F of subsets of S is up-directed if for up-directed any X, Y ∈ F, there exists Z ∈ F such that X ∪ Y ⊆ Z.

Lemma 7.3. Given a 2-structure σ, consider an up-directed family F of subsets of V (σ). If σ[X] is prime for each X ∈ F, then σ[ ⋃ X∈F X] is prime. Proof. Let M be a module of σ[⋃ X∈F X] such that M ≥ 2. We have to show that M = ⋃ X∈F X. Since M ≥ 2, consider distinct x, y ∈ M . Let v ∈ ⋃ X∈F X. Since F is up- directed, there exists X ∈ F such that x, y, v ∈ X. By Assertion (M2) of Propo- sition 2.5, M ∩ X is a module of σ[X]. Since x, y ∈ M ∩ X, we have M ∩ X ≥ 2. Since σ[X] is prime, we obtain M ∩ X = X. Hence v ∈ M . It follows that M = ⋃ X∈F X.
1 Ille [START_REF] Ille | Graphes indécomposables infinis[END_REF][START_REF] Ille | A characterization of the indecomposable and infinite graphs[END_REF] proved this theorem for digraphs.

99

Lemma 7.3 allows to prove one direction of the equivalence in Theorem 7.1. The use of the next result is decisive in the proof of the other direction. Furthermore, it is also significant in the study of infinite and prime 2-structures.

Theorem 7.4. Given a prime 2-structure σ, consider X ⊆ V (σ) such that σ[X] is prime. Suppose that V (σ) ∖ X is infinite. For each v ∈ V (σ) ∖ X, there exists a finite F ⊆ V (σ) ∖ X such that v ∈ F and σ[X ∪ F ] is prime. Proof. Consider the set W of v ∈ V (σ) ∖ X such that for every finite F ⊆ V (σ) ∖ X, we have σ[X ∪ F ] is decomposable whenever v ∈ W . We have to show that W = ∅.
Recall that

p (σ,X) = {Ext σ (X), ⟨X⟩ σ } ∪ {X σ (y) ∶ y ∈ X} (see Notation 3.12).
By Lemma 3.13, p (σ,X) is a partition of V (σ) ∖ X. Consequently, to prove that W = ∅, it suffices to show that

W ∩ Ext σ (X) = ∅, (7.2)
W ∩ ⟨X⟩ σ = ∅, (7.3)
and W ∩ X σ (y) = ∅ for each y ∈ X.

First, for v ∈ Ext σ (X), we have σ[X ∪ {v}] is prime. Thus v ∈ W . Therefore, (7.2) holds. Second, we verify that V (σ) ∖ (W ∩ ⟨X⟩ σ ) is a module of σ. Consider w ∈ W ∩ ⟨X⟩ σ . Since w ∈ ⟨X⟩ σ , we have w ←→ σ X (see Notation 2.1). Consequently, to prove that w ←→ σ (V (σ) ∖ (W ∩ ⟨X⟩ σ )), it suffices to verify that w ←→ σ X ∪ {v} for every v ∈ (V (σ) ∖ X) ∖ (W ∩ ⟨X⟩ σ ).

(7.5)

Given v ∈ (V (σ) ∖ X) ∖ (W ∩ ⟨X⟩ σ ), we distinguish the following two cases

• Suppose that v ∈ ⟨X⟩ σ . Since w ∈ W , σ[X ∪ {v, w}] is decomposable. It follows from Assertions (P1) and (P2) of Lemma 3.17 that X ∪ {v} is a module of σ[X ∪ {v, w}]. Hence, we obtain w ←→ σ X ∪ {v}.

• Suppose that v ∈ ⟨X⟩ σ . Since v ∈ (V (σ) ∖ X) ∖ (W ∩ ⟨X⟩ σ ), v ∈ W . Thus, there exists a finite F ⊆ V (σ) ∖ X such that v ∈ F and σ[X ∪ F ] is prime. Set Y = X ∪ F. Since w ∈ W , w ∈ Y . Moreover, since w ∈ W , we have σ[Y ∪ {w}] is decomposable. Thus, w ∈ Ext σ (Y ). For a contradiction, suppose that w ∈ Y σ (z), where z ∈ Y . If z ∈ X, then w ∈ X σ (z), which contradicts w ∈ ⟨X⟩ σ because p (σ,X) is a partition of V (σ) ∖ X by Lemma 3.13. Now, suppose that z ∈ Y ∖ X, that is, z ∈ F . Set F ′ = (F ∖ {z}) ∪ {w}. Since w ∈ Y σ (z), we have {z, w} is a module of σ[(X ∪ F ) ∪ {w}]. It follows that σ[X ∪ F ] and σ[X ∪ F ′ ] are isomorphic. Therefore, σ[X ∪ F ′ ]
is prime too, which contradicts w ∈ W . Consequently, we obtain w ∈ ⟨Y ⟩ σ .

In particular, we have w ←→ σ X ∪ {v}.

It follows from both cases above that (7.5) holds. Consequently,

V (σ) ∖ (W ∩ ⟨X⟩ σ ) is a module of σ. Since σ[X] is prime, we have X ≥ 3. Since X ⊆ (V (σ) ∖ (W ∩ ⟨X⟩ σ )), we obtain V (σ) ∖ (W ∩ ⟨X⟩ σ )) = V (σ), that is, (7.3) holds.
Third, we verify that (7.4) holds. Given y ∈ X, we show that {y} ∪ (W ∩ X σ (y)) is a module of σ. Let w ∈ W ∩ X σ (y). We have to verify that v ←→ σ {y, w} for every v ∈ V (σ) ∖ ({y} ∪ (W ∩ X σ (y))).

(7.6)

Since w ∈ X σ (y), we have v ←→ σ {y, w} for v ∈ X ∖ {y}. To continue, suppose that v ∈ (V (σ) ∖ X) ∖ (W ∩ X σ (y)).

We distinguish the following two cases.

• Suppose that v ∈ X σ (y). Since w ∈ W , σ[X ∪ {v, w}] is decomposable. It follows from Assertions (P1), (P3), and (P4) of Lemma 3.17 that {y, w} is a module of σ[X ∪ {v, w}]. In particular, we have v ←→ σ {y, w}.

• Suppose that v ∈ X σ (y). Since v ∈ W ∩X σ (y), we have v ∈ W . Thus, there exists a finite

F ⊆ V (σ) ∖ X such that v ∈ F and σ[X ∪ F ] is prime. Set Y = X ∪ F. Since w ∈ W , w ∈ Y . Since w ∈ W , we have σ[Y ∪ {w}] is decompos- able. Thus, w ∈ Ext σ (Y ). Furthermore, if w ∈ ⟨Y ⟩ σ , then w ∈ ⟨X⟩ σ , which contradicts w ∈ X σ (y) because p (σ,X) is a partition of V (σ) ∖ X by Lemma 3.13. It follows that w ∈ Y σ (z), where z ∈ Y . For a contradiction, suppose that z ∈ Y ∖ X, that is, z ∈ F . Set F ′ = (F ∖ {z}) ∪ {w}. Since w ∈ Y σ (z), we have {z, w} is a module of σ[(X ∪F )∪{w}]. It follows that σ[X ∪F ] and σ[X ∪F ′ ] are isomorphic. Therefore, σ[X ∪F ′ ] is prime too, which contradicts w ∈ W . Therefore, z ∈ X. We obtain w ∈ X σ (z) because w ∈ Y σ (z). It follows from Lemma 3.13 that z = y. Hence, {y, w} is a module of σ[Y ∪ {w}].
In particular, we have v ←→ σ {y, w}.

Consequently, {y} ∪ (W ∩ X σ (y)) is a module of σ. Since (X ∖ {y}) ∩ ({y} ∪ (W ∩ X σ (y))) = ∅, we have {y} ∪ (W ∩ X σ (y)) ⊊ V (σ). Since σ is prime, we obtain {y} ∪ (W ∩ X σ (y)) ≤ 1, that is, W ∩ X σ (y) = ∅.
Hence, (7.4) holds.

Finally, we prove Theorem 7.1 as follows.

Proof of Theorem 7.1. To begin, suppose that σ is prime. Consider a finite F ⊆ V (σ). By Theorem 3.102 , there exists X ⊆ V (σ) such that X = 3 or 4, and σ[X] is prime. By applying Theorem 7.4 several times from X together with the elements of F ∖ X, we obtain F ′ ⊆ V (σ) satisfying (7.1). Conversely, suppose that for every finite F ⊆ V (σ), there exists F ′ ⊆ V (σ) satisfying (7.1). (7.7)

Consider the family F of finite X ⊆ V (σ) such that σ[X] is prime. Since (7.7) holds, we have

⋃ X∈F X = V (σ)
and F is up-directed.

It follows from Lemma 7.3 that σ is prime.

Chapter 8

Critical and non finitely critical 2-structures

The path P Z is defined on Z as follows. Given v, w ∈ Z, with v ≠ w, {v, w} ∈ E(Z) if vw = 1. In the sequel, P Z [N] is denoted by P N .

For each finite subset F of Z, there exist m, n ∈ Z such that nm ≥ 4 and F ⊆ {m, . . . , n}. Since P Z [{m, . . . , n}] ≃ P n-m , it follows from Fact 2.6 that P Z [{m, . . . , n}] is prime. Hence, σ(P Z )[{m, . . . , n}] is prime too. Therefore, it follows from Theorem 7.1 that σ(P Z ) is prime. Furthermore, for each finite and nonempty subset F of Z, Z -F is disconnected. Therefore, Z -F and hence σ(P Z ) -F are decomposable. The properties of σ(P Z ) lead us to introduce the following definition. Definition 8.1. An infinite prime 2-structure σ is finitely critical if for each finitely critical finite and nonempty subset F of V (σ), σ -F is decomposable.

The next result is a direct consequence of Corollary 3.21.

Corollary 8.2. Given an infinite prime 2-structure σ, σ is critical and non finitely critical if and only if the following two assertions hold

• for each v ∈ V (σ), σ -v is decomposable (i.e. σ is critical);
• there exist x, y ∈ V (σ) such that x ≠ y and σ -{x, y} is prime (i.e. P(σ) is nonempty).

The next result provides a characterization of the nontrivial components of the primality graph of an infinite critical 2-structure. It is an easy consequence of Lemma 4.4 and Proposition 4.5. For instance, the usual linear order L Z defined on Z belongs to F Z .

Second, we denote by F N the family of the 2-structures τ defined on V (τ ) = N and satisfying

• {2, 3, . . .} is a module of τ -0;

• {0} ∪ {2, 3, . . .} is not a module of τ ;

• for every n ≥ 1, {n -1, n + 1} is a module of τn and not of τ .

For instance, the usual linear order L N defined on N belongs to F N .

In the next four lemmas, we examine the elements of F N ∪ F Z . Lemma 8.6. Given a 2-structure τ such that V (τ ) = Z, τ ∈ F Z if and only if the following two assertions hold

• [1, 0] τ ≠ [1, 2] τ ; • for m, n ∈ Z such that m < n, we have [2m, 2n] τ = [0, 2] τ , [2m, 2n -1] τ = [0, 1] τ , [2m + 1, 2n] τ = [1, 2] τ , and [2m + 1, 2n + 1] τ = [1, 3] τ .
Proof. To begin, suppose that τ ∈ F Z . In particular, {0, 2} is a module of τ -1 and not of τ . It follows that [1,[START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF] τ ≠ [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ . For the second assertion, consider m, n ∈ Z such that m < n. Since {n, n + 2} is a module of τ -(n + 1), we have

[m, n] τ = [m, n + 2] τ .
• every module of τ is a module of L Z ;

• if τ is decomposable and τ ≠ σ(L Z ), then one of the following two situations holds ▸ for each nontrivial module M of τ , there exists n ≥ 1 such that M = {n, n + 1, . . .};

▸ for each nontrivial module M of τ , there exists n ∈ Z such that M = {. . . , n -1, n}.

Proof. The first two assertions follow from the second assertion of Lemma 8.6.

For the third assertion, consider a module M of τ . Consider p, q ∈ M such that p + 2 ≤ q. We verify that if p ≡ q mod 2, then {p, . . . , q} ⊆ M .

(8.1)

For instance, suppose that p and q are even. To begin, consider r be an odd integer such that p < r < q. By the second assertion of Lemma 8.6,

[r, p] τ = [1, 0] τ and [r, q] τ = [1, 2] τ . Since [1, 0] τ ≠ [1, 2]
τ by the first assertion of Lemma 8.6, we obtain r ∈ M . Now, let r be an even integer such that p < r < q. We have r -1, r + 1 ∈ M . Moreover, by the second assertion of Lemma 8.6, we have

[r, r -1] τ = [2, 1] τ and [r, r + 1] τ = [0, 1] τ . Since [0, 1] τ ≠ [2, 1]
τ by the first assertion of Lemma 8.6, we obtain r ∈ M . The case p and q both odd follows similarly. Thus, (8.1) holds. Now, suppose that p ≡ q mod 2. For instance, suppose that p is even and q is odd. For a contradiction, suppose that q -1 ∈ M and q + 1 ∈ M . By the second assertion of Lemma 8.6, [q -1, p] τ = [2, 0] τ and [q -1, q] τ = [0, 1] τ . Since q -1 ∈ M , we obtain [2, 0] τ = [0, 1] τ . Furthermore, by the second assertion of Lemma 8.6,

[q + 1, p] τ = [2, 0] τ and [q + 1, q] τ = [2, 1] τ . Since q + 1 ∈ M , we obtain [2, 0] τ = [2, 1] τ . Therefore, we have [0, 1] τ = [2, 1] τ ,
which contradicts the first assertion of Lemma 8.6. Consequently, q -1 ∈ M or q + 1 ∈ M , and we conclude by using (8.1).

For the fourth assertion, suppose that there exist p, q ∈ Z, with p < q, such that {p, . . . , q} is a module of τ . We have to show that τ = σ(L Z ). It follows from the second assertion of Lemma 8.6 that

[0, 1] τ = [0, 2] τ = [1, 2] τ = [1, 3] τ . (8.2)
By the first assertion of Lemma 8.6, we have

[1, 0] τ ≠ [1, 2] τ . By (8.2), [0, 1] τ ≠ [1, 0] τ . Therefore, τ = σ(L Z ).
Example 8.9. We consider the tournament U Z obtained from the linear order L Z by reversing all the arcs between the even integers. By Lemma 8.6, σ(U Z ) ∈ F Z . It follows from Lemma 8.8 that σ(U Z ) is prime. We can also see that σ(U Z ) is prime by using Theorem 7.1 as follows. Let F be a finite subset of Z. There exists n ∈ Z such that F ⊆ {-n, . . . , n}. By Theorem 4.28, σ(U 2n+1 ) is prime (see Figure 4.5). Since (σ(U Z ))[{-n, . . . , n}] and σ(U 2n+1 ) are isomorphic, (σ(U Z ))[{-n, . . . , n}] is prime too. It follows from Theorem 7.1 that σ(U Z ) is prime. Now, we consider the tournament W Z obtained from the linear order L Z by reversing all the arcs between the even integers and all the arcs between the odd integers. As previously for σ(U Z ), it is not difficult to verify that σ(W Z ) is a prime element of F Z .

Finally, we consider the bipartite graph H Z defined on Z in the following way. For p, q ∈ Z, with p ≠ q, {p, q} ∈ E(H Z ) if there exist i, j ∈ Z, with i ≤ j, such that {p, q} = {2i, 2j + 1}. Once again, σ(H Z ) is a prime element of F Z . Lemma 8.10. Given τ ∈ F N , the following four assertions hold

• for each n ≥ 1, {n, n+1, . . .} is a module of τ if and only if [0, 1] τ = [0, 2] τ ;
• every module of τ is a module of L N ;

• if τ is decomposable and τ ≠ σ(L N ), then for each nontrivial module M of τ , there exists n ≥ 1 such that M = {n, n + 1, . . .}.

Proof. The first assertion follows from the last two assertions of Lemma 8.7.

We show the second assertion as in the proof of Lemma 8.8.

For the third assertion, suppose that there exist p ≥ 0 and q > p such that {p, . . . , q} is a module of τ . We have to show that τ = σ(L N ). It follows from the second assertion of Lemma 8.7 that [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] As in Example 8.9, we verify easily that σ(U N ) and σ(H N ) are prime elements of

F N . Similarly, σ(W N ) is prime, but σ(W N ) ∈ F N because [1, 2] σ(W N ) ≠ [1, 3] σ(W N ) .
In fact, σ(W N ) is also interesting because it shows that the analogue of Theorem 7.1, when the primality is replaced by the criticality, doest not hold. Indeed, σ(W N ) satisfies the second assertion of the analogue. Precisely, for each finite F ⊆ N, there exists n ≥ 2 such that F ⊆ {0, . . . , 2n}. Clearly, σ(W N )[{0, . . . , 2n}] = σ(W 2n+1 ) (see Figure 4.6). By Theorem 4.37, σ(W 2n+1 ) is critical. But, σ(W N ) does not satisfy the first assertion of the analogue. Clearly, the function

N → N ∖ {0, 1}, defined by n → n + 2 for each n ∈ N, is an isomorphism from σ(W N ) onto σ(W N ) -{0, 1}. Thus, σ(W N ) -{0, 1} is prime. Set X = N ∖ {0, 1}. Since (3, 1), (1, 2) ∈ A(W N ), we have 1 ∈ ⟨X⟩ σ(W N ) . Let n ≥ 1. Since W N [{1, 2n, 2n + 1}] is a 3-cycle, we have 1 ∈ (X σ(W N ) (2n) ∪ X σ(W N ) (2n + 1)).
By Lemma 3.13, 1 ∈ Ext σ(W N ) (X).

Hence, σ(W N )[X ∪ {1}], which is σ(W N ) -0, is prime. It follows that σ(W N ) is not critical, so it does not satisfy the first assertion of the analogue. For the opposite direction, we consider σ(P Z ). As seen at the beginning of this chapter, σ(P Z ) is critical. Hence, it satisfies the first assertion of the analogue. Nevertheless, consider {0, 4} for the finite subset F of Z. Let F ′ be any subset Z containing {0, 4} and such that σ

(P Z )[F ′ ] is prime. Since σ(P Z )[F ′ ] is prime, P Z [F ′
] is connected. Thus, there exists n ≥ 4 such that

F ′ = {0, . . . , n}.
Clearly, σ(P Z )[F ′ ] is not critical because σ(P Z )[F ′ ]n is prime. Consequently, σ(P Z ) does not satisfy the second assertion of the analogue.

Observation 8.4 leads us to introduce the following definition.

Definition 8.12. An infinite 2-structure σ is locally critical if there exist a locally critical partition Q of V (σ) and a function ϕ ∶ V (σ) → Z satisfying the following two assertions

(I1) for every Y ∈ Q such that Y > 1, ϕ ↾Y is an isomorphism from σ[Y ] onto an element of F N ∪ F Z ; (I2) there exists Y ∈ Q such that Y > 1.
Note that we do not require a locally critical 2-structure to be prime.

Lemma 8.13. Given an infinite 2-structure σ, if σ is critical and non finitely critical, then σ is locally critical 1 . Precisely, Assertions (I1) and (I2) hold for the partition Q of V (σ) and a function ϕ ∶ V (σ) → Z defined as in Observation 8.4.

Proof. Let Q be the partition of V (σ) constituted by the vertex sets of the components of P(σ). Using the axiom of choice, consider also a function ϕ ∶ V (σ) → Z defined as in Observation 8.4.

Consider Y ∈ Q such that Y > 1.
There exists a nontrivial component C of P(σ) such that Y = V (C). Denote by ρ the unique 2-structure defined on Z or N such that ϕ ↾Y is an isomorphism from σ[Y ] onto ρ. To verify that Assertion (I1) holds, we distinguish the following two cases.

• Suppose that ϕ ↾Y is an isomorphism from C onto P Z . We have to verify that ρ ∈ F Z . For each n ∈ Z, we have

N P(σ) ((ϕ ↾Y ) -1 (n)) = {(ϕ ↾Y ) -1 (n -1), (ϕ ↾Y ) -1 (n + 1)}.
1 We use the axiom of choice to prove Lemma 8.13.

It follows from Lemma 4.4 that {(ϕ ↾Y ) -1 (n-1), (ϕ ↾Y ) -1 (n+1)} is a module of σ -(ϕ ↾Y ) -1 (n). Thus, {n -1, n + 1} is a module of ρn. Since σ is prime, {(ϕ ↾Y ) -1 (n -1), (ϕ ↾Y ) -1 (n + 1)} is not a module of σ. Hence, we have

(ϕ ↾Y ) -1 (n) ←→ σ {(ϕ ↾Y ) -1 (n -1), (ϕ ↾Y ) -1 (n + 1)}.
It follows that n ←→ ρ {n-1, n+1}. Therefore, {n-1, n+1} is not a module of ρ. Consequently, ρ ∈ F Z .

• Suppose that ϕ ↾Y is an isomorphism from C onto P N . We have to verify that ρ ∈ F N . Let n ≥ 1. As seen in the first case, {n -1, n + 1} is a module of ρn, but not of ρ. Furthermore, we have

N P(σ) ((ϕ ↾Y ) -1 (0)) = {(ϕ ↾Y ) -1 (1)}. It follows from Lemma 4.4 that V (σ)∖{(ϕ ↾Y ) -1 (0), (ϕ ↾Y ) -1 (1)} is a mod- ule of σ -(ϕ ↾Y ) -1 (0). Thus, {2, 3, . . .} is a module of ρ -0. Since σ is prime, V (σ) ∖ {(ϕ ↾Y ) -1
(1)} is not a module of σ. Hence, we have

(ϕ ↾Y ) -1 (1) ←→ σ {(ϕ ↾Y ) -1 (0), (ϕ ↾Y ) -1 (2)}.
It follows that 1 ←→ ρ {0, 2}. Therefore, {0} ∪ {2, 3, . . .} is not a module of ρ. Consequently, ρ ∈ F N .

It follows that Assertion (I1) holds. By the second assertion of Corollary 8.2, P(σ) is nonempty. Thus, P(σ) admits a nontrivial component C. We obtain that V (C) ∈ Q and V (C) > 1. It follows that Assertion (I2) holds. Notation 8.14. Let σ be a locally critical 2-structure. Consider a partition Q of V (σ) and a function ϕ ∶ V (σ) → Z satisfying Assertions (I1) and (I2).

Let Y ∈ Q such that Y > 1. Since Assertion (I1) holds, ϕ ↾Y is an isomorphism from σ[Y ] onto an element of F N ∪ F Z . We denote ϕ ↾Y by ϕ Y . Also, we denote by τ Y the unique 2-structure defined on Z or N such that ϕ Y is an isomorphism from σ[Y ] onto τ Y . Moreover, we denote by C Y the unique component of

P(σ) such that Y = V (C Y ). Lastly, set V even (σ) = {v ∈ V (σ) ∶ ϕ(v) ≡ 0 mod 2} and V odd (σ) = {v ∈ V (σ) ∶ ϕ(v) ≡ 1 mod 2}.
We consider also the partition

P = {Y ∈ Q ∶ Y = 1} ∪ ( ⋃ {Y ∈Q∶ Y >1} {Y ∩ V even (σ), Y ∩ V odd (σ)}) of V (σ).

A generalized quotient

Observation 8.15. Let σ be an infinite, critical, and non finitely critical 2structure. Consider the partition Q of V (σ) and a function ϕ ∶ V (σ) → Z defined as in Observation 8.4.

Let Y ∈ Q such that Y > 1. For instance, suppose that τ Y ∈ F Z . Recall that ϕ Y is an isomorphism from C Y onto P Z . Therefore, for each n ∈ Z, we have

N P(σ) ((ϕ Y ) -1 (2n + 1)) = {(ϕ Y ) -1 (2n), (ϕ Y ) -1 (2n + 2)}. By Lemma 4.4, {(ϕ Y ) -1 (2n), (ϕ Y ) -1 (2n+2)} is a module of σ -(ϕ Y ) -1 (2n+1). In particular, for each v ∈ V (σ) ∖ Y , we have [v, (ϕ Y ) -1 (2n)] σ = [v, (ϕ Y ) -1 (2n + 2)] σ . It follows that Y ∩ V even (σ) is a module of σ[V (σ) ∖ (Y ∩ V odd (σ))]. Similarly, Y ∩ V odd (σ) is a module of σ[V (σ) ∖ (Y ∩ V even (σ))]. The same holds when τ Y ∈ F N .
Observation 8.15 leads us to introduce the following definition. Definition 8.16. Let σ be a 2-structure. Consider partitions P and Q of V (σ) such that P is finer than Q. Hence, for each X ∈ P , there exists Y (X) ∈ Q such that X ⊆ Y (X).

We say that P is a modular partition of σ according to

Q [6] if for any modular partition X, X ′ ∈ P such that Y (X) ≠ Y (X ′ ), X and X ′ are modules of σ[X ∪ X ′ ]. according to
The generalized quotient is defined in the following manner. Consider partitions P and Q of V (σ) such that P is a modular partition of σ according to Q. The generalized quotient σ Q P of σ by P according to Q is defined on generalized quotient V (σ Q P ) = P as follows. Given X 0 , X 1 , X 2 , X 3 ∈ V (σ Q P ), with X 0 ≠ X 1 and

X 2 ≠ X 3 , (X 0 , X 1 ) ≡ (σ Q P ) (X 2 , X 3 ) if Y (X 0 ) = Y (X 1 ) and Y (X 2 ) = Y (X 3 ) (8.3) or ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ Y (X 0 ) ≠ Y (X 1 ), Y (X 2 ) ≠ Y (X 3 ) and (x 0 , x 1 ) ≡ σ (x 2 , x 3 ), where x i ∈ X i for i ∈ {0, 1, 2, 3}.
A priori, (8.3) might appear arbitrary. In fact, it ensures the following property (see the second assertion of Lemma 8.17). Let R be a module of

σ Q P such that {Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅} ≥ 2. For each Y ∈ Q such that Y ∩ (∪R) ≠ ∅, we have Y ⊆ (∪R).
Two results on the generalized quotient follow. Lemma 8.17. Let σ be a 2-structure. Consider two partitions P and Q of V (σ) such that P is a modular partition of σ according to Q.

• For each Y ∈ Q, Y is a module of σ if and only if {X ∈ P ∶ X ⊆ Y } is a module of σ Q P . • For every R ⊆ P such that {Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅} ≥ 2, R is a module of σ Q P if and only if (∪{Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅}) = (∪R)
and ∪R is a module of σ (see Notation 2.9).

Proof. The first assertion follows from the definition of the generalized quotient.

For the second assertion, consider

R ⊆ P such that {Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅} ≥ 2.
To begin, suppose that (∪{Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅}) = (∪R) and ∪R is a module of σ. It follows from the definition of the generalized quotient that R is a module of σ Q P . Conversely, suppose that R is a module of σ Q P . Clearly, (∪R) ⊆ (∪{Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅}). For a contradiction, suppose that

(∪R) ⊊ (∪{Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅}). Let v ∈ (∪{Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅}) ∖ (∪R). There exist X 0 ∈ P ∖ R and Y 0 ∈ Q such that v ∈ X 0 , X 0 ⊆ Y 0 , and Y 0 ∩ (∪R) ≠ ∅. Since Y 0 ∩ (∪R) ≠ ∅, there exists X 1 ∈ R such that X 1 ⊆ Y 0 . Since {Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅} ≥ 2, there exist X 2 ∈ R and Y 1 ∈ Q ∖ {Y 0 } such that X 2 ⊆ Y 1 . Since X 0 ∪ X 1 ⊆ Y 0 , X 2 ⊆ Y 1 and Y 0 ≠ Y 1 , we have (X 0 , X 1 ) ≡ (σ Q P ) (X 0 , X 2 ),
which contradicts the fact that R is a module of σ Q P . Consequently, we have

(∪R) = (∪{Y ∈ Q ∶ Y ∩ (∪R) ≠ ∅}).
It follows from the definition of the generalized quotient that ∪R is a module of σ.

The next result follows easily from Lemma 8.17.

Corollary 8.18. Let σ be a prime 2-structure. Consider two partitions P and Q of V (σ) such that P is a modular partition of σ according to Q. For every nontrivial module R of σ Q P , there exists Y ∈ Q such that (∪R) ⊊ Y and there exists v ∈ Y ∖ (∪R) such that v ←→ σ (∪R).

8.3

The main theorem: Theorem 8.26

In the next lemmas, we pursue the examination of infinite, critical, and non finitely critical 2-structures.

(ϕ Y ) -1 (1) ←→ σ ({(ϕ Y ) -1 (2)} ∪ (V (σ) ∖ Y )).
Proof. It follows from Observation 8.15 that Assertion (I3) holds. For Assertion (I4), consider Y ∈ Q such that Y > 1 and τ Y ∈ F N . Since ϕ Y is an isomorphism from C Y onto P N , we have

N P(σ) ((ϕ Y ) -1 (0)) = {(ϕ Y ) -1 (1)}. By Lemma 4.4, V (σ) ∖ {(ϕ Y ) -1 (0), (ϕ Y ) -1 (1)} is a module of σ -(ϕ Y ) -1 (0).
In particular, we have

(ϕ Y ) -1 (1) ←→ σ ({(ϕ Y ) -1 (2)} ∪ (V (σ) ∖ Y )).
Y ∈ Q, {X ∈ P ∶ X ⊆ Y } = 1 or 2.
In the next two facts, we consider locally critical 2-structures.

Fact 8.21. Let σ be a locally critical 2-structure. Consider a partition Q of V (σ) and a function ϕ ∶ V (σ) → Z satisfying Assertions (I1) and (I2). Suppose also that Assertions (I3) and (I4) hold.

Let Q ′ ⊆ Q such that {Y ∈ Q ′ ∶ Y > 1} ≠ ∅. Set P ′ = {X ∈ P ∶ X ⊆ (∪Q ′ )}.
Suppose that σ[∪Q ′ ] admits a nontrivial module M . The following statements hold.

• If M Q ′ possesses a unique element Y , then {Y ∩ V even (σ), Y ∩ V odd (σ)} is a module of (σ Q P )[P ′ ]. • Suppose that M Q ′ ≥ 2. Given Y ∈ (M Q ′ ), if Y > 1, then (Y ∩ M ) ∩ V even (σ) ≠ ∅ and (Y ∩ M ) ∩ V odd (σ) ≠ ∅. It follows that M P ′ is a module of (σ Q P )[P ′ ]. Moreover, if M P ′ = P ′ , then there exists Y ∈ (M Q ′ ), with Y > 1, such that P ′ ∖ {Y ∩ V even (σ), Y ∩ V odd (σ)} is a module of (σ Q P )[P ′ ].
V even (σ), Y ∩ V odd (σ)} is a module of (σ Q P )[P ′ ]. Now, suppose that M Q ′ ≥ 2.
For a contradiction, suppose that there exists

Y ∈ M Q ′ such that Y > 1 and Y ∩ M = 1. We have ϕ Y ∶ Y → N or ϕ Y ∶ Y → Z.
Thus, there exists an integer n such that Y ∩ M = {(ϕ Y ) -1 (n)}. We distinguish the following two cases. In each of them, we obtain a contradiction.

• Suppose that ϕ Y ∶ Y → Z or ϕ Y ∶ Y → N and n ≥ 1. Since M ≥ 2, there exists x ∈ M ∖ Y .
Since Assertion (I3) holds, P is a modular partition of σ according to Q. Therefore, we have

x ←→ σ {(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)}. Since M is a module of σ[∪Q ′ ] such that {(ϕ Y ) -1 (n), x} ⊆ M and M ∩ {(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)} = ∅, we obtain (ϕ Y ) -1 (n) ←→ σ {(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)}. Since ϕ Y is an isomorphism from σ[Y ] onto τ Y , we obtain n ←→ τ Y {n -1, n + 1}, which contradicts τ Y ∈ F Z ∪ F N . • Suppose that ϕ Y ∶ Y → N and n = 0. By considering x ∈ M ∖ Y , we obtain (ϕ Y ) -1 (1) ←→ σ {(ϕ Y ) -1 (0), x}.
Since Assertion (I4) holds, we have

(ϕ Y ) -1 (1) ←→ σ ({(ϕ Y ) -1 (2), x}.
Hence,

(ϕ Y ) -1 (1) ←→ σ ({(ϕ Y ) -1 (0), (ϕ Y ) -1 (2)}. Since ϕ Y is an isomorphism from σ[Y ] onto τ Y , we obtain 1 ←→ τ Y {0, 2}, which contradicts τ Y ∈ F N . Consequently, for each Y ∈ (M Q ′ ), we have if Y > 1, then Y ∩ M ≥ 2. (8.4)
As above, when

M Q ′ = 1, we obtain (Y ∩ M ) ∩ V even (σ) ≠ ∅ and (Y ∩ M ) ∩ V odd (σ) ≠ ∅. Let Y ∈ (M Q ′ ) such that Y ≥ 2, we have Y ∩ M ≥ 2. We obtain (Y ∩ M ) ∩ V even (σ) ≠ ∅ and (Y ∩ M ) ∩ V odd (σ) ≠ ∅. It follows that M P ′ is the family of X ∈ P ′ such that there exists Y ∈ (M Q ′ ) satisfying Y ⊇ X. Since P is a modular partition of σ according to Q by Assertion (I3), M P ′ is a module of (σ Q P )[P ′ ].

Lastly, suppose that M P

′ = P ′ . Since M is a nontrivial module of σ[∪Q ′ ], there exists Y ∈ (M Q ′ ) such that Y ∖ M ≠ ∅. Moreover, since M P ′ = P ′ , we have Y > 1. We have ϕ Y ∶ Y → N or ϕ Y ∶ Y → Z. For convenience, set R = P ′ ∖ {Y ∩ V even (σ), Y ∩ V odd (σ)}.
Let y ∈ Y ∖ M . We obtain y ←→ σ M.

Since P is a modular partition of σ according to Q and M P ′ = P ′ , we obtain

{z ∈ Y ∶ ϕ Y (z) ≡ ϕ Y (y) mod 2} ←→ (σ Q P )[P ′ ] R. (8.5) Therefore, if (Y ∖ M ) ∩ V even (σ) ≠ ∅ and (Y ∖ M ) ∩ V odd (σ) ≠ ∅, then R is a module of (σ Q P )[P ′ ]. Thus, suppose that (Y ∖ M ) ∩ V even (σ) = ∅ or (Y ∖ M ) ∩ V odd (σ) = ∅. (8.6) By Assertion (M2) of Proposition 2.5, M ∩ Y is a module of σ[Y ]. By (8.4), M ∩ Y ≥ 2. Since Y ∖ M ≠ ∅, M ∩ Y is a nontrivial module of σ[Y ]. Thus, ϕ Y (M ∩ Y ) is a nontrivial module of τ Y . Since Assertion (I1) holds, τ Y ∈ F N ∪ F Z . It
∈ F N and M ∩ Y = Y ∖ {(ϕ Y ) -1 (0)}.
Since Assertion (I4) holds, we have

(ϕ Y ) -1 (1) ←→ σ ({(ϕ Y ) -1 (2)} ∪ (V (σ) ∖ Y )).
Since P is a modular partition of σ according Q, we obtain

(Y ∩ V odd (σ)) ←→ (σ Q P )[P ′ ] R.
Furthermore, since (ϕ Y ) -1 (0) ∈ Y ∖ M , it follows from (8.5) that

(Y ∩ V even (σ)) ←→ (σ Q P )[P ′ ] R. Therefore, R is a module of (σ Q P )[P ′ ].
The next fact follows easily from Fact 8.21.

Fact 8.22. Let σ be a locally critical 2-structure. Consider a partition Q of V (σ) and a function ϕ ∶ V (σ) → Z satisfying Assertions (I1) and (I2). Suppose also that Assertions (I3) and (I4) hold. Let Q ′ be a nonempty subset of Q such that

{Y ∈ Q ′ ∶ Y > 1} ≠ ∅. Set P ′ = {X ∈ P ∶ X ⊆ (∪Q ′ )}. Suppose that {Y ∈ Q ′ ∶ Y > 1} ≥ 2 or Q ′ ≥ 3. (8.7) If σ[∪Q ′ ] is decomposable, then (σ Q P )[P ′ ] is as well.
We go back to the study of infinite, critical, and non finitely critical 2structures. The next lemma follows easily from Fact 8.22.

Lemma 8.23. Let σ be an infinite, critical, and non finitely critical 2-structure. Consider the partition Q of V (σ) and a function ϕ ∶ V (σ) → Z defined as in Observation 8.4. The following assertion holds

(I6) If {Y ∈ Q ∶ Y > 1} ≥ 2 or Q ≥ 4, (8.8
)

then {v ∈ V (σ) ∶ {v} ∈ Q} ⊆ (P ∖ S (σ Q P )).
Proof. Consider v ∈ V (σ) such that {v} ∈ Q. Furthermore, suppose that (8.8) holds. Set

Q ′ = Q ∖ {{v}}. Since Assertion (I2) holds, {Y ∈ Q ∶ Y > 1} ≠ ∅. Furthermore,
. If V (σ) ∖ V (C) ≥ 2, then • V (σ) ∖ V (C) = 2; • there exists a unique v ∈ V (σ) ∖ V (C) such that v ←→ σ V (C); • σ[V (C)] is decomposable. Proof. To use Notation 8.14, set Y = V (C). Obviously, Y ∈ Q. To begin, we show that for each W ⊊ V (σ) ∖ V (C), we have σ[Y ∪ W ] is decomposable. (8.9) Otherwise, consider W ⊊ V (σ)∖Y such that σ[Y ∪W ] is prime. Since V (σ)∖(Y ∪ W
) is finite, it follows from Corollary 3.21 that there exist v, w ∈ V (σ)∖(Y ∪W ) such that σ -{v, w} is prime. We cannot have v = w because σ is critical. Moreover, we cannot have v ≠ w because v and w are isolated in P(σ). It follows that (8.9) holds. Set

S(Y ) = {v ∈ V (σ) ∖ Y ∶ v ←→ σ Y }.
We prove that either for every v ∈ S(Y ), we have

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [v, (ϕ Y ) -1 (0)] σ = [(ϕ Y ) -1 (0), (ϕ Y ) -1 (2)] σ and [v, (ϕ Y ) -1 (1)] σ = [(ϕ Y ) -1 (1), (ϕ Y ) -1 (3)] σ , (8.10)
or for every v ∈ S(Y ), we have

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [v, (ϕ Y ) -1 (0)] σ = [(ϕ Y ) -1 (2), (ϕ Y ) -1 (0)] σ and [v, (ϕ Y ) -1 (1)] σ = [(ϕ Y ) -1 (3), (ϕ Y ) -1 (1)] σ . ( 8
P ′ = {Y ∩ V even (σ), Y ∩ V odd (σ), {v}}. For a contradiction, suppose that M Q ′ = 1. We obtain M Q ′ = {Y }. It follows from Fact 8.21 that {Y ∩ V even (σ), Y ∩ V odd (σ)} is a module of (σ Q P )[P ′ ], which contradicts v ∈ S(Y ). Therefore, M Q ′ ≥ 2.
Hence, we have v ∈ M . Furthermore, it follows from Fact 8.21 that (Y ∩ M ) ∩ V even (σ) ≠ ∅ and (Y ∩ M ) ∩ V odd (σ) ≠ ∅. Thus, there exist p, q ∈ Z such that

(ϕ Y ) -1 (2p), (ϕ Y ) -1 (2q + 1) ∈ Y ∩ M.
In particular, we have

M ∩ Y ≥ 2. By Assertion (M2) of Proposition 2.5, M ∩ Y is a module of σ[Y ]. Since v ∈ M and M ⊊ (Y ∪ {v}), we have (M ∩ Y ) ≠ Y . Moreover, since M ∩ Y ≥ 2, M ∩ Y is a nontrivial module of σ[Y ]. It follows that ϕ Y (Y ∩ M ) is a nontrivial module of τ Y .
We distinguish the following two cases.

• Suppose that there exists n ∈ Z such that

ϕ Y (Y ∩ M ) ⊆ {. . . , n -1, n}. (8.12)
There exists m ≥ 0 such that

(ϕ Y ) -1 (2m), (ϕ Y ) -1 (2m+1) ∈ (Y ∖M ). Since (ϕ Y ) -1 (2p), (ϕ Y ) -1 (2q+1) ∈ Y ∩M and v ∈ M , we obtain [v, (ϕ Y ) -1 (2m)] σ = [(ϕ Y ) -1 (2p), (ϕ Y ) -1 (2m)] σ , and hence [v, (ϕ Y ) -1 (2m)] σ = [(ϕ Y ) -1 (0), (ϕ Y ) -1 (2)] σ .
Since P is a modular partition of σ according to Q by Assertion (I3) (see Lemma 8.19), we have v ←→ σ Y ∩ V even (σ). It follows that

[v, (ϕ Y ) -1 (0)] σ = [(ϕ Y ) -1 (0), (ϕ Y ) -1 (2)] σ .
Similarly, we obtain

[v, (ϕ Y ) -1 (1)] σ = [(ϕ Y ) -1 (1), (ϕ Y ) -1 (3)] σ .
Therefore, v satisfies (8.10). Consequently, if M satisfies (8.12), then v satisfies (8.10).

• Suppose that there exists n ∈ Z such that

{n, n + 1, . . .} ⊆ ϕ Y (Y ∩ M ). ( 8
′ ∈ Z such that ϕ Y (Y ∩ M ) = {n ′ , n ′ + 1, . . .}.
We verify that v satisfies (8.11). We distinguish the following two cases.

▸ Suppose that τ Y ∈ F Z or τ Y ∈ F N and n ′ ≥ 2. There exists m ≥ 0 such that (ϕ Y ) -1 (2m), (ϕ Y ) -1 (2m+1) ∈ (Y ∖M ). Since (ϕ Y ) -1 (2p) ∈ Y ∩ M and v ∈ M , we obtain [v, (ϕ Y ) -1 (2m)] σ = [(ϕ Y ) -1 (2p), (ϕ Y ) -1 (2m)] σ ,
and hence

[v, (ϕ Y ) -1 (2m)] σ = [(ϕ Y ) -1 (2), (ϕ Y ) -1 (0)] σ .
Since P is a modular partition of σ according to Q by Assertion (I3), we obtain v ←→ σ Y ∩ V even (σ). It follows that

[v, (ϕ Y ) -1 (0)] σ = [(ϕ Y ) -1 (2), (ϕ Y ) -1 (0)] σ .
Similarly, we obtain

[v, (ϕ Y ) -1 (1)] σ = [(ϕ Y ) -1 (3), (ϕ Y ) -1 (1)] σ .
Therefore, v satisfies (8.11).

▸ Suppose that τ Y ∈ F N and n ′ = 1. We have ϕ Y (Y ∩ M ) = {1, 2, . . .}.

Since v ∈ M , we obtain

[v, (ϕ Y ) -1 (0)] σ = [(ϕ Y ) -1 (2), (ϕ Y ) -1 (0)] σ . Since τ Y ∈ F N , ϕ Y is an isomorphism from C onto P N . Hence, we have N P(σ) ((ϕ Y ) -1 (0)) = {(ϕ Y ) -1 (1)}. By Lemma 4.4, V (σ) ∖ {(ϕ Y ) -1 (0), (ϕ Y ) -1 (1)} is a module of σ - (ϕ Y ) -1 (0). It follows that [v, (ϕ Y ) -1 (1)] σ = [(ϕ Y ) -1 (3), (ϕ Y ) -1 (1)] σ .
Thus, v satisfies (8.11).

Consequently, if M satisfies (8.13), then v satisfies (8.11).

It follows that for each v ∈ S(Y ), v satisfies (8.10) or (8.11). Let v ∈ S(Y ).

Since v ←→ σ Y , we obtain

[v, (ϕ Y ) -1 (0)] σ ≠ [v, (ϕ Y ) -1 (1)] σ .
It follows from (8.10) or (8.11) that

[(ϕ Y ) -1 (0), (ϕ Y ) -1 (2)] σ ≠ [(ϕ Y ) -1 (1), (ϕ Y ) -1 (3)] σ . Therefore, [0, 2] τ Y ≠ [1, 3] τ Y , so τ Y ≠ σ(L(N)
Q ′ = {Y, {v}, {w}}.
We obtain

P ′ = {Y ∩ V even (σ), Y ∩ V odd (σ), {v}, {w}}. By Fact 8.21, if M Q ′ = 1, then {Y ∩ V even (σ), Y ∩ V odd (σ)} is a module of (σ Q P )[P ′ ], which contradicts v ∈ S(Y ). Therefore, M Q ′ ≥ 2.
• if V (σ) = Y , then ϕ ↾Y is an isomorphism from σ[Y ] onto a prime element of F N ∪ F Z ; • if V (σ)∖Y = 1, then (V (σ)∖Y ) ←→ σ Y and ϕ ↾Y is an isomorphism from σ[Y ] onto a decomposable element of F N ∪ F Z ; • if V (σ) ∖ Y = 2, then there exists a unique v ∈ V (σ) ∖ Y such that v ←→ σ Y , and ϕ ↾Y is an isomorphism from σ[Y ] onto a decomposable element of F N ∪ F Z .
Proof. Suppose that there exists a unique [START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] 2 ). Consider an infinite, critical, and non finitely critical 2-structure σ. Let Q be the partition of V (σ) constituted by the vertex sets of the components of P(σ). Using the axiom of choice, consider also a function ϕ ∶ V (σ) → Z defined as in Observation 8.4. Then, Assertions (I1),...,(I7) hold.

Y ∈ Q such that Y > 1. If V (σ) = Y , then ϕ ↾Y is an isomorphism from σ[Y ] onto a prime element of F N ∪F Z because σ is prime. Suppose that V (σ)∖Y contains a unique element v. Since σ is prime, Y is not a module of σ, and hence v ←→ σ Y . Moreover, since σ is critical, σ -v is decomposable. Thus, ϕ ↾Y is an isomorphism from σ[Y ] onto a decomposable element of F N ∪F Z . Finally, when V (σ)∖Y ≥ 2,

Locally critical 2-structures

The purpose of this section is to establish the following theorem. Theorem 8.27 (Boubabbous and Ille [START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] 3 ). Let σ be a locally critical 2-structure. Consider a partition Q of V (σ) and a function ϕ ∶ V (σ) → Z satisfying Assertions (I1) and (I2). Suppose also that Assertions (I3),...,(I6) hold. If (8.8) holds, then σ is critical and non finitely critical.

Before proving Theorem 8.27, we establish the following three results. Lemma 8.28. Let σ be a locally critical 2-structure. Consider a partition Q of V (σ) and a function ϕ ∶ V (σ) → Z satisfying Assertions (I1) and (I2). Suppose also that Assertions (I3), (I4), and (I5) hold. If

{Y ∈ Q ∶ Y > 1} ≥ 2 or Q ≥ 3, (8.16)
then σ is prime.

Proof. We consider the partition P of V (σ) defined as in Notation 8.14. Suppose that (8.16) holds. We have to show that σ is prime. We utilize Fact 8.22 as follows. Set Q ′ = Q. We obtain P ′ = P . Since Assertion (I2) holds,

{Y ∈ Q ∶ Y > 1} ≠ ∅.
∈ Q such that Y > 1. For every v ∈ Y , σ -v is decomposable. Proof. Let v ∈ Y . Set n = ϕ Y (v).
Since Assertions (I1) and (I2) hold, we consider the partition P of V (σ) defined as in Notation 8.14. First, suppose that τ Y ∈ F Z or τ Y ∈ F N and n ≥ 1. We obtain that {n -

1, n + 1} is a module of τ Y -n. Since ϕ Y is an isomorphism from σ[Y ] onto τ Y , {(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)} is a module of σ[Y ] -v. Since Assertion (I3)
holds, P is a modular partition of σ according to Q. We obtain

{(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)} ←→ σ V (σ) ∖ Y. It follows that {(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)} is a module of σ -v.
Second, suppose that τ Y ∈ F N and n = 0. We obtain that {2, 3, .

. .} is a module of τ Y -0. Since ϕ Y is an isomorphism from σ[Y ] onto τ Y , Y ∖ {(ϕ Y ) -1 (0), (ϕ Y ) -1 (1)} is a module of σ[Y ] -(ϕ Y ) -1 (0). Since Assertion (I4) holds, we have (ϕ Y ) -1 (1) ←→ σ ({(ϕ Y ) -1 (2)} ∪ (V (σ) ∖ Y )). It follows that V (σ) ∖ {(ϕ Y ) -1 (0), (ϕ Y ) -1 (1)} is a module of σ -v.
Proposition 8.30. Let σ be a locally critical 2-structure. Consider a partition Q of V (σ) and a function ϕ ∶ V (σ) → Z satisfying Assertions (I1) and (I2). Suppose also that Assertions (I3), (I4), and (I5) hold. For every Y ∈ Q such that Y > 1, the following two assertions hold

(J1) for each n ∈ V (τ Y ) (see Notation 8.14), σ -{(ϕ Y ) -1 (n), (ϕ Y ) -1 (n + 1)} is isomorphic to σ; (J2) there exists a nontrivial component C of P(σ) such that Y = V (C), and ϕ Y is an isomorphism from C onto P N or P Z . Proof. Let Y ∈ Q such that Y > 1. Since Assertion (I1) holds, ϕ Y is an isomorphism from σ[Y ] onto τ Y ∈ F N ∪ F Z .
Furthermore, since Assertions (I3), (I4), and (I5) hold, it follows from Lemma 8.28 that σ is prime. We prove that

for each n ∈ V (τ Y ), {(ϕ Y ) -1 (n), (ϕ Y ) -1 (n + 1)} ∈ E(P(σ)). (8.17)
More strongly, we establish that

for each n ∈ V (τ Y ), σ -{(ϕ Y ) -1 (n), (ϕ Y ) -1 (n + 1)} is isomorphic to σ, (8.18)
that is, Assertion (J1) holds. Let n ∈ V (τ Y ). Consider the function

f ∶ V (τ Y ) → V (τ Y ) ∖ {n, n + 1} p ≤ n -1 → p, p ≥ n → p + 2. (8.19)
Clearly, f is strictly increasing and preserves the parity. Since τ Y ∈ F N ∪ F Z , it follows from Lemmas 8.6 and 8.

7 that f is an isomorphism from τ Y onto τ Y -{n, n + 1}. Since Assertion (I1) holds, ϕ Y is an isomorphism from σ[Y ] onto τ Y . Thus, ((ϕ Y ) -1 ) ↾V (τ Y )∖{n,n+1} ○ f ○ ϕ Y is an isomorphism from σ[Y ] onto σ[Y ] -{(ϕ Y ) -1 (n), (ϕ Y ) -1 (n + 1)}. For convenience, set ψ = ((ϕ Y ) -1 ) ↾V (τ Z )∖{n,n+1} ○ f ○ ϕ Y .
Consider the extension ψ ∪ Id (V (σ)∖Y ) of ψ by the identity function on V (σ) ∖ Y defined by

V (σ) → V (σ) ∖ {(ϕ Y ) -1 (n), (ϕ Y ) -1 (n + 1)} w ∈ Y → ψ(w), w ∈ (V (σ) ∖ Y ) → w. (8.20)
Since Assertion (I3) holds, P is a modular partition of σ according to Q. It follows that ψ∪Id (V (σ)∖Y ) is an isomorphism from σ onto σ-{(ϕ Y ) -1 (n), (ϕ Y ) -1 (n+ 1)}. Consequently, (8.18) holds, so Assertion (J1) holds. Moreover, (8.17) holds because σ is prime.

To prove that Assertion (J2) holds, we distinguish the following two cases.

• Suppose that ϕ Y ∶ Y → Z. Let n ∈ Z. Since (8.17) holds,

{(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)} ⊆ N P(σ) ((ϕ Y ) -1 (n)).
Since Assertions (I3) and (I4) hold, it follows from Lemma 8.29 that σ -(ϕ Y ) -1 (n) is decomposable. By Lemma 4.4,

N P(σ) ((ϕ Y ) -1 (n)) = {(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)}.
It follows that P(σ)[Y ] is a component of P(σ), and ϕ Y is an isomorphism from P(σ)[Y ] onto P Z .

• Suppose that ϕ Y ∶ Y → N. As previously, we have

N P(σ) ((ϕ Y ) -1 (n)) = {(ϕ Y ) -1 (n -1), (ϕ Y ) -1 (n + 1)}
for each n ≥ 1. Furthermore, since (8.17) holds, we have

(ϕ Y ) -1 (1) ∈ N P(σ) ((ϕ Y ) -1 (0)). (8.21)
Since Assertion (I4) holds, we have

(ϕ Y ) -1 (1) ←→ σ ({(ϕ Y ) -1 (2)} ∪ (V (σ) ∖ Y )). ( 8
(ϕ Y ) -1 (1) ←→ σ (Y ∖ {(ϕ Y ) -1 (0), (ϕ Y ) -1 (1)}). ( 8
(ϕ Y ) -1 (1) ←→ σ (V (σ) ∖ {(ϕ Y ) -1 (0), (ϕ Y ) -1 (1)}). Thus, v ∈ N P(σ) ((ϕ Y ) -1 (0)) for every v ∈ V (σ) ∖ {(ϕ Y ) -1 (0), (ϕ Y ) -1 (1)}. Since (ϕ Y ) -1 (1) ∈ N P(σ) ((ϕ Y ) -1 (0)
) by (8.21), we obtain To continue, we prove that σ is critical. Let v ∈ V (σ). We must verify that σv is decomposable. Denote by Y the unique element of Q containing v. To begin, suppose that Y > 1. Since Assertions (I3) and (I4) hold, it follows from Lemma 8.29 that σv is decomposable. Now, suppose that Y = {v}. Since Assertion (I6) and (8.8) hold, (σ Q P ) -{v} is decomposable. Let R be a nontrivial module of (σ Q P ) -{v}. Set

N P(σ) ((ϕ Y ) -1 (0)) = {(ϕ Y ) -1 (1)}. Consequently, P(σ)[Y ] is a component of P(σ),
Q ′ = Q ∖ {{v}} and P ′ = P ∖ {{v}}.
Clearly, P ′ is a modular partition of σv according to Q ′ . Moreover, we have

(σ Q P ) -{v} = (σ -v) (P ′ ) Q ′ .
We apply Lemma 8.17 to σv together with partitons P ′ and Q ′ as follows. We distinguish the following two cases.

• Suppose that (∪R) Q ′ = 1. Denote by Z the unique element of (∪R) Q ′ .

Since R ≥ 2 and {X ∈ P ′ ∶ X ⊆ Z} ≤ 2, we have R = {X ∈ P ′ ∶ X ⊆ Z}. It follows from the first assertion of Lemma 8.17 that Z is a module of σv.

• Suppose that (∪R) Q ′ ≥ 2. It follows from the second assertion of Lemma 8.17 that

(∪{Y ∈ Q ′ ∶ Y ∩ (∪R) ≠ ∅}) = (∪R) and (∪R) is a module of σ -v. Since (∪{Y ∈ Q ′ ∶ Y ∩ (∪R) ≠ ∅}) = (∪R) and R is a nontrivial module of (σ -v) (P ′ ) Q ′ , (∪R) is a nontrivial module of σ -v.
Consequently, σ is critical. Finally, we verify that σ is not finitely critical. Since Assertion (I2) holds, there exists Y ∈ Q such that Y > 1. Moreover, since Assertions (I3), (I4), and (I5) hold, it follows from Assertion (J2) of Proposition 8.30 that there exists a nontrivial component C of P(σ) such that Y = V (C). Hence, there exist distinct v, w ∈ Y such that σ -{v, w} is prime.

Remark 8.31. Consider the tournament T defined on V (T ) = Z × {0, 1} which satisfies

• for i = 0 or 1, the function ψ i ∶ Z → Z × {i}, defined by n → (n, i) for every n ∈ Z, is an isomorphism from U Z onto T [Z × {i}];
• for p, q ∈ Z, we have ((2p, 0), (2q, 1)) ∈ A(T ), ((2p+1, 0), (2q+1, 1)) ∈ A(T ), ((2p + 1, 1), (2q, 0)) ∈ A(T ), and ((2p, 1), (2q + 1, 0)) ∈ A(T ).

First, consider the partition

Q = {Z × {0}, Z × {1}} of V (σ(T )) and the function ϕ ∶ V (σ(T )) → Z defined by ϕ ↾(Z×{0}) = (ψ 0 ) -1
and ϕ ↾(Z×{1}) = (ψ 1 ) -1 . We obtain

P = {(2Z) × {0}, (2Z + 1) × {0}, (2Z) × {1}, (2Z + 1) × {1}}.
It is not difficult to verify that σ(T ) satisfies Assertions (I1),...,(I6) with Q and ϕ. Furthermore, since {Y ∈ Q ∶ Y > 1} = 2, (8.8) holds. By Theorem 8.27, σ(T ) is critical but not finitely critical. Moreover, it follows from Assertion (J2) of Proposition 8.30 that Z×{0} and Z×{1} are the vertex sets of the components of P(σ(T )).

Second, consider the partition

Q = {Z × {0}} ∪ {{(n, 1)} ∶ n ∈ Z} of V (σ(T )
) and the same function ϕ as before. We obtain

P = {(2Z) × {0}, (2Z + 1) × {0}} ∪ {{(n, 1)} ∶ n ∈ Z}.
Once again, σ(T ) satisfies Assertions (I1),...,(I6) with Q and ϕ. Furthermore, since Q is infinite, (8.8) holds. Nevertheless, it follows only from Proposition 8.30 that Z × {0} is the vertex set of a component of P(σ(T )). Consequently, it is not possible to determine the primality graph from Assertions (I1),...,(I6) only.

The next result follows from Theorems 8.26 and 8.27.

Corollary 8.32 (Boubabbous and Ille [6]4 ). Given an infinite 2-structure σ, if σ is critical and non finitely critical, then the following two assertions hold

• for any distinct v, w ∈ V (σ), σ -{v, w} is prime if and only if σ -{v, w} is isomorphic to σ;

• there exist distinct v, w ∈ V (σ) such that σ -{v, w} is isomorphic to σ.

Proof. Suppose that σ is an infinite, critical, and non finitely critical 2-structure. By the second assertion of Corollary 8.2, there exist distinct v, w ∈ V (σ) such that σ -{v, w} is prime. Now, consider any distinct v, w ∈ V (σ) such that σ -{v, w} is prime. We have to verify that σ -{v, w} is isomorphic to σ. Consider the partition Q of V (σ) and a function ϕ ∶ V (σ) → Z defined as in Observation 8.4. There exists Y ∈ Q such that v, w ∈ Y . We have {v, w} ∈ E(P(σ)). Since ϕ Y is an isomorphism from P(σ)[Y ] onto P N or P Z , there exists n ∈ Z such that

{v, w} = {(ϕ Y ) -1 (n), (ϕ Y ) -1 (n + 1)}.
Since Assertions (I3), (I4), and (I5) hold, it follows from Assertion (J1) of Proposition 8.30 that σ -{(ϕ Y ) -1 (n), (ϕ Y ) -1 (n + 1)} is isomorphic to σ.

The next result follows from Corollary 8.32 and Theorem 5.8. It is the analogue of Theorem 5.3 in the infinite case.

Corollary 8.33. Given an infinite and prime 2-structure σ, if there exists a finite subset F of V (σ) such that F ≥ 2 and σ -F is prime, then there exist distinct v, w ∈ V (σ) such that σ -{v, w} is prime.

Proof. It follows from Corollary 3.20 that there exists F ′ ⊆ F such that F ′ = 2 or 3 and σ -F ′ is prime. The conclusion is obvious when F ′ = 2. Hence, suppose that F ′ = 3. By Corollary 3.20 again, there exists x ∈ F ′ such that σx is prime. Set τ = σx.

Clearly, if τ is not critical, then we conclude directly. Thus, suppose that τ is critical. By denoting by y and z the two elements of F ′ ∖ {x}, we obtain τ -{y, z} is prime. Therefore, τ is not finitely critical. By applying three times the second assertion of Corollary 8.32 from τ , we obtain

F ′′ ⊆ V (τ ) such that F ′′ = 6 and τ -F ′′ is isomorphic to τ . Since τ -F ′′ = σ -({x} ∪ F ′′ ),
G defined on V (G) = Z ∪ {∞} by G[Z] = P Z and E(G) = E(P Z ) ∪ {{0, ∞}}.
As observed at the beginning of this chapter, G -∞ = P Z is prime. Hence, σ(G) -∞ is prime as well. Set

X = V (G) ∖ {∞}.
Since {0, ∞} ∈ E(G) and {1, ∞} ∈ E(G), we have

∞ ∈ ⟨X⟩ σ(G) . Furthermore, since d G (∞) = 1 and d (G-∞) (n) = 2 for every n ∈ Z, we obtain ∞ ∈ X σ(G) (n) for each n ∈ Z.
It follows from Lemma 3.13 that ∞ ∈ Ext σ(G) (X), so σ(G) is prime too. However, for each finite subset

F of Z, with F ≥ 2, G -F is disconnected. It follows that σ(G) -F is decomposable for each finite subset F of Z such that F ≥ 2.
We complete this section with the following example which is constructed from the graph H Z (see Example 8.9) and the graph G defined on V (G) = Z ∪ {∞} in Remark 8. [START_REF] Spinrad | P4-trees and substitution decomposition[END_REF]. It shows that Proposition 8.24 does not hold if we do not suppose that the primality graph admits finitely many trivial components. • for every p ∈ Z, {2p, ∞ 0 } ∈ E(H).

We prove that σ(H) is prime, critical, but not finitely critical. Precisely, we show that P(σ(H))

[Z] = P Z , P(σ(H))[Z] is a component of P(σ(H)), and ∞ n is isolated in P(σ(H)) for each n ∈ Z. Set R = {{∞ n } ∶ n ∈ Z}.

Consider the partition

Q = {Z} ∪ R of V (σ(H)) and the function ϕ ∶ V (σ(H)) → Z defined by ϕ ↾Z = Id Z and ϕ(∞ n ) = 0 for every n ∈ Z.
We verify that σ(H) satisfies Assertions (I1),...,(I6) with Q and ϕ. As seen at the end of Example 8.9, σ(H Z ) ∈ F Z . Hence, Assertion (I1) holds. Assertion (I2) holds because Z ∈ Q. For Assertion (I3), we obtain

P = {2Z, 2Z + 1} ∪ R.
It follows from the definition of H that for each n ∈ Z, we have ∞ n ←→ σ(H) (2Z) and ∞ n ←→ σ(H) (2Z + 1). Thus, Assertion (I3) holds. Obviously, Assertion (I4) holds. Clearly, (σ(H

) Q P )[R] ≃ σ(H)[{∞ n ∶ n ∈ Z}]. Since H[{∞ n ∶ n ∈ Z}] ≃ P Z , we obtain that (σ(H) Q P )[R] is prime. Clearly, 2Z + 1 ∈ ⟨R⟩ (σ(H) Q P ) (see Notation 3.12).
Furthermore, the function

Z ∪ {∞} → {0} ∪ {∞ n ∶ n ∈ Z} ∞ → 0, n ∈ Z → ∞ n , is an isomorphism from the graph G defined in Remark 8.34 onto H[{0} ∪ {∞ n ∶ n ∈ Z}]. Since P is a modular partition of σ(H) according to Q, we have (σ(H) Q P )[R ∪ {2Z}] ≃ σ(H)[{0} ∪ {∞ n ∶ n ∈ Z}]. As seen in Remark 8.34, σ(G) is prime. It follows that (σ(H) Q P )[R ∪ {2Z}] is prime, so 2Z ∈ Ext (σ(H) Q P ) (R).
It follows from the definition of the generalized quotient (see Definition 8.16)

that [2Z + 1, {∞ 0 }] (σ(H) Q P ) ≠ [2Z + 1, 2Z] (σ(H) Q P ) .
Hence, R ∪ {2Z} is not a module of σ(H) Q P . It follows from Assertion (P2) of Lemma 3.17 that σ(H) Q P is prime. Therefore, Assertion (I5) holds. For Assertion (I6), consider n ∈ Z. We must show that (σ(H) Q P ) -{∞ n } is decomposable. Suppose that n ≥ 0. We obtain that

{∞ n+1 , ∞ n+2 , . . .} is a component of H -(∞ n ). (8.24)
It follows that {{∞ n+1 }, {∞ n+2 }, . . .} is a nontrivial module of (σ(H) Q P ) -{∞ n }. Suppose that n ≤ 0. We obtain that

{. . . , ∞ n-2 , ∞ n-1 } is a component of H -(∞ n ). (8.25) It follows that {. . . , {∞ n-2 }, {∞ n-1 }} is a nontrivial module of (σ(H) Q P ) - {∞ n }.
V (σ(H)) ∖ {∞ n }) ∖ M are infinite. It follows that (H -(∞ n )) -v is decomposable for every v ∈ (V (σ(H)) ∖ {∞ n }). Hence, ∞ n is isolated in P(σ(H)).
Consequently, P(σ(H)) admits a unique nontrivial component and infinitely many trivial components.

Epilogue on Assertion (I7)

In the next four facts, we complete the study begun in Assertion (I7) of Theorem 8.26, and in Theorem 8.27 when (8.8) does not hold. Precisely, we are interested in the infinite, critical, and non finitely critical 2-structures the primality graph of which admits one nontrivial component and one or two trivial ones.

Fact 8.36. Given a 2-structure σ defined on V (σ) = Z ∪ {∞}, σ is critical, P(σ)[Z] = P Z , and ∞ is isolated in P(σ) if and only if the following assertions hold

• σ -∞ ∈ F Z ; • ∞ ←→ σ (2Z), ∞ ←→ σ (2Z + 1), and [0, ∞] σ ≠ [1, ∞] σ ;
• at least one of the following two cases occurs:

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [0, 1] σ = [0, 2] σ , [1, 2] σ = [1, 3] σ , and [0, 1] σ ≠ [0, ∞] σ or [1, 2] σ ≠ [1, ∞] σ , (8.26) or ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [1, 2] σ = [0, 2] σ , [0, 1] σ = [1, 3] σ , and [0, 2] σ ≠ [∞, 2] σ or [0, 1] σ ≠ [∞, 1] σ . (8.27)
Proof. To begin, suppose that σ is critical, P(σ)[Z] = P Z , and ∞ is isolated in P(σ). First, we verify that σ -∞ ∈ F Z . Let n ∈ Z. Since P(σ)[Z] = P Z and ∞ is isolated in P(σ), we have

N P(σ) (n) = {n -1, n + 1}.
Since σ is critical, it follows from Lemma 4.4 that {n -1, n + 1} is a module of σn. By Assertion (M2) of Proposition 2.5, {n -

1, n + 1} is a module of (σ-∞)-n. Since σ is prime, {n-1, n+1} is not a module of σ. Since {n-1, n+1} is a module of σ -n, we obtain n ←→ σ {n -1, n + 1}. It follows that {n -1, n + 1} is not a module of σ -∞. Consequently, σ -∞ ∈ F Z . Second, we show that ∞ ←→ σ (2Z), ∞ ←→ σ (2Z + 1), and [0, ∞] σ ≠ [1, ∞] σ . Let n ∈ Z. As seen above, {2n, 2n + 2} is a module of σ -(2n + 1). Hence, ∞ ←→ σ {2n, 2n + 2}. It follows that ∞ ←→ σ (2Z). Similarly, we have ∞ ←→ σ (2Z + 1). Since σ is prime, Z is not a module of σ. Since ∞ ←→ σ (2Z) and ∞ ←→ σ (2Z + 1), we obtain [0, ∞] σ ≠ [1, ∞] σ .
Third, we prove that (8.26) or (8.27) hold. Since σ is critical, σ -∞ is decomposable. Let M be a nontrivial module of σ -∞. By the third assertion of Lemma 8.8, M is a nontrivial module of L Z . Thus, M admits a least or a greatest element. In the first instance, there exists n ∈ Z such that n, n + 1 ∈ M and M ⊆ {n, n + 1, . . .}. We obtain [0, 1] σ = [0, 2] σ and [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] σ = [1,3] σ as in the first assertion of Lemma 8.8. Furthermore, since M ∪ {∞} is not a module of σ, we obtain [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF]1] (8.26) holds. Similarly, when M admits a greatest element, (8.27) holds.

σ ≠ [0, ∞] σ or [1, 2] σ ≠ [1, ∞] σ . Therefore,
Conversely, suppose that the three assertions above hold. To begin, we verify that Assertions (I1),...,(I5) hold. Set Q = {Z, {∞}} and consider the function ϕ ∶ V (σ) → Z defined by

ϕ ↾Z = Id Z and ϕ(∞) = 0.
We obtain

P = {2Z, 2Z + 1, {∞}}.
Since σ -∞ ∈ F Z , σ satisfies Assertion (I1) with Q and ϕ. Since Z ∈ Q, Assertion (I2) holds. It follows that σ is locally critical. Furthermore, since ∞ ←→ σ (2Z) and ∞ ←→ σ (2Z + 1), P is a modular partition of σ according to Q. Hence,

Assertion (I3) holds. Since [0, ∞] σ ≠ [1, ∞] σ , we have [2Z, {∞}] (σ Q P ) ≠ [2Z + 1, {∞}] (σ Q P ) .
It follows from the definition of the generalized quotient (see Definition 8.16) that σ Q P is prime. Thus, Assertion (I5) holds. Obviously, Assertion (I4) holds. For a contradiction, suppose that σ admits a nontrivial module M . We utilize Fact 8.21 with Q ′ = Q as follows. Since [0, ∞] σ ≠ [1, ∞] σ , {2Z, 2Z + 1} is not a module of σ Q P . It follows from Fact 8.21 that M ∩(2Z) ≠ ∅, M ∩(2Z+1) ≠ ∅, and ∞ ∈ M . Thus, M ∖ {∞} is a nontrivial module of σ -∞. For instance, assume that M ∖ {∞} admits a least element n. Hence, n, n + 1 ∈ M ∖ {∞} and M ∖ {∞} ⊆ {n, n + 1, . . .}. Since σ -∞ ∈ F Z , we obtain that {n, n + 1, . . .} is a module of σ -∞. By the first assertion of Lemma 8.8, we have [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF]1]

σ = [0, 2] σ and [1, 2] σ = [1, 3] σ . Since M is a module of σ, we obtain [0, 1] σ = [0, ∞] σ and [1, 2] σ = [1, ∞] σ . Hence, (8.26) does not hold. Since [0, ∞] σ ≠ [1, ∞] σ , we have [0, 1] σ ≠ [1, 2] σ . Since [0, 1] σ = [0, 2] σ , we obtain [0, 2] σ ≠ [1, 2] σ .
, . . .} is a module of σ -∞ or {. . . , 0, 1} is a module of σ -∞. Hence, σ -∞ is decomposable. Consequently, σ is critical.
Finally, since Assertions (I3), (I4), and (I5) hold, it follows from Assertion (J2) of Proposition 8.30 that P(σ)[Z] is a component of P(σ) and P(σ

)[Z] = P Z . Since P(σ)[Z] is a component of P(σ), ∞ is isolated in P(σ).
Example 8.37. We consider the tournament T Z defined on V (T Z ) = Z ∪ {∞} and satisfying

• T Z -∞ = L Z ; • for each n ∈ Z, (∞, 2n) ∈ A(T Z ) and (2n + 1, ∞) ∈ A(T Z ).
It follows from Fact 8.36 that σ(T Z ) is critical, P(σ(T Z ))[Z] = P Z , and ∞ is isolated in P(σ(T Z )). Observe that for n ∈ Z and p ≥ 1, T Z [{2n, . . . , 2n + 2p -1} ∪ {∞}] is isomorphic to T 2p+1 (see Figure 1.2). Fact 8.38. Given a 2-structure σ defined on V (σ) = N ∪ {∞}, σ is critical, P(σ)[N] = P N , and ∞ is isolated in P(σ) if and only if the following assertions hold

• σ -∞ ∈ F N ; • [1, 2] σ = [1, ∞] σ ; • ∞ ←→ σ (2N), ∞ ←→ σ (2N + 1), and [0, ∞] σ ≠ [1, ∞] σ ;
• at least one of the following two cases occurs:

[0, 1] σ = [0, 2] σ and [0, 1] σ ≠ [0, ∞] σ , (8.28
)

or ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [1, 2] σ = [0, 2] σ , [0, 1] σ = [1, 3] σ , and [0, 2] σ ≠ [∞, 2] σ or [0, 1] σ ≠ [∞, 1] σ . (8.29)
Although the proof of Fact 8.38 is close to that of Fact 8.36, we provide it because some differences desserve to be pointed out.

Proof of Fact 8.38. To begin, suppose that σ is critical, P(σ)[N] = P N , and ∞ is isolated in P(σ). First, we verify that σ -∞ ∈ F N . Let n ≥ 1. As seen in the proof of Fact 8.36, {n -1, n + 1} is a module of (σ -∞)n, but not of σ -∞. Now, we have to show that N ∖ {0, 1} is a module of (σ -∞) -0, but N ∖ {1} is not a module of σ -∞. Since P(σ)[N] = P N and ∞ is isolated in P(σ), we have

N P(σ) (0) = {1}.
Since σ is critical, it follows from Lemma 4.4 that V (σ) ∖ {0, 1} is a module of σ -0. In particular, we have

[1, 2] σ = [1, ∞] σ . Moreover, by Assertion (M2) of Proposition 2.5, N ∖ {0, 1} is a module of (σ - ∞)-0. Since σ is prime, V (σ)∖{1} is not a module of σ. Hence, [1, 2] σ ≠ [1, 0] σ . Therefore, {0} ∪ {2, 3, . . .} is not a module of σ -∞. Consequently, σ -∞ ∈ F N .
Second, we show that ∞ ←→ σ (2N), ∞ ←→ σ (2N + 1), and

[0, ∞] σ ≠ [1, ∞] σ . Let n ≥ 0. As seen above, {2n, 2n + 2} is a module of σ -(2n + 1). Hence, ∞ ←→ σ {2n, 2n + 2}. It follows that ∞ ←→ σ (2N). Similarly, we have ∞ ←→ σ (2N + 1). Since σ is prime, N is not a module of σ. Since ∞ ←→ σ (2N) and ∞ ←→ σ (2N + 1), we obtain [0, ∞] σ ≠ [1, ∞] σ .
Third, we prove that (8.28) or (8.29) hold. Since σ is critical, σ -∞ admits a nontrivial module M . By the second assertion of Lemma 8.10, M is a module of L N . Since M ≥ 2, M contains even and odd integers. We distinguish the following two cases.

• Suppose that 0 ∈ M . Since M contains even and odd integers, we obtain

[0, 1] σ = [0, 2] σ .
Since σ is prime, V (σ) ∖ {0} is not a module of σ. Thus,

[0, 1] σ ≠ [0, ∞] σ .
It follows that (8.28) holds.

• Suppose that 0 ∈ M . Since M is a nontrivial module of L N , there exists n ≥ 1 such that M = {0, . . . , n}.

We obtain that (8.29) holds.

Conversely, suppose that the four assertions above hold. To begin, we verify that Assertions (I1),...,(I5) hold. Set Q = {N, {∞}} and consider the function ϕ ∶ V (σ) → Z defined by ϕ ↾N = Id N and ϕ(∞) = 0.

We obtain

P = {2N, 2N + 1, {∞}}.
Since σ-∞ ∈ F N , σ satisfies Assertion (I1) with Q and ϕ. Since N ∈ Q, Assertion (I2) holds. It follows that σ is locally critical. Furthermore, since ∞ ←→ σ (2N) and ∞ ←→ σ (2N + 1), P is a modular partition of σ according to Q. Hence, Assertion (I3) holds. Moreover, since [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF]

σ = [1, ∞] σ , Assertion (I4) holds. Lastly, since [0, ∞] σ ≠ [1, ∞] σ , we have [2N, {∞}] (σ Q P ) ≠ [2N + 1, {∞}] (σ Q P ) .
It follows from the definition of the generalized quotient (see Definition 8.16) that σ Q P is prime. Thus, Assertion (I5) holds. For a contradiction, suppose that σ admits a nontrivial module M . We utilize Fact 8.21 with

Q ′ = Q as follows. Since [0, ∞] σ ≠ [1, ∞] σ , {2N, 2N + 1} is not a module of σ Q P . It follows from Fact 8.21 that M ∩(2Z) ≠ ∅, M ∩(2Z+1) ≠ ∅, and ∞ ∈ M . Thus, M ∖{∞} is a nontrivial module of σ-∞. Since σ-∞ ∈ F N , M ∖ {∞}
is a nontrivial module of L N by the second assertion of Lemma 8.10. In particular, M ∖ {∞} contains even and odd integers. We distinguish the following two cases. In each of them, we obtain a contradiction.

• Suppose that 0 ∈ M . Since M contains even and odd integers, we obtain

[0, 1] σ = [0, 2] σ .
Furthermore, since ∞ ∈ M , we obtain

[0, 1] σ = [0, ∞] σ ,
which contradicts the fact that (8.28) holds.

• Suppose that 0 ∈ M . Since M is a nontrivial module of L N , there exists n ≥ 1 such that M = {0, . . . , n}.

We obtain

[0, 2] σ = [1, 2] σ = [∞, 2] σ and [0, 1] σ = [1, 3] σ = [∞, 1] σ ,
which contradicts the fact that that (8.29) holds.

Consequently, σ is prime. Since Assertions (I3) and (I4) hold, it follows from Lemma 8.29 that σn is decomposable for every n ∈ N. Moreover, since (8.28) or (8.29)

hold, {1, 2, . . .} is a module of σ -∞ or {0, 1} is a module of σ -∞. Hence, σ -∞ is decomposable. Consequently, σ is critical.
Finally, since Assertions (I3), (I4), and (I5) hold, it follows from Assertion (J2) of Proposition 8.30 that P(σ)[N] is a component of P(σ) and P(σ)[N] = P N . Since P(σ)[N] is a component of P(σ), ∞ is isolated in P(σ). Fact 8.40. Given a 2-structure σ defined on V (σ) = Z ∪ {∞, ∞ ′ }, σ is critical, P(σ)[Z] = P Z , and ∞ and ∞ ′ are isolated in P(σ) if and only if, by exchanging ∞ and ∞ ′ if necessary, we have

• σ -{∞, ∞ ′ } ∈ F Z ; • ∞ ←→ σ (2Z), ∞ ←→ σ (2Z + 1), and [0, ∞] σ ≠ [1, ∞] σ ; • ∞ ′ ←→ σ Z and [0, ∞ ′ ] σ ≠ [∞, ∞ ′ ] σ ;
• at least one of the following two cases occurs:

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [0, 1] σ = [0, 2] σ , [1, 2] σ = [1, 3] σ , and [0, 1] σ = [0, ∞] σ , [1, 2] σ = [1, ∞] σ , (8.30) or ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [1, 2] σ = [0, 2] σ , [0, 1] σ = [1, 3] σ , and [0, 2] σ = [∞, 2] σ , [0, 1] σ = [∞, 1] σ . (8.31)
Proof. To begin, suppose that σ is critical, P(σ)[Z] = P Z , and ∞ and ∞ ′ are isolated in P(σ). By Proposition 8.24,

{v ∈ {∞, ∞ ′ } ∶ v ←→ σ Z} = 1.
By exchanging ∞ and ∞ ′ if necessary, we can assume that ∞ ←→ σ Z and ∞ ′ ←→ σ Z. Since σ is prime, Z ∪ {∞} is not a module of σ. Thus, we have

[0, ∞ ′ ] σ ≠ [∞, ∞ ′ ] σ .
First, we verify that σ -{∞, ∞ ′ } ∈ F Z . Let n ∈ Z. Since P(σ)[Z] = P Z and ∞ and ∞ ′ are isolated in P(σ), we have

N P(σ) (n) = {n -1, n + 1}. Since σ is critical, it follows from Lemma 4.4 that {n -1, n + 1} is a module of σ -n. By Assertion (M2) of Proposition 2.5, {n -1, n + 1} is a module of (σ -{∞, ∞ ′ }) -n. Since σ is prime, {n -1, n + 1} is not a module of σ. Hence, n ←→ σ {n -1, n + 1}. It follows that {n -1, n + 1} is not a module of σ -{∞, ∞ ′ }. Consequently, σ -{∞, ∞ ′ } ∈ F Z .
Second, we verify that ∞ ←→ σ (2Z) and ∞ ←→ σ (2Z+1). Let n ∈ Z. As seen above, {2n, 2n + 2} is a module of σ -(2n + 1). Hence, ∞ ←→ σ {2n, 2n + 2}. It follows that ∞ ←→ σ (2Z). Similarly, we have

∞ ←→ σ (2Z + 1). Since ∞ ←→ σ Z, we obtain [0, ∞] σ ≠ [1, ∞] σ .
Third, we prove that (8.30) or (8.31) hold. Since

σ is critical, σ -∞ ′ is decomposable. Consider a nontrivial module M of σ -∞ ′ . Since [0, ∞] σ ≠ [1, ∞] σ , we have ∞ ∈ M . By Assertion (M2) of Proposition 2.5, M ∩ Z is a module of σ -{∞, ∞ ′ }.
For a contradiction, suppose that M ∩ Z = 1. Denote by n the unique element of M ∩ Z. Since ∞ ←→ σ (2Z) and ∞ ←→ σ (2Z + 1), we obtain n ←→ σ {n -1, n + 1}, which contradicts σ -{∞, ∞ ′ } ∈ F Z . It follows that M ∩ Z ≥ 2. Hence, M ∩ Z is a nontrivial module of σ -{∞, ∞ ′ }. For instance, suppose that M ∩ Z admits a least element. Since ∞ ∈ M , we obtain that (8.30) holds.

Conversely, suppose that the four assertions above hold. To begin, we verify that Assertions (I1),...,(I5) hold. Set Q = {Z, {∞}, {∞ ′ }} and consider the function ϕ ∶ V (σ) → Z defined by

ϕ ↾Z = Id Z , ϕ(∞) = 0, and ϕ(∞ ′ ) = 0.
We obtain

P = {2Z, 2Z + 1, {∞}, {∞ ′ }}. Since σ -{∞, ∞ ′ } ∈ F Z , σ satisfies Assertion (I1) with Q and ϕ. Since Z ∈ Q, Assertion (I2) holds. It follows that σ is locally critical. Furthermore, since ∞ ′ ←→ σ Z, ∞ ←→ σ (2Z), and ∞ ←→ σ (2Z + 1), P is a modular partition of σ according to Q. Hence, Assertion (I3) holds. Since [0, ∞] σ ≠ [1, ∞] σ , {2Z, 2Z + 1} and {2Z, 2Z + 1, {∞ ′ }} are not modules of σ Q P . Moreover, since [0, ∞ ′ ] σ ≠ [∞, ∞ ′ ] σ , {2Z, 2Z + 1, {∞}} is not a module of σ Q P . Lastly, since [0, ∞] σ ≠ [1, ∞] σ and ∞ ′ ←→ σ Z, {{∞}, {∞ ′ }}
. .} ∪ {∞} is a module of σ -∞ or {. . . , 0, 1} ∪ {∞} is a module of σ -∞ ′ . Hence, σ -∞ ′ is decomposable. Lastly, since ∞ ′ ←→ σ Z, σ -∞ is decomposable. Consequently, σ is critical.
Finally, since Assertions (I3), (I4), and (I5) hold, it follows from Assertion (J2) of Proposition 8.30 that P(σ)[Z] is a component of P(σ) and P(σ)[Z] = P Z . Lastly, it follows from Corollary 8.3 that ∞ and ∞ ′ are isolated in P(σ). Fact 8.41. Given a 2-structure σ defined on V (σ) = N ∪ {∞, ∞ ′ }, σ is critical, P(σ)[N] = P N , and ∞ and ∞ ′ are isolated in P(σ) if and only if, by exchanging ∞ and ∞ ′ if necessary, we have

• σ -{∞, ∞ ′ } ∈ F N ; • ∞ ←→ σ (2N), ∞ ←→ σ (2N + 1), and [0, ∞] σ ≠ [1, ∞] σ ; • ∞ ′ ←→ σ N and [0, ∞ ′ ] σ ≠ [∞, ∞ ′ ] σ ; • [1, 2] σ = [1, ∞] σ and [1, 2] σ = [1, ∞ ′ ] σ ;
• at least one of the following two cases occurs:

[0, 1] σ = [0, 2] σ and [0, 1] σ = [0, ∞] σ , (8.32) or ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [1, 2] σ = [0, 2] σ , [0, 1] σ = [1, 3] σ , and [0, 2] σ = [∞, 2] σ , [0, 1] σ = [∞, 1] σ . (8.33)
Proof. To begin, suppose that σ is critical, P(σ)[N] = P N , and ∞ and ∞ ′ are isolated in P(σ). By Proposition 8.24,

{v ∈ {∞, ∞ ′ } ∶ v ←→ σ N} = 1.
By exchanging ∞ and ∞ ′ if necessary, we can assume that ∞ ←→ σ N and

∞ ′ ←→ σ N. Since σ is prime, N ∪ {∞} is not a module of σ. Thus, we have [0, ∞ ′ ] σ ≠ [∞, ∞ ′ ] σ . First, we verify that σ -{∞, ∞ ′ } ∈ F N . Let n ≥ 1.
As seen in the proof of Fact 8.40, {n -1, n + 1} is a module of (σ -{∞, ∞ ′ })n, but not of σ -{∞, ∞ ′ }. Since P(σ)[N] = P N and ∞ and ∞ ′ are isolated in P(σ), we have

N P(σ) (0) = {1}.
Since σ is critical, it follows from Lemma 4.4 that V (σ) ∖ {0, 1} is a module of σ -0. In particular, we obtain

[1, 2] σ = [1, ∞] σ and [1, 2] σ = [1, ∞ ′ ] σ . (8.34)
Furthermore, by Assertion (M2) of Proposition 2.5,

V (σ) ∖ {0, 1} is a module of (σ -{∞, ∞ ′ }) -0. Since σ is prime, V (σ) ∖ {1} is not a module of σ. Hence, 1 ←→ σ (V (σ) ∖ {1}). It follows that N ∖ {1} is not a module of σ -{∞, ∞ ′ }. Consequently, σ -{∞, ∞ ′ } ∈ F N . Second, we verify that ∞ ←→ σ (2N) and ∞ ←→ σ (2N+1). Let n ∈ N. As seen above, {2n, 2n + 2} is a module of σ -(2n + 1). Hence ∞ ←→ σ {2n, 2n + 2}. It follows that ∞ ←→ σ (2N). Similarly, we have ∞ ←→ σ (2N + 1). Since ∞ ←→ σ Z, we obtain [0, ∞] σ ≠ [1, ∞] σ .
Third, we prove that (8.32) or (8.33) hold. Since σ is critical, σ -∞ ′ admits a nontrivial module M . Since ∞ ←→ σ (2N), ∞ ←→ σ (2N + 1), and

[0, ∞] σ ≠ [1, ∞] σ , we have ∞ ∈ M . By Assertion (M2) of Proposition 2.5, M ∩ N is a module of σ -{∞, ∞ ′ }.
For a contradiction, suppose that M ∩ N = 1. Denote by n the unique element of M ∩ N. We distinguish the following two cases. In each of them, we obtain a contradiction.

• Suppose that n ≥ 1. Since ∞ ←→ σ (2N) and ∞ ←→ σ (2N + 1), we obtain n ←→ σ {n -1, n + 1}, which contradicts σ -{∞, ∞ ′ } ∈ F N .
• Suppose that n = 0. We have

[1, 0] σ = [1, ∞] σ .
By (8.34), we have [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF]

σ = [1, ∞] σ . It follows that [1, 0] σ = [1, 2] σ , which contradicts σ -{∞, ∞ ′ } ∈ F N .
It follows that M ∩N ≥ 2. Hence, M ∩N is a nontrivial module of σ-{∞, ∞ ′ }. By the second assertion of Lemma 8.10, M ∩ N is a nontrivial module of L N . Since M ∩ N ≥ 2, M contains even and odd integers. We distinguish the following two cases.

• Suppose that 0 ∈ M ∩ N. Since M ∩ N contains even and odd integers, we obtain

[0, 1] σ = [0, 2] σ .
Moreover, since ∞ ∈ M , we obtain

[0, 1] σ = [0, ∞] σ .
It follows that (8.32) holds.

• Suppose that 0 ∈ M . Since M ∩N is a nontrivial module of L N , there exists n ≥ 1 such that M ∩ N = {0, . . . , n}.

Since ∞ ∈ M , we obtain M = {0, . . . , n} ∪ {∞}.

We obtain that (8.33) holds.

Conversely, suppose that the five assertions above hold. To begin, we verify that Assertions (I1),...,(I5) hold. Set Q = {N, {∞}, {∞ ′ }} and consider the function ϕ ∶ V (σ) → Z defined by

ϕ ↾N = Id N , ϕ(∞) = 0 and ϕ(∞ ′ ) = 0.
We obtain

P = {2N, 2N + 1, {∞}, {∞ ′ }}. Since σ -{∞, ∞ ′ } ∈ F Z , σ satisfies Assertion (I1) with Q and ϕ. Since N ∈ Q, Assertion (
P . Moreover, since [0, ∞ ′ ] σ ≠ [∞, ∞ ′ ] σ , {2N, 2N + 1, {∞}} is not a module of σ Q P . Lastly, since [0, ∞] σ ≠ [1, ∞] σ and ∞ ′ ←→ σ N, {{∞}, {∞ ′ }}
σ = [1, ∞] σ and [1, 2] σ = [1, ∞ ′ ] σ . We obtain 1 ←→ σ {2, ∞, ∞ ′ }.
It follows that Assertions (I4) holds.

To verify that σ is prime, we utilize Fact 8.22 with Q ′ = Q as follows. Clearly, (8.7) holds. Moreover, we have P ′ = P . As previously observed, σ Q P is prime. Since Assertions (I3) and (I4) hold, it follows from Fact 8.22 that σ is prime.

Since Assertions (I3) and (I4) hold, it follows from Lemma 8.29 that σn is decomposable for each n ∈ N. Since (8.32) or (8.33)

hold, {1, 2, . . .} ∪ {∞} is a module of σ -∞ or {0, 1} ∪ {∞} is a module of σ -∞. Hence, σ -∞ ′ is decomposable. Lastly, since ∞ ′ ←→ σ N, σ -∞ is decomposable. Consequently, σ is critical.
Finally, since Assertions (I3), (I4), and (I5) hold, it follows from Assertion (J2) of Proposition 8.30 that P(σ)[N] is a component of P(σ) and P(σ)[N] = P N . Lastly, it follows from Corollary 8.3 that ∞ and ∞ ′ are isolated in P(σ).

Chapter 9

Partially critical 2-structures

We consider the following weakening of the notion of a critical structure (see Definition 4.1). Definition 9.1. Let σ be a prime 2-structure. Given W ⊆ V (σ), σ is W -critical W -critical if all the elements of W are critical vertices of σ.

A prime 2-structure σ is partially critical if there exists a proper subset X partially critical of V (σ) such that σ[X] is prime and σ is (V (σ) ∖ X)-critical.

Finite partially critical graphs were characterized by Breiner et al. [START_REF] Breiner | Partially critical indecomposable graphs[END_REF] . Furthermore, finite partially critical tournaments were characterized by Sayar [32] who adapted the examination of partial criticality presented in [START_REF] Breiner | Partially critical indecomposable graphs[END_REF] to tournaments. A nice presentation of finite and partially critical tournaments is provided in [START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] (see[2, Theorem 2 and Corollary 1] ). Lastly, Belkhechine et al. [3] characterized the finite or infinite partially critical 2-structures. In the finite case, they followed the same approach as that of [START_REF] Breiner | Partially critical indecomposable graphs[END_REF].

Theorem 3.19 leads us to introduce the outside graph as follows. The outside graph is the main tool to characterize the partially critical 2-structures. It is frequently used in the study of prime digraphs [START_REF] Ille | Indecomposable graphs[END_REF][START_REF] Ille | Recognition of prime graphs from a prime subgraph[END_REF]. We need the next notation.

Notation 9.2. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. The set of the nonempty subsets Y of V (σ) ∖ X, such that σ[X ∪ Y ]
is prime, is denoted by P (σ,X) (compare with Notation 3.1). Hence, we have Ext σ (X) = {v ∈ V (σ) ∖ X ∶ {v} ∈ P (σ,X) } (see Notation 3.12). Furthermore, suppose that V (σ) ∖ X ≥ 2. By Theorem 3.19, P (σ,X) contains an unordered pair.

Definition 9.3. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. The outside graph Γ (σ,X) is defined on V (Γ (σ,X) ) = V (σ) ∖ X by outside graph 137 E(Γ (σ,X) ) = {Y ∈ P (σ,X) ∶ Y = 2}.
By Theorem 3.19, the outside graph Γ (σ,X) is nonempty when σ is prime and V (σ) ∖ X ≥ 2.

Remark 9.4. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Given k ∈ {1, . . . , V (σ) ∖ X -1}, we consider the following statement

{Y ∈ P (σ,X) ∶ Y = k} = ∅. (Sk)
Clearly, Ext σ (X) = ∅ means that Statement (S1) holds. First, we make the following observation. Consider k ∈ {1, . . . , V (σ)∖X -1} and m ∈ {1, . . . , k -2} such that km ≡ 0 mod 2. If Statement (Sk) holds, then it follows from Corollary 3.20 that Statement (Sm) holds.

Second, suppose that σ is (V (σ) ∖ X)-critical and

V (σ) ∖ X is finite.
We verify that Statement (Sk) holds for each k ∈ {1, . . . , V (σ) ∖ X -1} such that k is odd.

To begin, we verify that

V (σ) ∖ X is even. (9.1)
Otherwise, it follows from Corollary 3.20 that σ admits a noncritical vertex v such that v ∈ V (σ) ∖ X, which contradicts the fact that σ is

(V (σ) ∖ X)-critical. Now, consider Y ∈ P (σ,X) such that Y ≠ V (σ) ∖ X. Since σ is (V (σ) ∖ X)-critical, σ is (V (σ) ∖ (X ∪ Y ))
-critical as well. It follows from (9.1) that V (σ) ∖ (X ∪ Y ) is even. Since V (σ) ∖ X is even, Y is even too. Consequently, Statement (Sk) holds for each k ∈ {1, . . . , V (σ) ∖ X -1} such that k is odd.

Main results

We begin with a hereditary property of primality through the components of the outside graph. Theorem 9.5 (Belkhechine et al. [3]). Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. The following three assertions are equivalent

1. σ is prime; 2. for each component C of Γ (σ,X) , σ[X ∪ V (C)] is prime; 3. for each component C of Γ (σ,X) , v(C) = 2 or v(C) ≥ 4 and C is prime.
Theorem 9.5 allows us to provide a simple and short proof of Theorem 5.8 (see Section 9.6). Furthermore, Theorem 9.5 is proved for finite graphs in [START_REF] Ille | Recognition of prime graphs from a prime subgraph[END_REF] (see [START_REF] Ille | Recognition of prime graphs from a prime subgraph[END_REF]Theorem 17] and [START_REF] Ille | Recognition of prime graphs from a prime subgraph[END_REF]Corollary 18]). We pursue with a hereditary property of partial criticality through the components of the outside graph. The next theorem also provides a characterization of partially critical 2-structures in terms of criticality of the components of their outside graph. Theorem 9.11. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S5) holds. Suppose also that σ is (V (σ) ∖ X)critical. For each v ∈ V (σ) ∖ X, there exists w ∈ (V (σ) ∖ X) ∖ {v} such that σ -{v, w} is ((V (σ) ∖ {v, w}) ∖ X)-critical. In particular, we obtain

for each v ∈ V (σ) ∖ X, N P(σ) (v) ∩ (V (σ) ∖ X) ≠ ∅ (see Definition 4.3).
We prove Theorem 9.11 at the end of Section 9.5.

Remark 9.12. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S5) holds. Suppose also that σ is (V (σ) ∖ X)-critical. Lastly, suppose that V (σ) ∖ X is infinite. Consider a finite and nonempty subset F of V (σ) ∖ X. By applying several times Theorem 9.11, we obtain a finite subset

F ′ of V (σ) ∖ X such that F ⊆ F ′ and σ -F ′ is ((V (σ) ∖ F ′ ) ∖ X)-critical.
Furthermore, it follows from Corollary 3.20 that F ′ is even.

Modules of the outside graph

We begin with two preliminary results on the isolated vertices of an outside graph. We utilize the following remark.

Remark 9.13. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. It follows from Remark 3.15 that for each B ∈ p (σ,X) ∖ {Ext σ (X)}, Γ (σ,X) [B] is empty. In other words, if Ext σ (X) = ∅, then Γ (σ,X
) is multipartite with partition p (σ,X) (see Lemma 3.13).

Lemma 9.14 (Breiner et al. 1 [5]). Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime.

1. If M is a module of σ such that X ⊆ M , then the elements of V (σ) ∖ M are isolated vertices of Γ (σ,X) .

2. Given y ∈ X, if M is a module of σ such that M ∩ X = {y}, then the elements of M ∖ {y} are isolated vertices of Γ (σ,X) .

Proof. For the first assertion, consider a module M of σ such that

X ⊆ M . Let v ∈ V (σ) ∖ M . Moreover, consider w ∈ (V (σ) ∖ X) ∖ {v}. We have to verify that σ[X ∪ {v, w}] is decomposable. By Remark 3.16, (V (σ) ∖ M ) ⊆ ⟨X⟩ σ . It follows from Remark 9.13 that σ[X ∪ {v, w}] is decomposable when w ∈ M . Now, suppose that w ∈ M ∖ X. By Assertion (M2) of Proposition 2.5, M ∩ (X ∪ {v, w}), which is X ∪ {w}, is a module of σ[X ∪ {v, w}]. Thus, σ[X ∪ {v, w}] is decomposable.
For the second assertion, consider y ∈ X and a module M of σ such that M ∩ X = {y}. Let v ∈ M ∖ {y}. Moreover, consider w ∈ (V (σ) ∖ X) ∖ {v}. We have to verify that σ[X ∪ {v, w}] is decomposable.

By Remark 3.16, M ∖ {y} ⊆ X σ (y). It follows from Remark 9.13 that σ[X ∪ {v, w}] is decomposable when w ∈ M . Now, suppose that w ∈ M . By Assertion (M2) of Proposition 2.5, M ∩ (X ∪ {v, w}), which is {y, v}, is a module of σ[X ∪ {v, w}]. Thus, σ[X ∪ {v, w}] is decomposable.

The next result is an immediate consequence of Remark 3.16 and Lemma 9.14.

Corollary 9.15. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. If σ admits a nontrivial module M such that M ∩ X ≠ ∅, then Γ (σ,X)
possesses isolated vertices.

Now, we study the modules of the outside graph. We need the following refinement of the outside partition (see Notation 3.12). Notation 9.16. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. We consider the following subsets of V (σ) ∖ X

• for e, f ∈ E(σ),

⟨X⟩ (e,f ) σ = ⟨X⟩ σ ∩ N (e,f ) σ (y) (see Notation 3.7),
where y ∈ X;

• for e, f ∈ E(σ) and y ∈ X, (y) ∶ e, f ∈ E(σ), y ∈ X} is denoted by q (σ,X) . Lemma 9.17. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S1) holds. Given M ⊆ (V (σ) ∖ X), if M is a module of σ, then M is a module of Γ (σ,X) , and there exist B p ∈ p (σ,X) and

X (e,f ) σ (y) = X σ (y) ∩ N (e,f )
B q ∈ q (σ,X) such that M ⊆ B q ⊆ B p and M is a module of σ[B p ].
Proof. Consider a module M of σ such that M ∩ X = ∅. Let v ∈ M . Denote by B q the unique block of q (σ,X) containing v. Consider w ∈ M ∖ {v}. Since M is a module of σ such that M ∩ X = ∅, we have y ←→ σ {v, w} for every y ∈ X. It follows that w ∈ B q . Consequently, M ⊆ B q . Denote by B p the unique block of p (σ,X) containing B q . We obtain

M ⊆ B q ⊆ B p . Since M is a module of σ, M is a module of σ[B p ] by Assertion (M2) of Propo- sition 2.5.
Lastly, we prove that M is a module of Γ (σ,X) . Let w ∈ (V (σ) ∖ X) ∖ M . Recall that Ext σ (X) = ∅ because Statement (S1) holds. If w ∈ B p , then it follows from Remark 9.13 that {v, w} ∈ E(Γ (σ,X) ) for every v ∈ M . Hence, suppose that w ∈ (V (σ) ∖ X) ∖ B p . Since Ext σ (X) = ∅, we distinguish the following two cases.

• Suppose that B p = ⟨X⟩ σ . Consider y ∈ X and u ∈ M . First, suppose that u ←→ σ {y, w}. Let v ∈ M . Since M is a module of σ, we obtain v ←→ σ {y, w}. Since v ←→ σ X, we obtain v ←→ σ X ∪ {w}. Hence, X ∪ {w} is a module of σ[X ∪ {v, w}]. It follows that {v, w} ∈ E(Γ (σ,X) ).

Second, suppose that u ←→ σ {y, w}. Let v ∈ M . Since M is a module of σ, we have v ←→ σ {y, w}. Thus, X ∪ {v} is not a module of σ[X ∪ {y, v}]. It follows from Assertion (Q1) of Corollary 3.18 that {v, w} ∈ E(Γ (σ,X) ).

• Suppose that B p = X σ (y), where y ∈ X. Consider u ∈ M .

First, suppose that w ←→ σ {y, u}. Let v ∈ M . Since M is a module of σ, we obtain w ←→ σ {y, v}. Since {y, v} is a module of σ[X ∪ {v}], {y, v} is a module of σ[X ∪ {v, w}]. It follows that {v, w} ∈ E(Γ (σ,X) ) for every v ∈ M .

Second, suppose that w ←→ σ {y, u}. Let v ∈ M . Since M is a module of σ, we obtain w ←→ σ {y, v}. Therefore, {y, v} is not a module of σ[X ∪ {v, w}]. It follows from Assertion (Q2) of Corollary 3.18 that {v, w} ∈ E(Γ (σ,X) ).

The opposite direction in Lemma 9.17 is false. Nevertheless, it is true for (finite) graphs (see the second assertion of [START_REF] Breiner | Partially critical indecomposable graphs[END_REF]Lemma 2.6]). Moreover, the opposite direction in Lemma 9.17 is true if we require that Statement (S3) holds (see Corollary 9.19 below). We need the following fact. Fact 9.18 (Breiner et al. 2 [5]). Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. Given distinct elements u, v, w of V (σ) ∖ X, if {u, v}, {u, w} ∈ E(Γ (σ,X) ), then {v, w} is a module of σ[X ∪ {u, v, w}], and hence there exists B q ∈ q (σ,X) such that v, w ∈ B q .

Proof. Since {u, v} ∈ E(Γ (σ,X) ), σ[X ∪ {u, v}] is prime. Set Y = X ∪ {u, v}. Since Statement (S3) holds, w ∈ Ext σ (Y ).
For a contradiction, suppose that w ∈ ⟨Y ⟩ σ . We obtain that . By Lemma 9.17, there exists B q ∈ q (σ,X) such that v, w ∈ B q .

The next result follows from Fact 9.18. Since u and v belong to the same block of q (σ,X) , (9.2) holds when w ∈ X. Now, suppose that

Corollary 9.19. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. Consider M ⊆ (V (σ) ∖ X) such that there exist B p ∈ p (σ,X) and B q ∈ q (σ,X) with M ⊆ B q ⊆ B p . Suppose that M is a module of σ[B p ]. If M is a module of Γ (σ,X) , then M is a module of σ. Proof. Suppose that M is a module of Γ (σ,X) . Consider u, v ∈ M and w ∈ V (σ) ∖ M . It
w ∈ V (σ) ∖ (X ∪ B p ).
Since M is a module of Γ (σ,X) , we have {u, w}, {v, w} ∈ E(Γ (σ,X) ) or (9.3) {u, w}, {v, w} ∈ E(Γ (σ,X) ).

Suppose that {u, w}, {v, w} ∈ E(Γ (σ,X) ). By Fact 9.18, {u, v} is a module of σ[X ∪ {u, v, w}], so (9.2) holds.

Lastly, suppose that {u, w}, {v, w} ∈ E(Γ (σ,X) ). Since Ext σ (X) = ∅, we distinguish the following two cases.

• Suppose that B p = ⟨X⟩ σ . Since {u, w}, {v, w} ∈ E(Γ (σ,X) ), it follows from Assertion (Q1) of Corollary 3.18 that X ∪{w} is a module of σ[X ∪{u, w}] and σ[X ∪ {v, w}]. Given y ∈ X, we obtain u ←→ σ {y, w} and v ←→ σ {y, w}. Since u, v ∈ B q and B q ⊆ ⟨X⟩ σ , y ←→ σ {u, v}. It follows that (9.2) holds.

• Suppose that B p = X σ (y), where y ∈ X. Since {u, w}, {v, w} ∈ E(Γ (σ,X) ), it follows fromAssertion (Q2) of Corollary 3.18 that {y, u} is a module of σ[X ∪ {u, w}], and {y, v} is a module of σ[X ∪ {v, w}]. Therefore, we have w ←→ σ {y, u} and w ←→ σ {y, w}. It follows that (9.2) holds.

The next fact follows from Lemma 9.17.

Fact 9.20 (Breiner et al. 3 [5]). Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. Given B p , D p ∈ p (σ,X) , consider u ∈ B p and v, w ∈ D p such that {u, v} ∈ E(Γ (σ,X) ) and {u, w} ∈ E(Γ (σ,X) ).

1. If D p = ⟨X⟩ σ , then X ∪ {u, v} is a module of σ[X ∪ {u, v, w}]. 2. If D p = X σ (y), where y ∈ X, then {y, w} is a module of σ[X ∪ {u, v, w}].
Proof. To begin, we establish two preliminary statements (see (9.4) and (9.5)).

Since

{u, v} ∈ E(Γ (σ,X) ), σ[X ∪ {u, v}] is prime. Set Y = X ∪ {u, v}. Since Statement (S3) holds, w ∈ Ext σ (Y ).
Since {u, v} ∈ E(Γ (σ,X) ), we have

B p ≠ D p (9.4)
by Remark 9.13. For a contradiction, suppose that w ∈ Y σ (u). Hence, {u, w} is a module of σ[X ∪ {u, v, w}]. By Remark 9.4, Statement (S1) holds because Statement (S3) holds. It follows from Lemma 9.17 applied to σ[X ∪ {u, v, w}] that B p = D p , which contradicts (9.4). Thus,

w ∈ Y σ (u).
Now, suppose for a contradiction that w ∈ Y σ (v). Hence, {v, w} is a module of σ[X ∪ {u, v, w}]. It follows from Lemma 9.17 applied to σ[X ∪ {u, v, w}] that {v, w} is a module of Γ (σ,X) , which is impossible because {u, v} ∈ E(Γ (σ,X) ) and {u, w} ∈ E(Γ (σ,X) ). Therefore,

w ∈ Y σ (v). Since w ∈ (Y σ (u) ∪ Y σ (v)), it follows from Lemma 3.13 applied to σ[Y ] that w ∈ ⟨Y ⟩ σ or w ∈ Y σ (y), where y ∈ X. (9.5)
First, suppose that D p = ⟨X⟩ σ . If w ∈ Y σ (y), where y ∈ X, then w ∈ X σ (y), and hence w ∈ Y σ (y) ∩ ⟨X⟩ σ , which contradicts Lemma 3.13. It follows from (9.5)

that w ∈ ⟨Y ⟩ σ , that is, X ∪ {u, v} is a module of σ[X ∪ {u, v, w}].
Second suppose that D p = X σ (y), where y ∈ X. If w ∈ ⟨Y ⟩ σ , then w ∈ ⟨X⟩ σ , and hence w ∈ X σ (y) ∩ ⟨X⟩ σ , which contradicts Lemma 3.13. It follows from (9.5) that w ∈ Y σ (z), where z ∈ X. Hence, we have w ∈ X σ (z). We obtain w ∈ X σ (y) ∩ X σ (z). By Lemma 3.13, we have y = z. Consequently, w ∈ Y σ (y), that is, {y, w} is a module of σ[X ∪ {u, v, w}].

The next two results follow from Fact 9.20.

Corollary 9.21. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. Let B q ∈ q (σ,X) . For each v ∈ (V (σ) ∖ X) ∖ B q , {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )} and {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )} are modules of σ[B q ]. Precisely, if {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )} ≠ ∅ and {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )} ≠ ∅, then the following two assertions hold. 1. If B q = ⟨X⟩ (e,f ) σ
, where e, f ∈ E(σ), then

[{u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )}, {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )}] σ = (f, e) (see Notation 2.1). 2. If B q = X (e,f ) σ (α), where α ∈ X and e, f ∈ E(σ), then [{u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )}, {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )}] σ = (e, f ). Proof. Let v ∈ (V (σ) ∖ X) ∖ B q . Suppose that {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )} ≠ ∅ and {u ∈ B q ∶ {u, v} ∈ E(Γ (σ,X) )} ≠ ∅. Consider u + , u -∈ B q such that {u + , v} ∈ E(Γ (σ,X)
) and {u -, v} ∈ E(Γ (σ,X) ). We distinguish the following two cases.

1. Suppose that B q = ⟨X⟩ (e,f ) σ

, where e, f ∈ E(σ). By the first assertion of Fact 9.20 applied to u

+ , u -, v, X ∪{u + , v} is a module of σ[X ∪{u + , u -, v}]. Since u -∈ ⟨X⟩ (e,f ) σ , we obtain [u -, u + ] σ = (f, e).
2. Suppose that B q = X (e,f ) σ (y), where y ∈ X and e, f ∈ E(σ). By the second assertion of Fact 9.20 applied to u + , u -, v, {y,

u -} is a module of σ[X ∪ {u + , u -, v}]. Hence, [u -, u + ] σ = [y, u + ] σ . Since u + ∈ X (e,f ) σ (y), we obtain [y, u + ] σ = (e, f ), so [u -, u + ] σ = (e, f ).
The proof of the next corollary follows from Corollary 3.18 and Fact 9.20.

Corollary 9.22 (Breiner et al. 4 [5]). Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. If σ is prime, then Γ (σ,X) has no isolated vertices.

Proof. We denote by I the set of the isolated vertices of Γ (σ,X) . By Remark 9.4, Statement (S1) holds because Statement (S3) holds. Therefore, we have Ext σ (X) = ∅. By Lemma 3.13, to show that I = ∅, it suffices to verify that

I ∩ ⟨X⟩ σ = ∅ (9.6) and I ∩ X σ (y) = ∅ (9.7)
for each y ∈ X.

To verify that (9.6) holds, we show that V (σ) ∖ (I ∩ ⟨X⟩ σ ) is a module of σ. Consider u ∈ I ∩ ⟨X⟩ σ and v ∈ V (σ) ∖ (I ∩ ⟨X⟩ σ ). We verify that X ∪ {v} is a module of σ[X ∪ {u, v}]. This is clear when v ∈ X because u ∈ ⟨X⟩ σ . Hence, suppose that v ∈ X. We distinguish the following two cases.

• Suppose that v ∈ ⟨X⟩ σ . Since u ∈ I, we have {u, v} ∈ E(Γ (σ,X) ). It follows from Assertion (Q1) of Corollary 3.18 that X ∪ {v} is a module of σ[X ∪ {u, v}].

• Suppose that v ∈ ⟨X⟩ σ . Since v ∈ V (σ) ∖ (I ∩ ⟨X⟩ σ ), we have v ∈ I. Since v ∈ I, there exists w ∈ V (σ) ∖ X such that {v, w} ∈ E(Γ (σ,X) ). Since u ∈ I, we have {u, w} ∈ E(Γ (σ,X) ). By the first assertion of Fact 9.20, X ∪ {v, w} is a module of σ[X ∪ {u, v, w}]. By Assertion (M2) of Proposition 2.5,

X ∪ {v} is a module of σ[X ∪ {u, v}]. In both cases above, X ∪ {v} is a module of σ[X ∪ {u, v}]. It follows that V (σ) ∖ (I ∩ ⟨X⟩ σ ) is a module of σ. Since σ is prime, V (σ) ∖ (I ∩ ⟨X⟩ σ ) is a trivial module of σ. Thus, we obtain V (σ) ∖ (I ∩ ⟨X⟩ σ ) = V (σ).
Hence, (9.6) holds.

To verify that (9.7) holds, consider y ∈ X. We show that {y} ∪ (I ∩ X σ (y)) is a module of σ. Consider u ∈ I ∩ X σ (y) and v ∈ V (σ) ∖ ({y} ∪ (I ∩ X σ (y))). We verify that {y, u} is a module of σ[X ∪ {u, v}]. This is clear when v ∈ X ∖ {y} because u ∈ X σ (y). Hence, suppose that v ∈ X. We distinguish the following two cases.

• Suppose that v ∈ X σ (y). Since u ∈ I, we have {u, v} ∈ E(Γ (σ,X) ). It follows from Assertion (Q2) of Corollary 3.18 that {y, u} is a module of σ[X ∪ {u, v}].

• Suppose that v ∈ X σ (y). Since v ∈ V (σ) ∖ ({y} ∪ (I ∩ X σ (y))), we have v ∈ I. Since v ∈ I, there exists w ∈ V (σ) ∖ X such that {v, w} ∈ E(Γ (σ,X) ).

Since u ∈ I, we have {u, w} ∈ E(Γ (σ,X) ). By the second assertion of Fact 9.20, {y, u} is a module of σ[X ∪ {u, v, w}]. By Assertion (M2) of Proposition 2.5, {y, u} is a module of σ[X ∪ {u, v}].

In both cases above, {y, u} is a module of σ[X ∪ {u, v}]. It follows that {y} ∪ (I ∩ X σ (y)) is a module of σ. Since σ is prime, {y} ∪ (I ∩ X σ (y)) is a trivial module of σ. Thus, we obtain I ∩ X σ (y) = ∅. Hence, (9.7) holds.

Blocks of the outside partition and of its refinement

Lemma 9.23. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. Consider e, f ∈ E(σ), and y ∈ X. If Γ (σ,X) does not have isolated vertices, then the following two assertions hold

1. if ⟨X⟩ (e,f ) σ ≠ ∅, then ⟨X⟩ (e ′ ,f ′ ) σ = ∅ for any e ′ , f ′ ∈ E(σ) such that {e ′ , f ′ } ≠ {e, f }; 2. if X (e,f ) σ (y) ≠ ∅, then X (e ′ ,f ′ ) σ (y) = ∅ for any e ′ , f ′ ∈ E(σ) such that {e ′ , f ′ } ≠ {e, f }.
Proof. Consider e, f, e ′ , f ′ ∈ E(σ). For the first assertion, suppose that there exist v ∈ ⟨X⟩

(e,f ) σ and v ′ ∈ ⟨X⟩ (e ′ ,f ′ ) σ . We have to prove that {e, f } = {e ′ , f ′ }. (9.8) Since v, v ′ ∈ ⟨X⟩ σ , we have {v, v ′ } ∈ E(Γ (σ,X)
) by Remark 9.13. Furthermore, since Γ (σ,X) does not have isolated vertices, there exist w, w

′ ∈ (V (σ) ∖ X) ∖ {v, v ′ } such that {v, w}, {v ′ , w ′ } ∈ E(Γ (σ,X) ). Suppose that w = w ′ . We obtain {w, v}, {w, v ′ } ∈ E(Γ (σ,X)
). It follows from Fact 9.18 that (e, f ) = (e ′ , f ′ ), so (9.8) holds. We obtain the same conclusion when {v, w ′ } ∈ E(Γ (σ,X) ) or {v ′ , w} ∈ E(Γ (σ,X) ). Thus, suppose that w ≠ w ′ , and {v, w ′ }, {v ′ , w} ∈ E(Γ (σ,X) ). It follows from the first assertion of Fact 9.

20 applied to v, v ′ , w ′ that X ∪ {v ′ , w ′ } is a module of σ[X ∪ {v, v ′ , w ′ }]. Since v ∈ ⟨X⟩ (e,f ) σ , we obtain [v, v ′ ] σ = (f, e).
Similarly, it follows from the first assertion of Fact 9.20 applied to v, v ′ , w that [v ′ , v] σ = (f ′ , e ′ ). Therefore, we have e = f ′ and e ′ = f . Consequently, (9.8) holds.

For the second assertion, suppose that there exist v ∈ X (e,f ) σ (y) and v ′ ∈ X (e ′ ,f ′ ) σ (y), where y ∈ X. We have to prove that (9.8) holds. Since v, v ′ ∈ X σ (y), we have {v, v ′ } ∈ E(Γ (σ,X) ) by Remark 9.13. Furthermore, since Γ (σ,X) does not have isolated vertices, there exist w, w

′ ∈ (V (σ) ∖ X) ∖ {v, v ′ } such that {v, w}, {v ′ , w ′ } ∈ E(Γ (σ,X) ). Suppose that w = w ′ . We obtain {w, v}, {w, v ′ } ∈ E(Γ (σ,X) )
. By Fact 9.18, (e, f ) = (e ′ , f ′ ), so (9.8) holds. We obtain the same conclusion when {v, w ′ } ∈ E(Γ (σ,X) ) or {v ′ , w} ∈ E(Γ (σ,X) ). Now, suppose that w ≠ w ′ , and {v, w ′ }, {v ′ , w} ∈ E(Γ (σ,X) ). It follows from the second assertion of Fact 9.20 applied to v, v ′ , w ′ that {y, v} is a module of σ

[X ∪ {v, v ′ , w ′ }]. We obtain [v ′ , y] σ = [v ′ , v] σ . Since v ′ ∈ X (e ′ ,f ′ ) σ (y), we have [y, v ′ ] σ = (e ′ , f ′ ). Therefore, we obtain [v ′ , v] σ = (f ′ , e ′ ).
Similarly, it follows from the second assertion of Fact 9.20 applied to v, v ′ , w that [v, v ′ ] σ = (f, e). Thus, we have e = f ′ and e ′ = f . Consequently, (9.8) holds. Lemma 9.24. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. Consider e, f ∈ E(σ) and y ∈ X. Suppose that e ≠ f. (

If Γ (σ,X) does not have isolated vertices, then the following two assertions hold

1. if ⟨X⟩ (e,f ) σ ≠ ∅ and ⟨X⟩ (f,e) σ ≠ ∅, then [⟨X⟩ (e,f ) σ , ⟨X⟩ (f,e) σ ] σ = (f, e); 2. if X (e,f ) σ (y) ≠ ∅ and X (f,e) σ (y) ≠ ∅, then [X (e,f ) σ (y), X (f,e) σ (y)] σ = (f, e).
Proof. For the first assertion, consider v ∈ ⟨X⟩

(e,f ) σ and v ′ ∈ ⟨X⟩ (f,e) σ . Since v, v ′ ∈ ⟨X⟩ σ , we have {v, v ′ } ∈ E(Γ (σ,X)
) by Remark 9.13. Furthermore, since Γ (σ,X) does not have isolated vertices, there exists

w ′ ∈ (V (σ) ∖ X) ∖ {v, v ′ } such that {v ′ , w ′ } ∈ E(Γ (σ,X)
). Suppose for a contradiction that {v, w ′ } ∈ E(Γ (σ,X) ). We obtain {v, w ′ }, {v ′ , w ′ } ∈ E(Γ (σ,X) ). It follows from Fact 9.18 that e = f , which contradicts (9.9). Therefore, we have {v, w ′ } ∈ E(Γ (σ,X) ). It follows from the first assertion of Fact 9.20 applied to

v, v ′ , w ′ that X ∪ {v ′ , w ′ } is a module of σ[X ∪ {v, v ′ , w ′ }]. Since v ∈ ⟨X⟩ (e,f ) σ , we obtain [v ′ , v] σ = (e, f ). For the second assertion, consider v ∈ X (e,f ) σ (y) and v ′ ∈ X (f,e) σ (y). Since v, v ′ ∈ X σ (y), we have {v, v ′ } ∈ E(Γ (σ,X)
) by Remark 9.13. Furthermore, since Γ (σ,X) does not have isolated vertices, there exists w ′ ∈ (V (σ) ∖ X) ∖ {v, v ′ } such that {v ′ , w ′ } ∈ E(Γ (σ,X) ). Suppose for a contradiction that {v, w ′ } ∈ E(Γ (σ,X) ). We obtain {v, w ′ }, {v ′ , w ′ } ∈ E(Γ (σ,X) ). It follows from Fact 9.18 that e = f , which contradicts (9.9). Therefore, we have {v,

w ′ } ∈ E(Γ (σ,X) ). It follows from the second assertion of Fact 9.20 applied to v, v ′ , w ′ that {y, v} is a module of σ[X ∪ {v, v ′ , w ′ }]. Thus, we obtain [v, v ′ ] σ = [y, v ′ ] σ . Since v ′ ∈ X (f,e) σ (y), we have [y, v ′ ] σ = (f, e), so [v, v ′ ] σ = (f, e).
To state the next result, we use the following notation and definition. Notation 9.25. Let σ be a 2-structure. For e ∈ E(σ) and W ⊆ V (σ), set

e[W ] = e ∩ (W × W ). Given e ∈ E(σ) and W ⊆ V (σ), we do not have e ∈ E(σ[W ]), but we have e[W ] ∈ E(σ[W ]) when e[W ] ≠ ∅. Lemma 9.26. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. If σ is prime, then the next two assertions hold. 1. Let e ∈ E(σ). If ⟨X⟩ (e,e) σ ≥ 2, then σ[⟨X⟩ σ ] is constant and E(σ[⟨X⟩ σ ]) = {e[⟨X⟩ σ ]}. Similarly, given y ∈ X, if X (e,e) σ (y) ≥ 2, then σ[X σ (y)] is constant and E(σ[X σ (y)]) = {e[X σ (y)]}. 2. Consider distinct e, f ∈ E(σ). If ⟨X⟩ (e,f ) σ ≥ 2, then σ[⟨X⟩ σ ] is linear and E(σ[⟨X⟩ σ ]) = {e[⟨X⟩ σ ], f [⟨X⟩ σ ]}. Similarly, given y ∈ X, if X (e,f ) σ (y) ≥ 2, then σ[X σ (y)] is linear and E(σ[X σ (y)]) = {e[X σ (y)], f [X σ (y)]}. Proof. Consider B q ∈ q (σ,X) , with B q ≥ 2. There exist e, f ∈ E(σ) such that B q = ⟨X⟩ (e,f ) σ or X (e,f ) σ (y), where y ∈ X. Consider C ∈ C {e,f } (σ[B q ]) (see Definition 2.
2). We prove that C is a module of σ. We utilize Corollary 9.19 in the following manner. Since B q ∈ q (σ,X) , there exists

B p ∈ p (σ,X) such that B q ⊆ B p . By Lemma 2.4, C is a module of σ[B q ]. Now, we show that C is a module of σ[B p ].
Suppose that e = f . It follows from Lemma 9.23 that B q = B p . Hence, C is a module of σ[B p ]. Suppose that e ≠ f . If B q = B p , then we proceed as previously. Hence, suppose that B q ≠ B p . It follows from Lemma 9.23 that B p ∖ B q ∈ q (σ,X) and

B p ∖ B q = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ⟨X⟩ (f,e) σ if B q = ⟨X⟩ (e,f ) σ or X (f,e) σ (y) if B q = X (e,f ) σ (y).
It follows from Lemma 9.24 that B q is a module of

σ[B p ]. Since C is a module of σ[B q ], it follows from Assertion (M3) of Proposition 2.5 that C is a module of σ[B p ].
Lastly, we prove that C is a module of Γ (σ,X) . Since C ⊆ B p , we have {c, v} ∈ E(Γ (σ,X) ) for c ∈ C and v ∈ B p ∖ C by Remark 9.13. Therefore, we have to verify that C is a module of Γ (σ,X)

[C ∪ {v}] for each v ∈ (V (σ) ∖ X) ∖ B p . Let v ∈ (V (σ) ∖ X) ∖ B p . Set C + = C ∩ N Γ (σ,X) (v) and C -= C ∖ N Γ (σ,X) (v). For a contradiction, suppose that C -≠ ∅ and C + ≠ ∅. It follows from Corol- lary 9.21 that [C -, C + ] σ = (e, f ) or (f, e), which contradicts C ∈ C {e,f } (σ[B q ]). Therefore, C -= ∅ or C + = ∅, that is, C is a module of Γ (σ,X) [C ∪ {v}] for each v ∈ (V (σ) ∖ X) ∖ B p . Thus, C is a module of Γ (σ,X) .
Consequently, C is a module of σ[B p ] and C is a module of Γ (σ,X) . It follows from Corollary 9.19 that C is a module of σ. Since σ is prime, C is trivial. Hence, we obtain C = 1 because C ≠ ∅ and C ∩ X = ∅. We conclude as follows by distinguishing the following two cases.

• Suppose that e = f . Recall that B q = B p by Lemma 9. [START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF]

(σ[B p ]) = {e[B p ]}.
• Suppose that e ≠ f . For instance, suppose that

B q = ⟨X⟩ (e,f ) σ . All the {e, f }-components of σ[⟨X⟩ (e,f ) σ ] are reduced to singletons. It follows from Proposition 2.8 that σ[B p ] is linear. Precisely, it follows from Lemma 2.4 that (v, w) σ = e or f for distinct v, w ∈ B p . In other words, σ[⟨X⟩ (e,f ) σ ] is linear and E(σ[⟨X⟩ (e,f ) σ ]) = {e[⟨X⟩ (e,f ) σ ], f [⟨X⟩ (e,f ) σ ]}. Lastly, suppose that B q ⊊ B p . It follows from Lemma 9.23 that B p ∖ B q = ⟨X⟩ (f,e) σ . Similarly, we have σ[⟨X⟩ (f,e) σ ] is linear and E(σ[⟨X⟩ (f,e) σ ]) = {e[⟨X⟩ (f,e) σ ], f [⟨X⟩ (f,e) σ ]}. Moreover, we have [⟨X⟩ (e,f ) σ , ⟨X⟩ (f,e) σ ] σ = (f, e)
by the first assertion of Lemma 9.24. Consequently, σ[⟨X⟩ σ ] is linear and

E(σ[⟨X⟩ σ ]) = {e[⟨X⟩ σ ], f [⟨X⟩ σ ]}.
We complete Section 9.3 with a result on the components of the outside graph, which follows from Fact 9.18 and the following easy consequence of Fact 9.20. We use the following notation. Notation 9.27. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. First, the set {⟨X⟩ (e,e) σ ∶ e ∈ E(σ)} ∪ {X (e,e) σ (α) ∶ e ∈ E(σ), α ∈ X} is denoted by q s (σ,X) . Second, the set q (σ,X) ∖ (q s (σ,X) ∪ {Ext σ (X)}) is denoted by q a (σ,X) .

Fact 9.28. Given a 2-structure σ, consider

X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. Consider distinct v, v ′ , w, w ′ ∈ V (σ) ∖ X such that {v, w}, {v ′ , w ′ } ∈ E(Γ (σ,X) ) and {v, w ′ }, {v ′ , w} ∈ E(Γ (σ,X)
). If there exist B q ∈ q (σ,X) such that w, w ′ ∈ B q , then B q ∈ q s (σ,X) .

Proof. Since w and w ′ belong to the same block of p (σ,X) , we have {w, w ′ } ∈ E(Γ (σ,X) ) by Remark 9.13. Besides, there exist e, f ∈ E(σ) such that B q = ⟨X⟩ (e,f ) σ

or B q = X (e,f ) σ (y), where y ∈ X. First, suppose that B q = ⟨X⟩ (e,f ) σ

. By the first assertion of Fact 9.20 applied to

v, w, w ′ , X ∪ {v, w} is a module of σ[X ∪ {v, w, w ′ }]. Since w ′ ∈ ⟨X⟩ (e,f ) σ , we have [w ′ , X] σ = (f, e). It follows that [w ′ , w] σ = (f, e).
Similarly, it follows from the first assertion of Fact 9.20 applied to v ′ , w, w ′ that [w ′ , w] σ = (e, f ). Thus, we obtain e = f , and hence B q ∈ q s (σ,X) . Second, suppose that B q = X (e,f ) σ (y), where y ∈ X. By the second assertion of Fact 9.20 applied to v, w, w ′ , {y, w ′ } is a module of σ[X ∪{v, w, w ′ }]. Thus, we have [w, w ′ ] σ = [w, y] σ . Since w ∈ X (e,f ) σ (y), we have [w, y] σ = (f, e). We obtain [w, w ′ ] σ = (f, e). Similarly, it follows from the second assertion of Fact 9.20 applied to v ′ , w, w ′ that [w ′ , w] σ = (f, e). Therefore e = f , so B q ∈ q s (σ,X) .

Proposition 9.29. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S3) holds. If Γ (σ,X) does not have isolated vertices, then the following two assertions hold.

1. For each component C of Γ (σ,X) , there exist distinct B p , D p ∈ p (σ,X) and B q , D q ∈ q (σ,X) such that B q ⊆ B p , D q ⊆ D p , and C is bipartite with bipartition

{V (C) ∩ B q , V (C) ∩ D q }. 2. For a component C of Γ (σ,X) and for B q ∈ q a (σ,X) , if V (C) ∩ B q ≠ ∅, then B q ⊆ V (C).
Proof. For the first assertion, consider a component C of Γ (σ,X) . Since Γ (σ,X) does not have isolated vertices, v(C) ≥ 2. Hence, there exist distinct c, d ∈ V (C) such that {c, d} ∈ E(Γ (σ,X) ). Furthermore, there exist B p , D p ∈ p (σ,X) and B q , D q ∈ q (σ,X) such that c ∈ B q , d ∈ D q , B q ⊆ B p and D q ⊆ D p . Since {c, d} ∈ E(Γ (σ,X) ), we have B p ≠ D p by Remark 9.13. Let v ∈ V (C) ∖ {c, d}. Since C is a component of Γ (σ,X) , there exists a sequence v 0 , . . . , v n of vertices of C satisfying

• v 0 ∈ {c, d}; • v n = v; {v 0 , . . . , v n } ∩ {c, d} = {v 0 }; • for i, j ∈ {0, . . . , n}, {v i , v j } ∈ E(Γ (σ,X) ) if and only if i -j = 1.
Since v 0 ∈ {c, d} and v n ∈ V (C) ∖ {c, d}, we have n ≥ 1. We distinguish the following two cases.

1. Suppose that n is even. It follows from Fact 9.18 that v 0 , v 2 , . . . , v n belong to the same block of q (σ,X) . Since v 0 ∈ {c, d} and v n = v, we obtain v ∈ B q ∪ D q .

2. Suppose that n is odd. Set

v -1 = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ d if v 0 = c and c if v 0 = d.
We have v -1 ∈ B q ∪ D q . By considering the sequence v -1 v 0 . . . v n , it follows from Fact 9.18 that v and v -1 belong to the same block of q (σ,X) . Hence, v ∈ B q ∪ D q .

Therefore, we obtain V (C) ∖ {c, d} ⊆ B q ∪ D q , and hence V (C) ⊆ B q ∪ D q . By Remark 9.13, C is bipartite with bipartition {V (C) ∩ B q , V (C) ∩ D q }. For the second assertion, consider a component C of Γ (σ,X) and an element B q of q a (σ,X) such that V (C) ∩ B q ≠ ∅. Consider v ∈ V (C) ∩ B q . For a contradiction, suppose that

B q ∖ V (C) ≠ ∅,
and consider v ′ ∈ B q ∖ V (C). Since Γ (σ,X) does not have isolated vertices, there

exist u ∈ (V (σ) ∖ X) ∖ {v} and u ′ ∈ (V (σ) ∖ X) ∖ {v ′ } such that {u, v}, {u ′ , v ′ } ∈ E(Γ (σ,X) ). Furthermore, since C is a component of Γ (σ,X) , with v ∈ V (C) and v ′ ∈ V (C), we obtain u ∈ V (C) and u ′ ∈ V (C). Therefore, u ≠ u ′ , and {u, v ′ }, {u ′ , v} ∈ E(Γ (σ,X)
). It follows from Fact 9.28 that B q ∈ q s (σ,X) , which contradicts B q ∈ q a (σ,X) . Consequently, we have B q ⊆ V (C).

9.4 Proofs of Theorems 9.5 and 9.6

We use the following notation.

Notation 9.30. Given a graph Γ, C(Γ) denotes the set of the components of Γ.

Proof of Theorem 9.5. To begin, suppose that σ is not prime. We prove that there exists C ∈ C(Γ (σ,X) ) such that σ[X ∪ V (C)] is not prime. First, suppose that Γ (σ,X) admits isolated vertices. Hence, consider v ∈ V (σ) ∖ X such that {v} ∈ C(Γ (σ,X) ). Since Statement (S3) holds, Ext σ (X) = ∅ by Remark 9.4. Thus σ[X ∪ {v}] is not prime. Second, suppose that Γ (σ,X) does not have isolated vertices. Since σ is not prime, σ admits a nontrivial module M . It follows from Corollary 9.15 that M ∩ X = ∅. By Lemma 9.17, there exists B p ∈ p (σ,X) such that M ⊆ B p and M is a module of Γ (σ,X) . Let u ∈ M . Since Γ (σ,X) does not have isolated vertices, there exists v ∈ (V (σ) ∖ X) ∖ {u} such that {u, v} ∈ E(Γ (σ,X) ).

Since M ⊆ B p , we have v ∈ M by Remark 9.13. Denote by C the component of

Γ (σ,X) containing u. We obtain v ∈ V (C) because {u, v} ∈ E(Γ (σ,X) ). Since M is a module of Γ (σ,X) , we obtain {u ′ , v} ∈ E(Γ (σ,X) ) for every u ′ ∈ M . Therefore, we have M ⊆ V (C). It follows that M is a nontrivial module of σ[X ∪ V (C)]
. Now, we suppose that there exists

C ∈ C(Γ (σ,X) ) such that σ[X ∪ V (C)] is not prime. Since σ[X ∪ V (C)] is not prime, we have v(C) ≠ 2. Assume that v(C) ≥ 4. We have to prove that C is not prime. Consider a nontrivial module M of σ[X ∪ V (C)]. Clearly, σ[X ∪ V (C)] satisfies Statement (S3). Moreover, Γ (σ[X∪V (C)],V (C)) = C. Since v(C) ≥ 4, it follows from Corollary 9.15 applied to σ[X ∪ V (C)] that M ⊆ V (C)
. By Lemma 9.17 applied to σ[X ∪ V (C)], there exists

B p ∈ p (σ[X∪V (C)],V (C)) such that M ⊆ B p and M is a module of C. We have to verify that M ≠ V (C). Let u ∈ M . Since v(C) ≥ 4, there exists v ∈ V (C) ∖ {u} such that {u, v} ∈ E(C).
In particular, we have v ∈ V (C). Since u ∈ B p , we have v ∈ B p by Remark 9.13 applied to σ

[X ∪ V (C)]. Since M ⊆ B p , we obtain v ∈ V (C) ∖ M .
Lastly, we suppose that there exists C ∈ C(Γ (σ,X) ) such that v(C) = 1 or v(C) ≥ 3 and C is not prime. We have to prove that σ is not prime. Therefore, by Corollary 9.22, we can suppose that Γ (σ,X) does not have isolated vertices.

(9.10)

In particular, we obtain v(C) ≥ 3. Consider a nontrivial module M of C.

Clearly, M is a module of Γ (σ,X) because C is a component of Γ (σ,X) . Since Γ (σ,X) does not have isolated vertices by (9.10), it follows from the first assertion of Proposition 9.29 that there exist distinct B p , D p ∈ p (σ,X) and B q , D q ∈ q (σ,X) such that B q ⊆ B p , D q ⊆ D p , and C is bipartite with bipartition

{V (C) ∩ B q , V (C)∩D q }. Since C is connected, we have M ⊆ V (C)∩B q or M ⊆ V (C)∩D q .
For instance, assume that M ⊆ V (C) ∩ B q . To conclude, we distinguish the following two cases.

1. Suppose that B q ∈ q s (σ,X) . There exists e ∈ E(σ) such that B q = ⟨X⟩ (e,e) σ or X (e,e) σ (y), where y ∈ X. If σ[B p ] is not constant, then it follows from the first assertion of Lemma 9.26 that σ is not prime. Thus, suppose that

σ[B p ] is constant. It follows that any subset of B p is a module of σ[B p ].
In particular, M is a module of σ[B p ]. Since M is a module of Γ (σ,X) , it follows from Corollary 9.19 that M is a module of σ.

2. Suppose that B q ∈ q a (σ,X) . Since Γ (σ,X) does not have isolated vertices by (9.10), it follows from the second assertion of Proposition 9.29 that

B q ⊆ V (C).
In general, M is not a module of σ[B q ], and hence M is not a module of σ[B p ]. Therefore, we cannot apply Corollary 9.19 to M . Nevertheless, we construct a superset of M , which is a module of Γ (σ,X) and a module of

σ[B p ]. Consider the set M of the nontrivial modules M ′ of C such that M ⊆ M ′ . Set M = ⋃ M.
Clearly, M ∈ M. Since M ≠ ∅ and all the elements of M contain M , it follows from Assertion (M5) of Proposition 2.

5 that M is a module of C. Since C is a component of Γ (σ,X) , M is a module of Γ (σ,X) . As previously seen for M , M ⊆ V (C) ∩ B q or M ⊆ V (C) ∩ D q . Since M ⊆ M and M ⊆ V (C) ∩ B q , we have M ⊆ V (C) ∩ B q . Therefore, M ⊆ B q . Set N = {v ∈ B q ∖ M ∶ v ←→ σ M }.
We verify that M ∪ N is a module of C. It suffices to show that for any

w ∈ V (C) ∩ D q , u ∈ M and v ∈ N , we have {u, w}, {v, w} ∈ E(Γ (σ,X) ) or {u, w}, {v, w} ∈ E(Γ (σ,X) ). Since v ∈ N , there exist u ′ , u ′′ ∈ M such that v ←→ σ {u ′ , u ′′ }.
Furthermore, since M is a module of C, we have {u, w}, {u ′ , w}, {u ′′ , w} ∈ E(Γ (σ,X) ) or {u, w}, {u ′ , w}, {u ′′ , w} ∈ E(Γ (σ,X) ).

For instance, suppose that {u, w}, {u ′ , w}, {u ′′ , w} ∈ E(Γ (σ,X) ). By Corollary 9.21,

{z ∈ B q ∶ {z, v} ∈ E(Γ (σ,X) )} is a module of σ[B q ]. Since u, u ′ , u ′′ ∈ {z ∈ B q ∶ {z, v} ∈ E(Γ (σ,X) )} and v ←→ σ {x ′ , x ′′ }, we obtain v ∈ {z ∈ B q ∶ {z, v} ∈ E(Γ (σ,X) )}.
Hence {u, w}, {u ′ , w}, {u ′′ , w}, {v, w} ∈ E(Γ (σ,X) ). Similarly, if {u, w}, {u ′ , w}, {u ′′ , w} ∈ E(Γ (σ,X) ), then if follows from Corollary 9.21 that {u, w}, {u ′ , w}, {u ′′ , w}, {v, w} ∈ E(Γ (σ,X) ). Consequently, M ∪ N is a module of C. It follows from the definition of M that N ⊆ M . Therefore, we have N = ∅, and hence, M is a module of σ[B q ]. Since Γ (σ,X) does not have isolated vertices by (9.10), it follows from Lemmas 9.23 and 9.24 that M is a module of σ[B p ]. Lastly, since M is a module of Γ (σ,X) , it follows from Corollary 9.19 that M is a module of σ.

The next result is an easy consequence of Theorem 9.5.

Corollary 9.31. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. If Statement (S5) holds, then Γ (σ,X) does not embed P 5 (see Figure 1.1).

Proof. For a contradiction, suppose that there exists Y ⊆ (V (σ) ∖ X) such that Γ (σ,X) [Y ] ≃ P 5 . Since P 5 is connected, there exists a component C of Γ (σ,X) such that Y ⊆ V (C). We have

Γ (σ[X∪Y ],X) = Γ (σ,X) [Y ]. Since Γ (σ,X) [Y ] = C[Y ], Γ (σ[X∪Y ],X) is prime. It follows from Theorem 9.5 applied to σ[X ∪ Y ] that σ[X ∪ Y ]
is prime, which contradicts the fact that Statement (S5) holds.

Since the proof of the next observation is obvious, we omit it.

Observation 9.32. Given a connected and bipartite graph Γ, Γ embeds K 2 ⊕K 2 if and only if Γ embeds P 5 .

Proof of Theorem 9.6. We make a preliminary observation. Since Statement (S5) holds, it follows from Remark 9.4 that Statement (S3) holds as well.

To begin, suppose that the first assertion holds, that is, σ is (V (σ) ∖ X)critical. We have to prove that the second assertion holds. Consider C ∈ C(Γ (σ,X) ). By Theorem 9.

5 applied to σ, σ[X ∪ V (C)] is prime. We have to show that σ[X ∪ V (C)] is V (C)-critical. Let c ∈ V (C). Since σ is (V (σ) ∖ X)- critical, σ -c is not prime. We have Γ (σ-c,X) = Γ (σ,X) -c.
Therefore, we obtain

C(Γ (σ-c,X) ) = (C(Γ (σ,X) ) ∖ {C}) ∪ C(C -c). (9.11)
Since σc is not prime, it follows from Theorem 9.5 applied to σc that there exists

C ′ ∈ C(Γ (σ-c,X∖{c}) ) such that σ[X ∪ V (C ′ )] is not prime. By (9.11), C ′ ∈ (C(Γ (σ,X) ) ∖ {C}) ∪ C(C -c)
. By Theorem 9.5 applied to σ, σ[X ∪ V (D)] is prime for every D ∈ (C(Γ (σ,X) ) ∖ {C}). Thus, we obtain

C ′ ∈ C(C -c). Finally, since Γ (σ[X∪V (C)]-c,V (C)∖{c}) = C -c, it follows from Theorem 9.5 applied to σ[X ∪ V (C)] -c that σ[X ∪ V (C)] -c is not prime. Consequently, σ[X ∪ V (C)] is V (C)-critical.
To continue, suppose that the second assertion holds. We have to prove that the third assertion holds. Consider C ∈ C(Γ (σ,X) ). By Theorem 9.5 applied to σ, v(C) = 2 or v(C) ≥ 4 and C is prime. Suppose that v(C) ≥ 4 and C is prime. We have to show that

C is critical. Let c ∈ V (C). We have to show that C -c is not prime. If C -c is disconnected, then C -c is not prime. Thus, suppose that C -c is connected. It follows that Γ (σ[X∪V (C)]-c,V (C)∖{c}) = C -c. Since the second assertion holds, σ[X ∪ V (C)] -c is not prime. It follows from Theorem 9.5 applied to σ[X ∪ V (C)] -c that C -c is not prime.
Lastly, suppose that the third assertion holds. Hence, for every C ∈ C(Γ (σ,X) ), we have v(C) = 2 or v(C) ≥ 4 and C is critical. (9.12)

We have to prove that σ is (V (σ) ∖ X)-critical. By Theorem 9.5 applied to σ, σ is prime. Let v ∈ V (σ) ∖ X. We have to prove that σv is not prime.

Denote by C the component of Γ (σ,X) containing v. Since σ is prime, it follows from Corollary 9.22 that Γ (σ,X) has no isolated vertices. By the first assertion of Proposition 9.29, C is bipartite. Moreover, C does not embed P 5 by Corollary 9.31. It follows from Observation 9.32 that C does not embed

K 2 ⊕ K 2 . Therefore, C -v does not embed K 2 ⊕ K 2 . ( 9.13)
As seen in (9.11),

C(C -v) ⊆ C(Γ (σ-v,X) ). (9.14)
Suppose that Cv admits isolated vertices. By (9.14), Γ (σ-v,X) admits isolated vertices as well. It follows from Corollary 9.22 that σv is not prime. Finally, suppose that Cv does not admit isolated vertices. Hence, v(C ′ ) ≥ 2 for each

C ′ ∈ C(C -v).
In particular, we do not have v(C) = 2. It follows from (9.12) that v(C) ≥ 4 and C is critical.

(9.15) Since v(C ′ ) ≥ 2 for each C ′ ∈ C(C -v), it follows from (9.13) that C -v is connected. By (9.14), C -v ∈ C(Γ (σ-x,X)
). Furthermore, it follows from (9.15) that v(Cv) ≥ 3 and Cv is not prime. By Theorem 9.5 applied to σv, σv is not prime.

Outside graph and half graph

Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S5) holds. Suppose also that σ is Definition 9.33. We extend to the infinite case the definition of the half graph H 2m (see Figure 4.1). Given a bipartite graph Γ, with bipartition {X, Y }, Γ is a half graph [START_REF] Erdős | Chromatic number of finite and infinite graphs and hypergraphs[END_REF] if there exist a linear order L defined on X, and a bijection ϕ from X onto Y such that

(V (σ)∖X)-critical. Consider a component C of Γ (σ,X) such that v(C) ≥ 4.
E(Γ) = {{x, ϕ(x ′ )} ∶ x ≤ L x ′ }. (9.16)
Clearly, a finite half graph is isomorphic to the graph H 2m , where m ≥ 1 (see Figure 4.1).

Remark 9.34. Given a bipartite graph Γ, with bipartition {X, Y }. Suppose that Γ is a half graph. There exist a linear order L defined on X, and a bijection ϕ from X onto Y such that (9.16) holds. Given x, y ∈ X, we obtain that

x ≤ L y if and only if N Γ (x) ⊇ N Γ (y).

2. Γ does not embed P 5 and Γ is critical. Now, we examine Theorem 9.38 in the finite case (see Proposition 9.41). We need the following result which follows from the characterization of finite critical 2-structures done in Section 4.2.

Corollary 9.39. Given a finite and symmetric 2-structure τ , with v(τ ) ≥ 5, τ is critical if and only if τ is isomorphic to σ(H 2n ), where n ≥ 3.

Proof. To begin, suppose that τ is isomorphic to σ(H 2n ), where n ≥ 3. By Corollary 4.20, τ is critical.

Conversely, suppose that τ is critical. By Corollary 4.6, there exists n ≥ 3 such that P(τ ) is isomorphic to P 2n or there exists n ≥ 3 such that P(τ ) is isomorphic to P 2n ⊕K {2n} , P 2n+1 , or C 2n+1 . Since τ is symmetric, it follows from Propositions 4.23, 4.27, and 4.36 that P(τ ) is not isomorphic to P 2n ⊕ K {2n} , P 2n+1 , or C 2n+1 . Consequently, P(τ ) is isomorphic to P 2n . It follows from Corollary 4.20 that τ is isomorphic to σ(H 2n ), where n ≥ 3.

The next result is an immediate consequence of Corollary 9.39. Corollary 9.40. Given a finite and bipartite graph Γ, with v(Γ) ≥ 5, Γ is critical if and only if Γ is a half graph. Proof. To begin, suppose that Γ is critical. We obtain that σ(Γ) is symmetric and critical. By Corollary 9.39 that σ(Γ) is isomorphic to σ(H 2n ), where n ≥ 3. We obtain that Γ is isomorphic to H 2n or its complement H 2n . Since n ≥ 3, H 2n embeds the complete graph K 3 . Since Γ is bipartite, we obtain that Γ is isomorphic to H 2n . As seen at the end of Definition 9.33, H 2n is a half-graph.

Conversely, suppose that Γ is a half graph. As seen at the end of Definition 9.33, Γ is isomorphic to H 2n , where n ≥ 3. By Corollary 9.39, σ(H 2n ) is critical. Hence, H 2n is critical too. Therefore, Γ is critical. Proposition 9.41. For a finite and bipartite graph Γ, with v(Γ) ≥ 4, the following assertions are equivalent 1. Γ does not embed P 5 and Γ is prime;

2. Γ is critical; 3. Γ is a half graph.
Proof. First, suppose that v(Γ) = 4. We have Γ is prime if and only if Γ is isomorphic to P 4 , which is isomorphic to the half graph H 4 . Therefore, the three assertions above are equivalent when v(Γ) = 4.

Second, suppose that v(Γ) = 5. The first assertion does not hold because a prime and bipartite graph defined on 5 vertices 6 is isomorphic to P 5 . (9.17) Furthermore, by Corollary 9.40, the last two assertions do not hold because v(Γ) is odd. Thus, the three assertions above are equivalent when v(Γ) = 5. Now, suppose that v(Γ) ≥ 6. By Corollary 9.40, the last two assertions are equivalent. To begin, suppose that the first assertion holds. By (9.17), Γ does not embed a prime graph of size 5.

(9.18)

It follows from Theorem 5.3 that v(Γ) is even. If v(Γ) = 6, then Γ is critical by (9.18). Hence, suppose that v(Γ) ≥ 7. Since v(Γ) is even, it follows from Theorem 5.3 and (9.18) that Γ is critical. Consequently, the first assertion implies the second one. Lastly, suppose that Γ is both critical and a half graph. If v(Γ) = 6, then Γ does not embed P 5 . Suppose that v(Γ) ≥ 7. Since Γ is a half graph, v(Γ) is even. Since Γ is critical, it follows from Corollary 3.20 that Γ does not embed P 5 .

The next result is a consequence of Proposition 9.41 and Theorem 7.1.

Corollary 9.42. A half graph Γ, with v(Γ) ≥ 4, is prime.

Proof. There exists a bipartition {X, Y } of V (Γ), a linear order L defined on X, and a bijection ϕ from X onto Y such that

E(Γ) = {{x, ϕ(x ′ )} ∶ x ≤ L x ′ }. By Proposition 9.41, we can suppose that Γ is infinite. Consider a finite subset F of V (Γ). Let X ′ be a finite subset of X such that F ∩ X ⊆ X ′ , ϕ -1 (F ∩ Y ) ⊆ X ′ , and X ′ ≥ 2. Set F ′ = X ′ ∪ ϕ(X ′ ).
Clearly, we have

F ⊆ F ′ . By considering Y ′ = ϕ(X ′ ), the linear order L ′ = L[X ′ ],
and the bijection ϕ ↾X ′ ∶ X ′ → Y ′ , we obtain that Γ[F ′ ] is a half graph. By Proposition 9.41, Γ[F ′ ] is prime. To conclude, it suffices to use Theorem 7.1. Now, we are ready to demonstrate Theorem 9.38.

Proof of Theorem 9.38. By Proposition 9.41, we can suppose that Γ is infinite.

To begin, suppose that Γ is a discrete half graph. There exists a bipartition {X, Y } of V (Γ), a discrete linear order L defined on X, and a bijection ϕ from X onto Y such that E(Γ) = {{x, ϕ(x ′ )} ∶ x ≤ L x ′ }. By Corollary 9.42, Γ is prime. Hence, Γ is connected. Since Γ is a half graph, Γ does not embed K 2 ⊕ K 2 . It follows from Observation 9.32 that Γ does not embed P 5 . Now, we have to verify that for every x ∈ X, Γx is not prime. (

First, suppose that x is not the least element of L. Since L is discrete, x admits an immediate predecessor x -. It is easy to verify that {ϕ(x -), ϕ(x)} is a module of Γx. Second, suppose that x is the least element of L. Clearly, ϕ(x) is an isolated vertex of Γx, so Γx is not prime. Thus (9.19) holds. Similarly, it follows from Remark 9.34 that Γy is not prime for each y ∈ Y . Consequently, Γ is critical. Conversely, suppose that Γ does not embed P 5 and Γ is critical. Since Γ is bipartite, there exists a bipartition {X, Y } of V (Γ) such that X and Y are stable sets of Γ. To complete the proof, we establish the next claims. To begin, we define a linear order L on X as follows.

Proof. Let M be a nontrivial module of Γx. Since Γx is connected, we have M ⊆ X ∖ {x} or M ⊆ Y . In the first instance, M is a module of Γ. Therefore, we have M ⊆ Y . Set M -= {y ∈ M ∶ {x, y} ∈ E(Γ)} and M + = {y ∈ M ∶ {x, y} ∈ E(Γ)}. Clearly, M -and M + are modules of Γ. Since Γ is prime and M ≥ 2, we obtain M -= 1 and M + = 1. Denote by x -the unique element of M -and denote by x + the unique element of M + . We obtain M = {x -, x + }. Furthermore, we have {x, x -} ∈ E(Γ) and {x, x + } ∈ E(Γ).

Claim 9.46. Given x ∈ X, if Γx is connected, then there exist x -, x + ∈ Y satisfying the following assertions 1. {x -, x + } is the only nontrivial module of Γx; 2. {x, x -} ∈ E(Γ) and {x, x + } ∈ E(Γ); 3. for every u ∈ X, if u < L x, then {u, x -} ∈ E(Γ); 4. for every u ∈ X, if x < L u, then {u, x + } ∈ E(Γ); 5. Γ -{x, x -} and Γ -{x, x + } are prime; 6. x + is the unique element of V (Γ) ∖ {x} such that {x, x + } ∈ E(Γ) and Γ -{x, x + } is prime.

Proof. Since Γ is critical, Γx admits a nontrivial module M . By Claim 9.45, there exist x -, x + ∈ Y such that M = {x -, x + }, {x, x -} ∈ E(Γ), and {x, x + } ∈ E(Γ). Hence, {x -, x + } is a nontrivial module of Γx.

For a contradiction, suppose that M is not the only nontrivial module of Γx. Thus, there exists a nontrivial module N of Γx such that N ≠ M . By Claim 9.45, there exist z -, z + ∈ Y such that N = {z -, z + }, {x, z -} ∈ E(Γ), and {x, z + } ∈ E(Γ). If M ∩ N ≠ ∅, then M ∪ N is a nontrivial module of Γx of size 3, which contradicts Claim 9.45. Hence, we have M ∩ N = ∅. We show that M ∪ N is a module of Γx. Let u ∈ (X ∖ {x}). It suffices to verify that M ∪ N is a module of Γ[M ∪ N ∪ {u}]. Suppose that there exists v ∈ M ∪ N such that {u, v} ∈ E(Γ). For instance, suppose that v ∈ M . Since M is a module of Γx, we have {u, x -}, {u, x + } ∈ E(Γ). We obtain {u, x -} ∈ E(Γ), {x, x -} ∈ E(Γ), and {x, z + } ∈ E(Γ). Since Γ does not embed K 2 ⊕ K 2 , we obtain {u, z + } ∈ E(Γ). Since {z -, z + } is a module of Γx, we have {u, z -} ∈ E(Γ). Therefore, {u, w} ∈ E(Γ) for every w ∈ M ∪N . It follows that M ∪N is a module of Γx, which contradicts Claim 9.45 because M ∪ N = 4. Consequently, {x -, x + } is the only nontrivial module of Γx. It follows that Γ -{x, x -} and Γ -{x, x + } are prime.

Let u ∈ X such that u < L x. Since u < L x, we have N Γ (u) ⊇ N Γ (x). Hence, we have {u, x + } ∈ E(Γ) because {x, x + } ∈ E(Γ). Since {x -, x + } is a module of Γx, we obtain {u, x -} ∈ E(Γ).

Let u ∈ X such that x < L u. Since x < L u, we have N Γ (x) ⊇ N Γ (u). Hence, we have {u, x -} ∈ E(Γ) because {x, x -} ∈ E(Γ). Since {x -, x + } is a module of Γx, we obtain {u, x + } ∈ E(Γ).

As previously seen, Γ -{x, x -} and Γ -{x, x + } are prime. Now, consider v ∈ V (Γ) ∖ {x, x -, x + }. Clearly, {x -, x + } is a nontrivial module of Γ -{x, v}, so Γ -{x, v} is not prime. Since {x, x -} ∈ E(Γ), x + is the unique element of V (Γ) ∖ {x} such that {x, x + } ∈ E(Γ) and Γ -{x, x + } is prime. Definition 9.47. We define a function ϕ ∶ X → Y as follows. Given x ∈ X,

ϕ(x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ i x if Γ -x is disconnected (see Claim 9.44), or x + if Γ -x is connected (see Claim 9.46).
The next claim follows easily from Claims 9.44 and 9.46.

Claim 9.48. For every x ∈ X, ϕ(x) is the unique element of V (Γ) ∖ {x} such that {x, ϕ(x)} ∈ E(Γ) and Γ -{x, ϕ(x)} is prime.

In the next two claims, we verify that ϕ is bijective.

Claim 9.49. ϕ is injective.

Proof. Consider distinct u, v ∈ X. For instance, suppose that u < L v. In particular, v is not the least element of L. It follows from Claim 9.44 that Γv is connected. By Claim 9.46, there exist v -, v + ∈ Y such that {v, v -} ∈ E(Γ), {v, v + } ∈ E(Γ), and {v -, v + } is the only nontrivial module of Γv. We have ϕ(v) = v + . First, suppose that Γu is disconnected. We have ϕ(u) = i u , where i u is the unique isolated vertex of Γu by Claim 9.44. We obtain {v, ϕ(u)} ∈ E(Γ). Thus, we have ϕ(u) ≠ ϕ(v) because {v, ϕ(v)} ∈ E(Γ) (see Claim 9.48).

Second, suppose that Γ-u is connected. By Claim 9.46, there exist u -, u + ∈ Y such that {u, u -} ∈ E(Γ), {u, u + } ∈ E(Γ), and {u -, u + } is the only nontrivial module of Γu. We have ϕ(u) = u + . Since u < L v, it follows from the fourth assertion of Claim 9.46 applied to u that {v, ϕ(u)} ∈ E(Γ). Since {v, ϕ(v)} ∈ E(Γ) (see Claim 9.48), ϕ(u) ≠ ϕ(v). Claim 9.50. ϕ is surjective. [START_REF] Given X ∈ X | if Γ-x is disconnected, then the following assertions hold 1. Γx admits a unique isolated vertex i x and i x ∈ Y ; 2. N Γ (x) = Y , so x is the least element of L; 3. i x is the unique element of V (Γ) ∖ {x} such that Γ -{x[END_REF], we obtain that Γv admits an isolated vertex i v . Thus, we have

Proof. Let v ∈ Y . Since Γ is critical, Γ -v is not prime. First, suppose that Γ -v is disconnected. As in Claim 9.
N Γ (i v ) = {v}. Since {i v , ϕ(i v )} ∈ E(Γ), we obtain ϕ(i v ) = (i v ) + = v.
Second, suppose that Γ-v is connected. As in Claim 9.46, there exist v -, v + ∈ X such that {v -, v + } is the only nontrivial module of Γv, {v, v -} ∈ E(Γ), and {v, v + } ∈ E(Γ). Furthermore, Γ -{v, v -} and Γ -{v, v + } are prime. Thus, we obtain Γ -{v, v + } is prime and {v, v + } ∈ E(Γ). It follows from Claim 9.48 that v = ϕ(v + ).

It follows from Claims 9.49 and 9.50 that ϕ is bijective. Claim 9.51. Γ is the half graph defined from the linear order L, and the bijection ϕ.

Proof. Consider distinct u, x ∈ X. We have to verify that {u, ϕ(x)} ∈ E(Γ) if and only if u ≤ L x.

Suppose that u ≤ L x. We obtain N Γ (x) ⊆ N Γ (u). By Claim 9.48, we have ϕ(x) ∈ N Γ (x). Hence, we obtain ϕ(x) ∈ N Γ (u). Conversely, suppose that x < L u. In particular, u is not the least element of L. It follows from Claim 9.44 that Γu is connected. By the fourth assertion of Claim 9.46 applied to x, {u, x + } ∈ E(Γ), that is, {u, ϕ(x)} ∈ E(Γ).

Claim 9.52. Given x ∈ X, if x is not the least element of L, then x admits an immediate predecessor in L.

Proof. Let x ∈ X. Suppose that x is not the least element of L. It follows from Claim 9.44 that Γx is connected. By Claim 9.46, there exist x -, x + ∈ Y such that {x -, x + } is the only nontrivial module of Γx, {x, x -} ∈ E(Γ), and {x, x + } ∈ E(Γ). Furthermore, for every u ∈ X, we have

if u < L x, then {u, x -} ∈ E(Γ), (9.20)
by the third assertion of Claim 9.46 applied to x. Set

t = ϕ -1 (x -).
By Claim 9.48, {t, ϕ(t)} ∈ E(Γ), that is, {t, x -} ∈ E(Γ). We obtain x -∈ N Γ (t) ∖ N Γ (x). Hence, we have N Γ (t) ⊋ N Γ (x), so t < L x. We prove that t is the immediate predecessor of x. We must verify that

{u ∈ X ∶ t < L u < L x} = ∅.
First, suppose that Γt is disconnected. By Claim 9.44, there exists i t ∈ Y such that i t is an isolated vertex of Γt. Since ϕ(t) = i t , i t = x -. We obtain that {u, x -} ∈ E(Γ) for every u ∈ V (Γ) ∖ {t, x -}. It follows from (9.20)

that {u ∈ X ∶ t < L u < L x} = ∅.
Second, suppose that Γt is connected. By Claim 9.46, there exist t -, t + ∈ Y such that {t -, t + } is the only nontrivial module of Γt, {t, t -} ∈ E(Γ), and {t, t + } ∈ E(Γ). Furthermore, for every u ∈ X such that t < L u, we have {u, t + } ∈ E(Γ) by the fourth assertion of Claim 9.46 applied to t. Recall that t + = ϕ(t). Since t = ϕ -1 (x -), we obtain t + = x -. Therefore, for every u ∈ X such that t < L u, we have {u, x -} ∈ E(Γ). It follows from (9.20)

that {u ∈ X ∶ t < L u < L x} = ∅.
By Remark 9.34, Γ is also the half graph defined from the linear order ϕ(L) ⋆ defined on Y , and the bijection ϕ -1 ∶ Y → X. The analogue of Claim 9.52 for ϕ(L) ⋆ follows.

Claim 9.53. Given y ∈ Y , if y is not the least element of ϕ(L) ⋆ , then y admits an immediate predecessor in ϕ(L) ⋆ .

The next claim is an immediate consequence of Claims 9.53. Claim 9.54. Given x ∈ X, if x is not the greatest element of L, then x admits an immediate successor in L.

It follows from Claims 9.52 and 9.54 that L is discrete, which completes the proof of Theorem 9.38.

The next theorem follows from Theorems 9.6 and 9.38. Theorem 9.55. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S5) holds. Suppose also that σ is (V (σ) ∖ X)critical. For each component C of Γ (σ,X) , with v(C) ≥ 3, C is a discrete half graph.

Proof. Let C be a component of Γ (σ,X) such that v(C) ≥ 3. By Theorem 9.6, v(C) ≥ 4 and C is critical. Furthermore, since Statement (S5) holds, C does not embed P 5 by Corollary 9.31. Finally, to use Theorem 9.38, we must verify that C is bipartite. Indeed, since σ is prime, it follows from Corollary 9.22 that Γ (σ,X) has no isolated vertices. Furthermore, since Statement (S5) holds, Statement (S3) holds too by Remark 9.4. Therefore, it follows from Proposition 9.29 that there exist distinct B p , D p ∈ p (σ,X) and B q , D q ∈ q (σ,X) such that B q ⊆ B p , D q ⊆ D p , and C is bipartite with bipartition {V (C)∩B q , V (C)∩D q }. Therefore, it follows from Theorem 9.38 that C is a discrete half graph.

The next result follows from Theorems 9.5 and 9.6, Proposition 9.41, and Corollary 9.31. It is the finite version of Theorem 9.10. Moreover, we use it in the proof of Theorem 9.10. Corollary 9.56. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that V (σ) ∖ X is finite.

The following two assertions are equivalent 1. Statement (S5) holds and σ is prime;

2. σ is (V (σ) ∖ X)-critical.
Proof. To begin, suppose that σ is (V (σ)∖X)-critical. In particular, σ is prime. Furthermore, by Remark 9.4, Statement (S5) holds. Conversely, suppose that Statement (S5) holds and σ is prime. Since Statement (S5) holds, we can use Theorem 9.6 to prove that σ is (V (σ) ∖ X)-critical. Consider a component C of Γ (σ,X) such that v(C) ≥ 3. We have to show that C is critical. Since σ is prime, it follows from Theorem 9.5 that v(C) ≥ 4 and C is prime. Moreover, since Statement (S5) holds, it follows from Corollary 9.31 that C does not embed P 5 . By Proposition 9.41, Γ is critical. It follows from Theorem 9.6 that σ is (V (σ) ∖ X)-critical.

Proof of Theorem 9.10. To begin, suppose that σ is finitely (V (σ) ∖ X)-critical. Let v ∈ V (σ)∖X. Since σ is finitely (V (σ)∖X)-critical, there exists a finite subset F of V (σ)∖X such that σ[X ∪F ] is F -critical. It follows from Corollary 9.56 that Statement (S5) holds and σ is prime.

Conversely, suppose that Statement (S5) holds and σ is prime. We prove that σ is finitely (V (σ) ∖ X)-critical. Let F be a finite subset of V (σ) ∖ X. We have to find a finite subset

F ′ of V (σ) ∖ X such that F ⊆ F ′ and σ[X ∪ F ′ ] is (F ′ )-critical.
We distinguish the following two cases.

• Suppose that V (σ) ∖ X is finite. It follows from Corollary 9.56 that σ is (V (σ) ∖ X)-critical. Hence, we can consider V (σ) ∖ X for F ′ .

• Suppose that V (σ) ∖ X is infinite. By Corollary 9.8, there exists a finite subset

F ′ of V (σ) ∖ X such that F ⊆ F ′ and σ[X ∪ F ′ ] is prime. Since Statement (S5) holds, it follows from Corollary 9.56 applied to σ[X ∪ F ′ ] that σ[X ∪ F ′ ] is (F ′ )-critical.
As announced in Subsection 9.1, we discuss Theorem 9.10 in the next remark by using Theorems 9.6 and 9.38. Remark 9.57. We denote by L Q the usual linear order on the set of rational numbers. Obviously, L Q is not discrete. We consider the graph G defined on {0, 1, 2, 3} ∪ ({0, 1} × Q) by E(G) = {{0, 1}, {1, 2}, {2, 3}} ∪ {{1, (1, q)} ∶ q ∈ Q} ∪ ( ⋃ q∈Q {{(0, q), (1, r)} ∶ r ≥ q}). Fact 2.6). We consider the 2-structure σ(G) associated with G. Since G[X] is prime, σ(G)[X] is prime too. We have [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF], and p (σ(G),X) = {Y, Z}.

Set X = {0, 1, 2, 3}, Y = {0} × Q and Z = {1} × Q. We have G[X] is prime because G[X] = P 4 (see
Y = ⟨X⟩ σ(G) , Z = X σ(G)
Furthermore, it follows from Corollary 3.18 that

Γ (σ(G),X) = G[Y ∪ Z]. (9.21)
We verify that σ(G) is finitely (V (σ) ∖ X)-critical (see Definition 9.9) without being (V (σ) ∖ X)-critical. First, we show that Statement (Sk) holds for every odd integer k ≥ 1. Let W be a finite and nonempty subset of Y ∪ Z such that W ∈ P (σ,X) (see Notation 9.2). We have to show that W is even.

If W ∩ Y = ∅, then {0} ∪ W is a module of σ(G)[X ∪ W ] because Z = X σ G (0)
. Hence, we have W ∩ Y ≠ ∅. We denote the elements of W ∩ Y by (0, q 0 ), . . . , (0, q m ), where m ≥ 0, in such a way that q 0 < ⋯ < q m , when m ≥ 1. Set

Z -= {j < q 0 ∶ (1, j) ∈ W }. Since Z = X σ(G) (0), {0} ∪ ({1} × Z -) is a module of σ(G)[X ∪ W ]. Hence, we have Z -= ∅. Set Z + = {j ≥ q m ∶ (1, j) ∈ W }.
We obtain that {1} × Z + is a module of σ(G)[X ∪ W ]. Hence, we have

Z + ≤ 1. If Z + = ∅, then (X ∪ W ) ∖ {(0, q m )} is a module of σ(G)[X ∪ W ] because (0, q m ) ∈ ⟨X⟩ σ(G)
. Thus, we obtain Z + = 1. Therefore, W = 2 if m = 0. Now, suppose that m ≥ 1. Set

Z i = {q i ≤ j < q i+1 ∶ (1, j) ∈ W } for i = 0, . . . , m -1. Given i = 0, . . . , m -1, we have {1} × Z i is a module of σ(G)[X ∪ W ]. Hence, we have Z i ≤ 1. Moreover, {(0, q i ), (0, q i+1 )} is a module of σ(G)[X ∪ W ] if Z i = ∅
. Therefore, we obtain Z i = 1. Consequently, Z -= ∅, Z + = 1, and Z i = 1 for i = 0, . . . , m -1. Thus, W ∩ Z = m + 1, and hence

W = 2m + 2.
Second, we prove that σ G is finitely (V (σ) ∖ X)-critical. Let F be a finite subset of Y ∪ Z. There exists a finite subset Third, we verify that σ(G) is not (V (σ(G)) ∖ X)-critical. To begin, we verify that G[Y ∪ Z] is a non discrete half graph. Clearly, G[Y ∪ Z] is bipartite with bipartition {Y, Z}. Consider the bijection ϕ ∶ Y → Z, which maps (0, q) to (1, q) for each q ∈ Q. Moreover, consider the linear order L Y defined on Y as follows. Given distinct q, r ∈ Q, (0, q) < L Y (0, r) if q < L Q r. Clearly, G[Y ∪ Z] is the half graph defined from L Y and ϕ. Recall that the linear order L Y is unique by Remark 9.34. Since

F ′ of Q such that F ′ ≥ 2 and F ⊆ ({0, 1} × F ′ ). We have G[{0, 1} × F ′ ] ≃ H 2× F ′ (see
L Y ≃ L Q , G[Y ∪ Z] is not discrete.
Since Statement (S5) holds, Γ (σ(G),X) does not embed P 5 by Corollary 9.31. Since G[Y ∪Z] is a non discrete half graph, Γ (σ(G),X) is a non discrete half graph by (9.21). It follows from Theorem 9.38 that Γ (σ(G),X) is not critical. Clearly, G[Y ∪ Z] is connected. Therefore, Γ (σ(G),X) is connected by (9.21). Since Statement (S5) holds, it follows from Theorem 9.6 that σ(G) is not (V (σ) ∖ X)critical. Since σ(G) is finitely (V (σ(G)) ∖ X)-critical, it follows from Theorem 9.10 that σ(G) is prime. Consequently, there exists v ∈ V (σ(G)) ∖ X such that σ(G)v is prime. In fact, we have σ(G)w is prime for every w ∈ V (σ(G)) ∖ X.

Proof of Theorem 9.11. Since Statement (S5) holds, we can use Theorem 9.6 as follows. Let v ∈ V (σ) ∖ X. Denote by C the component of Γ (σ,X) such that v ∈ V (C). By Theorem 9.6 applied to σ, v(C) = 2 or v(C) ≥ 4 and C is critical.

First, suppose that v(C) = 2. We have

Γ (σ-V (C),X) = Γ (σ,X) -V (C).
Therefore, the components of Γ (σ-V (C),X) are the components of Γ (σ,X) that are distinct from C. Let D be a component of Γ (σ,X) such that D ≠ C. By Theorem 9.6 applied to σ, v(D) = 2 or v(D) ≥ 4 and D is critical. It follows from Theorem 9.6 applied to σ -

V (C) that σ -V (C) is ((V (σ) ∖ X) ∖ V (C))- critical.
Hence, we consider for w the unique element of V (C) ∖ {v}.

Second, suppose that v(C) ≥ 4 and C is critical. By Theorem 9.55, C is a discrete half graph. As seen in the proof of Theorem 9.55, there exist distinct B q , D q ∈ q (σ,X) such that C is bipartite with bipartition {V (C)∩B q , V (C)∩D q }. For instance, assume that v ∈ V (C) ∩ B q . Since C is a discrete half graph, there exists a discrete linear order L defined on V (C) ∩ B q and a bijection ϕ ∶ V (C) ∩ B q → V (C) ∩ D q such that C is defined from L and ϕ (see Definition 9.33). By Claim 9.48, C -{v, ϕ(v)} is prime. Hence, C -{v, ϕ(v)} is connected. Consequently, the components of Γ (σ-{v,ϕ(v)},X) are the components of Γ (σ,X) that are distinct from C and C -{v, ϕ(v)}. Let D be a component of Γ (σ,X) such that D ≠ C. By Theorem 9.6 applied to σ, v(D) = 2 or v(D) ≥ 4 and D is critical. If v(C) = 4, then v(C -{v, ϕ(v)}) = 2 and it follows from Theorem 9.6 applied to σ -{v, ϕ(v)} that σ -{v, ϕ(v)} is ((V (σ) ∖ X) ∖ {v, ϕ(v)})-critical. Lastly, suppose that v(C) ≥ 5. To apply Theorem 9.6 to σ -{v, ϕ(v)}, we must verify that v(C-{v, ϕ(v)}) ≥ 4 and C-{v, ϕ(v)} is critical. Since C is a half graph with v(C) ≥ 5, we have v(C) ≥ 6, and hence v(C -{v, ϕ(v)}) ≥ 4. Clearly, Lv is a discrete linear order. Moreover, C -{v, ϕ(v)} is the half graph defined from L-x and the bijection ϕ ↾(V (C)∩Bq)∖{v} ∶ (V (C) ∩ B q ) ∖ {v} → (V (C) ∩ D q ) ∖ {ϕ(v)}. Therefore, C -{v, ϕ(v)} is a discrete half graph. By Theorem 9.38, C -{v, ϕ(v)} is critical. Consequently, it follows from Theorem 9.6 applied to σ -{v, ϕ(v)} that σ -{v, ϕ(v)} is ((V (σ) ∖ X) ∖ {v, ϕ(v)})-critical. 9.6 Proofs of Theorems 5.8 and 5.9

Proof of Theorem 5.8. Let σ be a prime 2-structure. Consider X ⊊ V (σ) such that σ[X] is prime. Suppose that V (σ) ∖ X is finite and V (σ) ∖ X ≥ 6.

For a contradiction, suppose that for each proper subset

Y of V (σ) ∖ X, we have if σ[X ∪ Y ] is prime, then V (σ) ∖ (X ∪ Y ) is odd. (9.23)
For Y = ∅ in (9.23), we obtain V (σ) ∖ X is odd. Hence, we have V (σ) ∖ X ≥ 7.

For Y ⊊ (V (σ) ∖ X), with Y = 5, it follows from (9.23) that σ[X ∪ Y ] is not prime. Consequently, Statement (S5) holds. Since V (σ) ∖ X is odd, there exists a component C of Γ (σ,X) such that v(C) is odd. Since Statement (S5) holds, Statement (S3) holds too by Remark 9.4. Since σ is prime, it follows from Theorem 9.5 that σ[X ∪ V (C)] is prime. We have

V (σ) ∖ X = V (C) ∪ (V (σ) ∖ (X ∪ V (C))).
Since V (σ) ∖ X and v(C) are odd, we obtain that V (σ) ∖ (X ∪ V (C)) is even. It follows from (9.23) that V (C) = V (σ) ∖ X. Thus, Γ (σ,X) is connected. Since σ is prime, it follows from Theorem 9.5 that Γ (σ,X) is prime. Furthermore, since σ is prime, it follows from Corollary 9.22 that Γ (σ,X) has no isolated vertices. Since Statement (S3) holds, it follows from Proposition 9.29 that C is bipartite. Finally, since Statement (S5) holds, Γ (σ,X) does not embed P 5 by Corollary 9.31. It follows from Proposition 9.41 that Γ (σ,X) is a half graph, which is impossible because v(Γ (σ,X) ) = V (σ) ∖ X and V (σ) ∖ X is odd. Consequently (9.23) does not hold. Therefore, there exists

Y ⊊ (V (σ) ∖ X) such that σ[X ∪ Y ] is prime and V (σ) ∖ (X ∪ Y ) is even. Recall that V (σ) ∖ X is finite, so V (σ) ∖ (X ∪ Y ) is as well. It follows from Corollary 3.20 applied to σ[X ∪ Y ] that there exist distinct v, w ∈ V (σ) ∖ (X ∪ Y ) such that σ -{v, w} is prime.
Proof of Theorem 5.9. Since Statement S 1 or Statement S 2 hold, we have q a (σ,X) ≠ ∅. (9.24)

By Theorem 5.8, we can assume that V (σ) ∖ X = 4 or 5. If V (σ) ∖ X = 4, then it suffices to apply Theorem 3.19. Hence, suppose that V (σ) ∖ X = 5.

For a contradiction, suppose that Statement (S3) holds. It follows from Theorem 9.5 that for each component C of Γ (σ,X) , we have v(C) = 2 or v(C) ≥ 4 and C is prime. Since V (σ) ∖ X = 5, we obtain that Γ (σ,X) is connected. Thus, Γ (σ,X) is prime. Since σ is prime, it follows from Corollary 9.22 that Γ (σ,X) has no isolated vertices. Since Statement (S3) holds, it follows from the first assertion of Proposition 9.29 that p (σ,X) = q (σ,X) , and q (σ,X) has two elements, denoted by B q and D q . Moreover, Γ (σ,X) is bipartite, with bipartition {B q , D q }. Since Γ (σ,X) is prime and bipartite, we have Γ (σ,X) ≃ P 5 . Hence, Γ (σ,X) embeds K 2 ⊕ K 2 . Thus, there exists distinct v, v ′ ∈ B q and distinct w, w ′ ∈ D q such that {v, w}, {v ′ , w ′ } ∈ E(Γ (σ,X) ) and {v, w ′ }, {v ′ , w} ∈ E(Γ (σ,X) ). It follows from Fact 9.28 that B q , D q ∈ q s (σ,X) , which contradicts (9.24). Consequently, Statement (S3) does not hold. Therefore, there exists Y ⊆ (V (σ) ∖ X) such that Y = 3 and σ[X ∪ Y ] is prime, which completes the proof because X = 5.

Chapter 10

The Rigollet-Thomassé theorem

The aim of this chapter is to demonstrate the following theorem.

Theorem 10.1 (Rigollet and Thomassé1 [START_REF] Rigollet | Relations infinies indécomposables critiques[END_REF]). Given an infinite prime 2structure σ, there exists

X ⊆ V (σ) such that 2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ X ≠ V (σ),
X is equipotent to V (σ), and σ[X] is prime.

(RT) Observation 10.2. Let σ be an infinite prime 2-structure. Suppose that σ is not finitely critical (see Definition 8.1). Hence, there exists a finite and nonempty subset F of V (σ) such that σ -F is prime. Clearly, V (σ) ∖ F is a proper subset of V (σ) and V (σ) ∖ F is equipotent to V (σ). Therefore, Theorem 10.1 holds for infinite prime 2-structure that are not finitely critical.

Rigollet and Thomassé [START_REF] Rigollet | Relations infinies indécomposables critiques[END_REF] associated the following digraph with a critical 2-structure. Proof. Since σ is connected, it follows from Proposition 2.12 that σ is uncuttable.

First, we prove that ⋃ Υ(σ) = V (σ).

(10.1)

Consider M ∈ Υ(σ). Let v ∈ V (σ) ∖ M. Consider the family N v of the proper modules of σ containing v. Set N = ⋃ N v .
It is easy to verify that N is a proper module of σ. Indeed, consider x, y ∈ N and w ∈ V (σ) ∖ N . We have to verify that

w ←→ σ {x, y}. Since x, y ∈ N , there exist N, N ′ ∈ N v such that x ∈ N and y ∈ N ′ . Since N, N ′ ∈ N v , we have v ∈ N ∩ N ′ . By Assertion (M5) of Proposition 2.5, N ∪ N ′ is a module of σ. Since w ∈ N , we have w ∈ N ∪ N ′ . It follows that w ←→ σ N ∪ N ′ .
In particular, we have w ←→ σ {x, y}. Therefore, N is a module of σ. Hence, (10.1) holds. Second, we show that Υ(σ) is a modular partition of σ. Since (10.1) holds, it suffices to verify that the elements of Υ(σ) are pairwise disjoint. Consider M, N ∈ Υ(σ) such that M ∩ N ≠ ∅. By Assertion (M5) of Proposition 2.5, M ∪ N is a module of σ. For a contradiction, suppose that M ∪ N = V (σ). Since N ≠ V (σ), we have M ∖ N ≠ ∅. By Assertion (M6) of Proposition 2.5, N ∖ M is a module of σ. Since M ∪ N = V (σ), we have N ∖ M = V (σ) ∖ M. Thus, M is a nontrivial modular cut of σ, which contradicts the fact that σ is uncuttable. It follows that M ∪ N ≠ V (σ).

It follows from the maximality of M and N that M = M ∪ N and N = M ∪ N . Consequently, we have M = N . It follows that Υ(σ) is a modular partition of σ.

Third, we prove that σ Υ(σ) is prime. Since σ is uncuttable, we have Υ(σ) ≥ 3. Let Ψ be a module of σ Υ(σ) such that Ψ ≥ 2. We must verify that Ψ = Υ(σ). By the second assertion of Lemma 2.10, ⋃ Ψ is a module of σ. Let M ∈ Ψ. Since Ψ ≥ 2, we have M ⊊ ⋃ Ψ. It follows from the maximality of M that ⋃ Ψ = V (σ). Hence, we obtain Ψ = Υ(σ).

The following fact is useful to utilize Proposition 10.5.

Fact 10.6. Let σ be a 2-structure. Consider X ⊊ V (σ) such that σ[X] is prime. Let M be a module of σ. If X ⊆ M , then Therefore, v ∈ M . Consequently, (V (σ) ∖ ⟨X⟩ σ ) ⊆ M . Now, we prove that σ -⟨X⟩ σ is uncuttable. Consider a modular cut C of σ -⟨X⟩ σ . By exchanging C and (V (σ) ∖ ⟨X⟩ σ ) ∖ C if necessary, we can assume that C ∩ X ≥ 2. Since C ∩ X is a module of σ[X] by Assertion (M2) of Proposition 2.5, we obtain X ⊆ C. It follows from the first assertion above that (V (σ) ∖ ⟨X⟩ σ ) ⊆ C. Hence, C is a trivial modular cut of σ -⟨X⟩ σ . It follows that σ -⟨X⟩ σ is uncuttable. By Proposition 2.12, σ -⟨X⟩ σ is connected.

(V (σ) ∖ ⟨X⟩ σ ) ⊆ M (see Notation 3.12). Consequently, σ -⟨X⟩ σ is connected. Proof. To begin, consider a module M of σ such that X ⊆ M . Let v ∈ (V (σ) ∖ ⟨X⟩ σ ). By Lemma 3.13, v ∈ Ext σ (X) or v ∈ X σ (y), where y ∈ X. First, suppose that v ∈ Ext σ (X). Set Y = X ∪ {v}. Since v ∈ Ext σ (X), σ[Y ] is prime. By Assertion (M2) of Proposition 2.5, M ∩ Y is a module of σ[Y ]. Since X ⊆ (M ∩ Y ), we obtain M ∩ Y = Y . Thus, v ∈ M . Second, suppose that v ∈ X σ (y),

Extreme vertices

Rigollet and Thomassé [START_REF] Rigollet | Relations infinies indécomposables critiques[END_REF] introduced the following definition. Definition 10.7. Consider a critical 2-structure σ. A vertex v of σ is extreme extreme if there exists w ∈ V (σ) ∖ {v} such that V (σ) ∖ {v, w} is a module of σv. The set of the extreme vertices of σ is denoted by E (σ).

For instance, as seen in Example 8.11, σ(H N ) is a prime element of F N . Hence, σ(H N ) is critical. Furthermore, {2, 3, . . .} is a module of σ(H N ) -0. Therefore, 0 is an extreme vertex of σ(H N ).

We use the next notation to prove Proposition 10.9. Notation 10.8. Consider an infinite critical 2-structure σ. By using the axiom of choice, we obtain a function

F E (σ) ∶ E (σ) → V (σ) satisfying for each v ∈ E (σ), v ≠ F E (σ) (v) and F E (σ) (v) ←→ σ V (σ) ∖ {v, F E (σ) (v)} (see Notation 2.1), that is, V (σ) ∖ {v, F E (σ) (v)} is a module of σ -v.
Lastly, observe that

F E (σ) is injective. Indeed, consider distinct v, v ′ ∈ E (σ). If F E (σ) (v) = F E (σ) (v ′ ), then V (σ) ∖ {F E (σ) (v)
} is a module of σ, which contradicts the fact that σ is prime. Consequently, F E (σ) is injective. Proposition 10.9. Given an infinite critical 2-structure σ, V (σ) and V (σ) ∖ E (σ) are equipotent.

Proof. Clearly, if E (σ) is finite, then V (σ) and V (σ) ∖ E (σ) are equipotent because V (σ) is infinite. Thus, suppose that E (σ) is infinite.

To begin, we show that F E (σ) does not contain cycles. Otherwise, there exists extreme vertices v 0 , . . . , v n of σ, where n ≥ 1, such that

F E (σ) (v 0 ) = v 1 ,. . . , F E (σ) (v n-1 ) = v n , and F E (σ) (v n ) = v 0 . We obtain that V (σ) ∖ {F E (σ) (v i ) ∶ i ∈ {0, .
. . , n}} is a module of σ, which contradicts the fact that σ is prime. Consequently, F E (σ) does not contain cycles. Now, given v ∈ E (σ), we prove that

if F E (σ) (v) ∈ E (σ) and (F E (σ) ) 2 (v) ∈ E (σ), then (F E (σ) ) 3 (v) ∈ E (σ). (10.2)
For a contradiction, suppose that there exists v ∈ E (σ) such that

F E (σ) (v), (F E (σ) ) 2 (v), (F E (σ) ) 3 (v) ∈ E (σ). Set v = (F E (σ) ) 0 (v) and F E (σ) (v) = (F E (σ) ) 1 (v).
Since F E (σ) does not contain cycles, (F E (σ) ) 0 (v), (F E (σ) ) 1 (v), (F E (σ) ) 2 (v), (F E (σ) ) 3 (v) and (F E (σ) ) 4 (v) are pairwise distinct. For i = 0, 1, 2, 3, there exists e i+1 , f i+1 ∈ E(σ) such that

[(F E (σ) ) i+1 (v), V (σ) ∖ {(F E (σ) ) i (v), (F E (σ) ) i+1 (v)}] σ = (e i+1 , f i+1 ). (10.3)
Moreover, for i = 0, 1, 2, 3, we have

(e i+1 , f i+1 ) ≠ [(F E (σ) ) i+1 (v), (F E (σ) ) i (v)] σ (10.4) because V (σ) ∖ {(F E (σ) ) i+1 (v)
} is not a module of σ. Using (10.3) and (10.4), we obtain

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ [(F E (σ) ) 1 (v), (F E (σ) ) 3 (v)] σ = (e 1 , f 1 ), [(F E (σ) ) 2 (v), (F E (σ) ) 3 (v)] σ = (e 2 , f 2 ), [(F E (σ) ) 3 (v), (F E (σ) ) 1 (v)] σ = (e 3 , f 3 ), and
[(F E (σ) ) 3 (v), (F E (σ) ) 2 (v)] σ ≠ (e 3 , f 3 ).
Therefore, we have (e 1 , f 1 ) ≠ (e 2 , f 2 ). (10.5) Using ( 10.3), we obtain

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ [(F E (σ) ) 1 (v), (F E (σ) ) 4 (v)] σ = (e 1 , f 1 ), [(F E (σ) ) 2 (v), (F E (σ) ) 4 (v)] σ = (e 2 , f 2 ), [(F E (σ) ) 4 (v), (F E (σ) ) 1 (v)] σ = (e 4 , f 4 ), and
[(F E (σ) ) 4 (v), (F E (σ) ) 2 (v)] σ = (e 4 , f 4 ).
It follows that (e 1 , f 1 ) = (e 2 , f 2 ), which contradicts (10.5). Consequently, (10.2) holds.

To conclude, set

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ E 0 (σ) = {v ∈ E (σ) ∶ F E (σ) (v) ∈ E (σ)}, E 1 (σ) = {v ∈ E (σ) ∖ E 0 (σ) ∶ (F E (σ) ) 2 (v) ∈ E (σ)}, and E 2 (σ) = {v ∈ E (σ) ∖ (E 0 (σ) ∪ E 1 (σ)) ∶ (F E (σ) ) 3 (v) ∈ E (σ)}. By (10.2), {E 0 (σ), E 1 (σ), E 2 (σ)} is a partition of E (σ). Since E (σ) is infinite, we obtain E (σ) = max( E 0 (σ) , E 1 (σ) , E 2 (σ) ). (10.6)
We obtain

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ F E (σ) (E 0 (σ)) ⊆ V (σ) ∖ E (σ), F E (σ) (E 1 (σ)) ⊆ E 0 (σ), and
F E (σ) (E 2 (σ)) ⊆ E 1 (σ).
Since F E (σ) is injective, we obtain

E 2 (σ) ≤ E 1 (σ) ≤ E 0 (σ) ≤ V (σ) ∖ E (σ) .
It follows from (10.6) that E (σ) ≤ V (σ) ∖ E (σ) . Therefore, we have

V (σ) = V (σ) ∖ E (σ) .
The next result follows from Proposition 10.5.

Corollary 10.10. Let σ be an infinite critical 2-structure.

Consider distinct v, w ∈ V (σ). If v ∈ E (σ) and (w, v) ∈ A(C(σ)), then {w} ∈ Υ(σ -v), Υ(σ -v)
is a modular partition of σv, and (σv) Υ(σv) is prime.

Proof. Since (w, v) ∈ A(C(σ)), we have {w} ∈ Υ(σv). For a contradiction, suppose that σv is not connected. There exist e, f ∈ E(σv) such that σv is not {e, f }-connected. Consider X ∈ C {e,f } (σv) (see Definition 2.2) such that w ∈ X. It follows from Lemma 2.4 that X is a module of σv. Since (w, v) ∈ A(C(σ)), we have X = {w}.

Using Proposition 2.8, we distinguish the following two cases. In each of them, we obtain a contradiction.

• Suppose that e = f . By Proposition 2.8,

(σ -v) C {e} (σ -v) is constant. Since {w} ∈ C {e} (σ -v), we obtain C {e} (σ -v) ∖ {{w}} is a module of (σ - v) C {e} (σ -v).
By the second assertion of Lemma 2.10, (V (σ)∖{v})∖{w} is a module of σv, which contradicts v ∈ E (σ).

• Suppose that e ≠ f . By Proposition 2.8,

(σ -v) C {e,f } (σ -v) is linear. Given ε ∈ E((σ -v) C {e,f } (σ -v)), (σ -v) C {e,f } (σ -v) is the 2-structure associated to the linear order (C {e,f } (σ -v), ε) (see Remark 1.3). Set X -= {Y ∈ C {e,f } (σ -v) ∶ (Y, X) ∈ ε}.
Clearly, X -∪ {{w}} is a module of (σv) C {e,f } (σv). By the second assertion of Lemma 2.10,

⋃(X -∪ {{w}}) is a module of σ -v. Since (w, v) ∈ A(C(σ)), we obtain ⋃(X -∪ {{w}}) = {w} or ⋃(X -∪ {{w}}) = V (σ) ∖ {v}.
We obtain that {w} is the least vertex or the greatest vertex of the linear order (C {e,f } (σv), ε). In both cases, it follows from the second assertion of Lemma 2.10 that (V (σ) ∖ {v}) ∖ {w} is a module of σv, which contradicts v ∈ E (σ).

Consequently, σv is connected. It follows from Proposition 10.5 that Υ(σv) is a modular partition of σv and (σv) Υ(σv) is prime.

Corollary 10.10 leads us to introduce the following notation.

Notation 10.11. Consider an infinite critical 2-structure σ. Set

W (σ) = V (σ) ∖ E (σ).
We consider the following subsets of W (σ). First, we denote by

W ∅ (σ) the set of v ∈ W (σ) such that Υ(σ -v) = ∅. Second, we denote by W δ (σ) the set of v ∈ W (σ) such that σ -v is not connected. Third, we denote by W π (σ) the set of v ∈ W (σ) such that Υ(σ -v) is a modular partition of σ -v and (σ -v) Υ(σ -v) is prime. Let v ∈ W δ (σ). Since σ -v is not connected, there exist e v , f v ∈ E(σ -v) such that σ -v is not {e v , f v }-connected. Hence, there exist e, f ∈ E(σ) such that e v = e ∩ (V (σ -v) × V (σ -v)) and f v = f ∩ (V (σ -v) × V (σ -v)). We denote {e, f } by λ(v).
Observation 10.12. Consider an infinite critical 2-structure σ. It follows from Proposition 10.5 that

W (σ) = W ∅ (σ) ∪ W δ (σ) ∪ W π (σ).
Furthermore, we have

(W ∅ (σ) ∪ W δ (σ)) ∩ W π (σ) = ∅. Clearly, we have W ∅ (σ) ∩ W π (σ) = ∅. To verify that W δ (σ) ∩ W π (σ) = ∅, it suffices to show that if v ∈ W δ (σ) and Υ(σ -v) is a modular partition of σ -v, then Υ(σ -v) = 2. Indeed, consider v ∈ W δ (σ) and Υ(σ -v) is a modular partition of σ -v. Since v ∈ W δ (σ), σ -v is not connected. It follows from Proposition 2.
12 that σv admits a nontrivial modular cut C. Set

P = {M ∈ Υ(σ -v) ∶ M ∩ C ≠ ∅}.
Since C is a nontrivial modular cut, C ≠ ∅, and hence, P ≠ ∅. By the first assertion of Lemma 2.10, P is a module of (σv) Υ(σv). By the second assertion of Lemma 2.10, ∪P is a module of (σv). It follows from the maximality of the elements of Υ(σv) that P = 1 or P = Υ(σv). For a contradiction, suppose that

P = Υ(σ -v). Since C is a nontrivial modular cut of σ -v, there exists M ∈ Υ(σ -v) such that M ∖ C ≠ ∅. Consider N ∈ Υ(σ -v) ∖ {M }. Since P = Υ(σ -v), N ∩ C ≠ ∅. By Assertion (M5) of Proposition 2.5, C ∪ N is a module of σ -v. Since P = Υ(σ -v), M ∩ C ≠ ∅. Thus, we have N ⊊ C ∪ N . Furthermore, since M ∖ C ≠ ∅, we obtain N ⊊ (C ∪ N ) ⊊ V (σ -v),
which contradicts the maximality of N . It follows that P = 1. Therefore, there exists M ∈ Υ(σv) such that C ⊆ M . Similarly, there exists

N ∈ Υ(σ -v) such that (V (σ -v) ∖ C) ⊆ N . Since Υ(σ -v) is a modular partition of σ -v, we obtain Υ(σ -v) = {M, N }.
Lastly, note that we can have

W ∅ (σ) ∩ W δ (σ) ≠ ∅.
Consider the 2-structure σ(T Z ) introduced in Example 8.37. We have

σ(T Z ) -∞ = σ(L Z ). Consequently, we have ∞ ∈ W ∅ (σ(T Z )) ∩ W δ (σ(T Z )).

The criticality digraph

Fact 10.13. Let σ be an infinite prime 2-structure. Consider distinct v, w ∈ V (σ). Suppose that σv admits a nontrivial module M v and σw admits a nontrivial module

M w . If w ∈ M v and v ∈ M w , then M v ∩ M w ≤ 1.
Proof. Suppose that w ∈ M v and v ∈ M w . We obtain that M v ∩ M w is a module of σ. Since σ is prime, we have M v ∩ M w ≤ 1.

Fact 10.14. Let σ be an infinite prime 2-structure. Consider distinct v, w ∈ V (σ). Suppose that σv admits a nontrivial module M v and σw admits a nontrivial module M w . If w ∈ M v and v ∈ M w , then M v ∩ M w ≠ ∅.

Proof. For a contradiction, suppose that M v ∩ M w = ∅. We verify that M w is a module of σ. Since M w is a module of σw, we have only to verify that w ←→ σ M w (see Notation 2.1). Thus, consider x, y ∈ M w . Since M v is a nontrivial module of σv, M v ≥ 2, and hence there exists

w ′ ∈ M v ∖ {w}. Since M v is a module of σ -v, we have [w, x] σ = [w ′ , x] σ and [w, y] σ = [w ′ , y] σ . Furthermore, we have [w ′ , x] σ = [w ′ , y] σ because M w is a module of σ -w.
Therefore, we obtain [w, x] σ = [w, y] σ . It follows that M w is a module of σ, which contradicts the fact that σ is prime. Consequently, M v ∩ M w ≠ ∅.

Fact 10.15. Let σ be an infinite prime 2-structure. Consider distinct u, v, w ∈ V (σ). Suppose that u, v ∈ W (σ) and (v, u) ∈ A(C(σ)). If σv admits a nontrivial module M v such that w ∈ M v and u ∈ M v , then σu admits a nontrivial module containing v and w.

Proof. Since (v, u) ∈ A(C(σ)), σu admits a nontrivial module M u containing v. We can conclude if w ∈ M u . Hence, suppose that w ∈ M u . Thus, we have

w ∈ M v ∖ M u . By Fact 10.14, M u ∩ M v ≠ ∅. Since u ∈ (M u ∪ M v )
, we obtain that M u ∪ M v is a module of σu. We distinguish the following two cases to verify that M u ∪ M v is a nontrivial module of σu.

• Suppose that M u ∖ M v ≥ 2. We show that M v ∖ M u is a module of σ.

Clearly, for every

x ∈ V (σ) ∖ (M v ∪ {v}), we have x ←→ σ M v ∖ M u (see Notation 2.1). Consider x ∈ (M u ∩ M v ) ∪ {v}. Since M u ∖ M v ≥ 2, there exists x ′ ∈ (M u ∖ M v ) ∖ v. Let y, z ∈ M v ∖ M u . Since M u is a module of σ -u, we have [x, y] σ = [x ′ , y] σ and [x, z] σ = [x ′ , z] σ . Furthermore, since M v is a module of σ -v, we have [x ′ , y] σ = [x ′ , z] σ . It follows that [x, y] σ = [x, z] σ . Thus, x ←→ σ M v ∖ M u for every x ∈ (M u ∩ M v ) ∪ {v}. Consequently, M v ∖ M u is a module of σ. Since σ is prime, M v ∖ M u is a trivial module of σ. Hence, we obtain M v ∖ M u = {w}. If M u ∪ M v is a trivial module of σ -u, then M u ∪ M v = V (σ) ∖ {u}, and hence M u = V (σ) ∖ {u, w}, which contradicts u ∈ E (σ). Therefore, M u ∪ M v is a nontrivial module of σ -u. • Suppose that M u ∖ M v ≤ 1. Since v ∈ M u ∖ M v , we obtain M u ∖ M v = {v}. If M u ∪ M v is a trivial module of σ -u, then M u ∪ M v = V (σ) ∖ {u}, and hence M v = V (σ)∖{u, v}, which contradicts v ∈ E (σ). Therefore, M u ∪M v is a nontrivial module of σ -u.
Consequently, M u ∪ M v is a nontrivial module of σu containing v and w.

The next result follows from Fact 10.15.

v ∈ W (σ). If v ∈ W ∅ (σ) ∪ W δ (σ), then (w, v) ∈ A(C(σ)) for every w ∈ V (σ) ∖ {v}.
W (σ). If v ∼ σ w and v ∈ W π (σ), then w ∈ W π (σ). Proof. Since (w, v) ∈ A(C(σ)) and v ∈ W π (σ), there exists M v ∈ Υ(σ -v) such that w ∈ M v and M v ≥ 2. Consider the set X of X ⊆ (V (σ) ∖ {v, w}) such that X ∩ M = 1 for each M ∈ Υ(σ -v).
Using the axiom of choice, we obtain

X ≠ ∅. Since v ∈ W π (σ), (σ -v) Υ(σ -v) is prime. It follows that σ[X] is prime for each X ∈ X .
Let X ∈ X . We show that

(V (σ) ∖ (X ∪ {v})) ⊆ ⋃ z∈X X σ (z) (see Notation 3.12). (10.8) Let u ∈ (V (σ) ∖ (X ∪ {v})). Since u ≠ v, there exists N v ∈ Υ(σ -v) such that u ∈ N v .
By denoting by z the unique element of N v ∩ X, we obtain u ∈ X σ (z). Hence, (10.8) holds. Consider a nontrivial module M w of σw containing v. For each X ∈ X , we show that M w ∩ X ≤ 1. (10.9)

As seen in Remark 3.16, we have M w ∩ X ≤ 1 or X ⊆ M w . It follows from Fact 10.6 and (10.8) that M w ∩ X ≤ 1. Hence, (10.9) holds. Given X ∈ X , we show that there exists y ∈ X such that v ∈ X σ (y).

(10.10)

Since (v, w) ∈ A(C(σ)), σw admits a nontrivial module M w containing v. By (10.9), M w ∩ X ≤ 1. Clearly, if M w ∩ X = 1, then (10.10) holds by denoting the unique element of M w ∩ X by y. Hence, suppose that M w ∩ X = ∅. Let u ∈ M w ∖ {v}. Since u ≠ v and u ≠ w, it follows from (10.8) that u ∈ X σ (y), where y ∈ X. We verify that v ∈ X σ (y) too. Let z ∈ X ∖ {y}. Since u ∈ X σ (y), we have z ←→ σ {y, u} (see Notation 2.1). Furthermore, since M w ∩ X = ∅, we have z ∈ M w . It follows that z ←→ σ {u, v}. Therefore, we obtain z ←→ σ {y, v}. Consequently, {y, v} is a module of σ[X ∪ {v}]. Hence, v ∈ X σ (y), so (10.10) holds. Now, we show that Υ(σ-w) ≠ ∅. Consider any nontrivial module M w of σ-w containing v. By (10.9), we have M w ∩ X ≤ 1. By (10.10), there exists y ∈ X such that v ∈ X σ (y). It follows from Remark 3.16 that M w ⊆ ({y} ∪ X σ (y)). Consequently, there exists M w ∈ Υ(σw) such that v ∈ M w .

Lastly, we prove that w ∈ W π (σ). Consider again M v ∈ Υ(σv) such that w ∈ M v and M v ≥ 2. There exists X ∈ X such that w ∈ X ∩ M v . It follows from (10.8) and (10.10) that

(V (σ) ∖ X) ⊆ ⋃ z∈X X σ (z).
Therefore, it follows from Fact 10.6 that σw is connected. Since Υ(σw) ≠ ∅, it follows from Proposition 10.5 that w ∈ W π (σ).

The next result follows from Lemma 10.20.

Corollary 10.21. Let σ be an infinite critical 2-structure. Given distinct v, w ∈ W π (σ), if v ∼ σ w, then there exists X ⊆ (V (σ) ∖ {v, w}) satisfying • σ[X] is prime; • there exist distinct y, z ∈ X such that v ∈ X σ (y) and w ∈ X σ (z); • Υ(σ -v) = {({y} ∪ X σ (y)) ∖ {v}, {z} ∪ X σ (z))} ∪ {{u} ∶ u ∈ X ∖ {y, z}}; • Υ(σ -w) = {{y} ∪ X σ (y)), ({z} ∪ X σ (z)) ∖ {w}} ∪ {{u} ∶ u ∈ X ∖ {y, z}}; • p (σ,X) = {X σ (y), X σ (z)} and E(Γ (σ,X) ) = {{v, w}}. • if M w ∩ X = 1, then M w ∩ Y = M w ∩ X; • if v ∈ M w , then M w ≥ 2 because (v, w) ∈ A(C(σ)), so that we can require that Y ∩ (M w ∖ {v}) = 1. It follows that X ⊆ Y ⊆ (V (σ) ∖ {v, w}). Since (σ -w) Υ(σ -w) is prime, σ[Y ] is prime. We show that X = Y. (10.15)
For a contradiction, suppose that there exists u ∈ Y ∖ X. There exists M u w ∈ Υ(σw) such that u ∈ M u w . Since M u w ∩ Y = {u} and u ∈ X, we have M u w ∩ X = ∅. It follows from (10.11) that M u w ⊆ X σ (t), where t ∈ X. Moreover, it follows from (10.12) and (10.13) that ({t} ∪ X σ (t)) ∖ {v} is a module of σv. By Assertion (M2) of Proposition 2.5, (10.15) holds. The analogues of (10.12) and (10.13) for Υ(σw) follow. They are proved as previously. For each t ∈ X ∖ {z}, we have ({t} ∪ X σ (t)) ∈ Υ(σw).

(({t} ∪ X σ (t)) ∖ {v}) ∩ Y is a module of σ[Y ]. Since X ⊆ Y ⊆ (V (σ) ∖ {v, w}) and M u w ⊆ X σ (t), we obtain (({t} ∪ X σ (t)) ∖ {v}) ∩ Y = ({t} ∪ X σ (t)) ∩ Y and u, t ∈ ({t} ∪ X σ (t)) ∩ Y . Furthermore, given t ′ ∈ (X ∖ {t}), we have t ′ ∈ Y ∖ ({t} ∪ X σ (t)). Therefore, ({t} ∪ X σ (t)) ∩ Y is a nontrivial module of σ[Y ], which contradicts the fact that σ[Y ] is prime. Consequently,
(10.16)

Similarly, we have

(({z} ∪ X σ (z)) ∖ {w}) ∈ Υ(σ -w). ( 10
(σ -v) = {({y} ∪ X σ (y)) ∖ {v}, {z} ∪ X σ (z))} ∪ {{u} ∶ u ∈ X ∖ {y, z}}.
Similarly, we have

Υ(σ -w) = {{y} ∪ X σ (y)), ({z} ∪ X σ (z)) ∖ {w}} ∪ {{u} ∶ u ∈ X ∖ {y, z}}.
It follows from (10.11) that p (σ,X) = {X σ (y), X σ (z)}.

Proof. To begin, suppose that there exist distinct v, w ∈ W ∅ (σ) ∖ W δ (σ). Let x ∈ V (σ) ∖ {v, w}. Since Υ(σv) = ∅, there exists a proper module M v of σv such that x, w ∈ M v . Similarly, there exists a proper module M w of σw such that x, v ∈ M w . It is easy to verify that M v ∪ M w is a module of σ. Since σ is prime, we obtain

M v ∪ M w = V (σ). (10.19) We show that M v is a nontrivial modular cut of σ -v. Recall that M v is a proper module of σ -v. By (10.19), V (σ -v) ∖ M v = (M w ∖ {v}) ∖ M v . ( 10
∖ {v}) ∖ (M v ∖ {w}), which is (M w ∖ {v}) ∖ M v , is a module of σ -{v, w}. To prove that (M w ∖ {v}) ∖ M v is a module of σ -v, it remains to verify that w ←→ σ ((M w ∖ {v}) ∖ M v ) (see Notation 2.1). Let y, z ∈ ((M w ∖ {v}) ∖ M v ). Since M v is a module of σ -v and y ≠ v, we have [w, y] σ = [x, y] σ . Similarly, we have [w, z] σ = [x, z] σ . Since (M w ∖ {v}) ∖ M v is a module of σ -{v, w}, we have [x, y] σ = [x, z] σ . It follows that [w, y] σ = [w, z] σ . Thus, (M w ∖ {v}) ∖ M v is a module of σ -v. It follows from (10.20) that M v is a modular cut of σ -v. Since M v is a proper module of σ -v containing x and w, M v is a nontrivial modular cut of σ -v. Therefore, v ∈ W δ (σ). It follows that W ∅ (σ) ∖ W δ (σ) ≤ 1.
Now, suppose that there exists v ∈ W ∅ (σ) ∖ W δ (σ). Suppose for a contradiction that W δ (σ) ≠ ∅, and consider w ∈ W δ (σ). It follows from Corollary 10.18 that v ∼ σ w, which contradicts Lemma 10.22. Consequently, we have

W δ (σ) = ∅,
σ -v) Υ(σ -v) is prime. Using the axiom of choice, consider X ⊆ V (σ) ∖ {v} such that X ∩ M = 1 for each M ∈ Υ(σ -v). We have σ[X] ≃ (σ -v) Υ(σ -v).
Hence, σ[X] is prime. Consequently, (RT) holds when X is equipotent to V (σ). Now, suppose that X is strictly subpotent to V (σ). We have Proof. First, suppose for a contradiction that there exist a sequence

V (σ) = sup{ M ∶ M ∈ Υ(σ -v)}. ( 10
(v n ) n≥0 of elements of V (σ) ∖ E (σ) and v ∞ ∈ V (σ) ∖ E (σ) such that the bijection N ∪ {∞} → {v n ∶ n ≥ 0} ∪ {v ∞ } n ≥ 0 → v n , ∞ → v ∞ is an isomorphism from L N onto C(σ)[{v n ∶ n ≥ 0} ∪ {v ∞ }]. Let n ≥ 2. We have (v n , v ∞ ), (v 0 , v n ) ∈ A(C(σ)) and (v ∞ , v n ) ∈ A(C(σ)). By Corollary 10.16, σ-v n admits a nontrivial module M n containing v 0 and v n-1 . Note that v ∞ ∈ M n because (v ∞ , v n ) ∈ A(C(σ)). Set M = ⋃ n≥2 M n . Since v 0 ∈ ⋂ n≥2 M n and {v n ∶ n ≥ 2} ⊆ M , it is not difficult to verify that M is a module of σ. Since v ∞ ∈ M n for every n ≥ 2, we have M ≠ V (σ). Moreover, since M 2 ⊆ M , we have M ≥ 2.
It follows that M is a nontrivial module of σ, which contradicts the fact that σ is prime. Consequently, C(σ) -E (σ) does not embed L N . Second, suppose for a contradiction that there exist a sequence

(v n ) n≥0 of elements of V (σ) ∖ E (σ) and v ∞ ∈ V (σ) ∖ E (σ) such that the bijection N ∪ {∞} → {v n ∶ n ≥ 0} ∪ {v ∞ } n ≥ 0 → v n , ∞ → v ∞ is an isomorphism from ( L N ) ⋆ onto C(σ)[{v n ∶ n ≥ 0}∪{v ∞ }]. Since σ is critical, σ -v ∞ admits a nontrivial module M ∞ . Let w ∈ M ∞ . We have (w, v ∞ ) ∈ A(C(σ)). Moreover, for each n ≥ 0, we have (v ∞ , v n ) ∈ A(C(σ)) and (v n , v ∞ ) ∈ A(C(σ)). By Corollary 10.16, σ -v n admits a nontrivial module M n containing v ∞ and w. Set M = ⋂ n≥2 M n . It is not difficult to verify that M is a module of σ. Since v 0 ∈ M 0 , we have v 0 ∈ M , so M ≠ V (σ). Moreover, since w, v ∞ ∈ M , we have M ≥ 2. It follows that M is a nontrivial module of σ, which contradicts the fact that σ is prime. Consequently, C(σ) -E (σ) does not embed ( L N ) ⋆ .
Remark 10.27. Let L be an infinite linear order. Suppose that L embeds neither L N nor its dual ( L N ) ⋆ . We show that L is isomorphic to L N , (L N ) ⋆ , or L Z . Indeed, using the axiom of countable choice, we obtain a countable subset W of V (L). It follows from infinite Ramsey's theorem that L[W ] embeds L N or (L N ) ⋆ . By exchanging L and L ⋆ if necessary, we can assume that L embeds L N . Hence, there exists a sequence

(v n ) n≥0 of vertices of L N → {v n ∶ n ≥ 0} n ≥ 0 → v n is an isomorphism from L N onto L[{v n ∶ n ≥ 0}]. Set V -= {v ∈ V (σ) ∖ {v n ∶ n ≥ 0} ∶ v < v 0 mod L}
and for each n ≥ 0, set

V n = {v ∈ V (σ) ∖ {v n ∶ n ≥ 0} ∶ v n < v < v n+1 mod L}. Let v ∈ V (σ) ∖ {v n ∶ n ≥ 0}. Since L does not embed L N , there exists n ≥ 0 such that v < v n mod L. Therefore, we have (V (σ) ∖ {v n ∶ n ≥ 0}) = V -∪ ( ⋃ n≥0 V n ).
For a contradiction, suppose that there exists n ≥ 0 such that V n is infinite. As previously, L[V n ] embeds L N or (L N ) ⋆ , which contradicts the fact that L embeds neither L N nor its dual ( L N ) ⋆ . Therefore, V n is finite for every n ≥ 0. Set

V + = {v n ∶ n ≥ 0} ∪ ( ⋃ n≥0 V n ). It follows that L[V + ] ≃ L N . (10.23)
Moreover, we have

V (L) = V -∪ V + . (10.24) If V -is finite, then L ≃ L N too. Hence, suppose that V -is infinite. As previ- ously, L[V -] embeds L N or (L N ) ⋆ . Since L does not embed L N , L[V -] embeds (L N ) ⋆ . Therefore, there exists a sequence (w n ) n≥0 of element of V - N → {v n ∶ n ≥ 0} n ≥ 0 → w n is an isomorphism from (L N ) ⋆ onto L[{w n ∶ n ≥ 0}]. Set W 0 = {v ∈ V -∶ w 0 < v mod L}
and for each n ≥ 1, set

W n = {v ∈ V -∶ w n < v < w n-1 mod L}.
Since L does not embed neither L N nor its dual ( L N ) ⋆ , we have Proof. It follows from Corollary 10.21 that there exists X ⊆ V (σ) ∖ {v, w} satisfying

V -= ⋃ n≥0 W n and W n is finite for each n ≥ 0. Consequently, L[V -] ≃ (L N ) ⋆ .
• σ[X] is prime;

• there exist distinct y, z ∈ X such that v ∈ X σ (y) and w ∈ X σ (z);

• Υ(σ -v) = {({y} ∪ X σ (y)) ∖ {v}, {z} ∪ X σ (z))} ∪ {{u} ∶ u ∈ X ∖ {y, z}}; • Υ(σ -w) = {{y} ∪ X σ (y)), ({z} ∪ X σ (z)) ∖ {w}} ∪ {{u} ∶ u ∈ X ∖ {y, z}}; • p (σ,X) = {X σ (y), X σ (z)} and E(Γ (σ,X) ) = {{v, w}}.
We verify that

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ({y} ∪ X σ (y)) ←→ σ (({z} ∪ X σ (z)) ∖ {w}) (see Notation 2.1) and (({y} ∪ X σ (y)) ∖ {v}) ←→ σ ({z} ∪ X σ (z)). (10.25)
Indeed, consider u ∈ X σ (y) and u ′ ∈ X σ (z). Suppose that u ≠ v or u ′ ≠ w.

Since E(Γ (σ,X) ) = {{v, w}}, we have {u, u ′ } ∈ E(Γ (σ,X) ). It follows from Assertion (P3) of Lemma 3.17 that {y, u} and {z,

u ′ } are modules of σ[X ∪ {u, u ′ }]. Therefore, we obtain [u, u ′ ] σ = [y, z] σ . Moreover, since u ∈ X σ (y), we have [u, z] σ = [y, z] σ . Similarly, we have [y, u ′ ] σ = [y, z] σ because u ′ ∈ X σ (z).
It follows that (10.25) holds. Moreover, consider W ⊆ ({y} ∪ X σ (y)) and Clearly, if X is equipotent to V (σ), then (RT) holds. Hence, suppose that X is strictly subpotent to V (σ). Consequently, {y} ∪ X σ (y) or {z} ∪ X σ (z) are equipotent to V (σ). For instance, assume that {y} ∪ X σ (y) is equipotent to V (σ).

W ′ ⊆ ({z} ∪ X σ (z)) such that v ∈ W and w ∈ W ′ . If W ≥ 2 or W ′ ≥ 2, then W ←→ σ W ′ . ( 10
We prove that X σ (z) ≥ 2. (10.27)

Otherwise, suppose that X σ (z) = {w}. We verify that z ∈ E (σ). Set In particular, we have

Y = (X ∖ {z}) ∪ {w}. Since w ∈ X σ (z), σ[Y ] is isomorphic to σ[X], so σ[Y ] is prime. Let u ′ ∈ (X σ (y) ∖ {v}). Since E(Γ (σ,X) ) = {{v, w}}, we have {w, u ′ } ∈ E(Γ (σ,X) ). It follows from Assertion (P3) of Lemma 3.17 that {y, u ′ } is a module of σ[X ∪ {u ′ , w}]. By Assertion (M2) of Proposition 2.5, {y, u ′ } is a module of σ[Y ∪{u ′ }]. Therefore, u ′ ∈ Y σ (y). It follows that (X σ (y) ∖ {v}) ⊆ Y σ (y). Since {v, w} ∈ E(Γ (σ,X) ), we obtain [v, w] σ ≠ [y, w] σ . Thus, v ∈ Y σ (y). It follows that ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ Y σ (y) = X σ (
v ←→ σ (({y} ∪ X σ (y)) ∖ {v}).
Clearly, we have t ←→ σ (({y} ∪ X σ (y)) ∖ {v})

for every t ∈ X ∖ {y, z}. Moreover, by (10.25), we have w ←→ σ (({y} ∪ X σ (y)) ∖ {v}).

It follows that ({y} ∪ X σ (y)) ∖ {v} is a module of σ, which contradicts the fact that σ is prime. Therefore, (10.29) holds. It follows that σz is uncuttable. In particular, we have z ∈ E (σ).

To obtain a contradiction when X σ (z) = {w}, we distinguish the following two cases.

1. Suppose that (v, z) ∈ A(C(σ)). Consider a nontrivial module M z of σz.

Since (v, z) ∈ A(C(σ)), it follows from (10.29) that M z ⊆ ({y} ∪ (X σ (y) ∖ {v})). Since M z ⊆ ({y} ∪ X σ (y)), we have z ←→ σ M z . It follows that M z is a nontrivial module of σ, which contradicts the fact that σ is prime.

Suppose that

(v, z) ∈ A(C(σ)). Since (z, v) ∈ A(C(σ)), we have v ∼ σ z.
By Lemma 10.20, z ∈ W π (σ). Furthermore, by Corollary 10.21, there exists Z ⊆ (V (σ) ∖ {v, z}) satisfying

• σ[Z] is prime; • there exist distinct y ′ , z ′ ∈ Z such that v ∈ Z σ (y ′ ) and z ∈ Z σ (z ′ ); • Υ(σ-v) = {({y ′ }∪Z σ (y ′ ))∖{v}, {z ′ }∪Z σ (z ′ )}∪{{u} ∶ u ∈ Z∖{y ′ , z ′ }}; • Υ(σ -z) = {{y ′ } ∪ Z σ (y ′ )), ({z ′ } ∪ Z σ (z ′ )) ∖ {z}} ∪ {{u} ∶ u ∈ Z ∖ {y ′ , z ′ }}; • p (σ,Z) = {Z σ (y ′ ), Z σ (z ′ )} and E(Γ (σ,Z) ) = {{v, z}}. Recall that Υ(σ -v) = {({y} ∪ X σ (y)) ∖ {v}, {z} ∪ X σ (z))} ∪ {{u} ∶ u ∈ X ∖ {y, z}}. Since X σ (z) = {w}, we have {z, w} ∈ Υ(σ -v). Since {y} ∪ X σ (y) is equipotent to V (σ), ({y} ∪ X σ (y)) ∖ {v} is an infinite element of Υ(σ -v). It follows that {z ′ } ∪ Z σ (z ′ ) = {z, w}. Moreover, since ({y} ∪ X σ (y)) ∖ {v} is an infinite element of Υ(σ -v) ∖ {{z ′ } ∪ Z σ (z ′ ) = {z, w}}, we have ({y} ∪ X σ (y)) ∖ {v} = ({y ′ } ∪ Z σ (y ′ )) ∖ {v}.
Therefore, we obtain Consequently, (10.27) holds. Consider u ′ ∈ X σ (z) ∖ {w}. We show that u

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ y ′ ∈ (({y} ∪ X σ (y)) ∖ {v}), v ∈ Z σ (y ′ ), and
′ ∈ W π (σ) and {{u} ∶ u ∈ {y} ∪ X σ (y)} ⊆ Υ(σ -u ′ ), (10.30)
which allows us to conclude because {y}∪X σ (y) is equipotent to V (σ). Let M u ′ be a nontrivial module of σu ′ . For a contradiction, suppose that

X ⊆ M u ′ . Since M u ′ is a nontrivial module of σ -u ′ , there exists u ∈ (V (σ) ∖ {u ′ }) ∖ M u ′ . Since p (σ,X) = {X σ (y), X σ (z)}, we get u ∈ X σ (t), where t = y or z. Set Y = (X ∖ {t}) ∪ {u}. Since {t, u} is a module of σ[X ∪ {u}], σ[Y ] is isomorphic to σ[X]. It follows that σ[Y ] is prime. By Assertion (M2) of Proposition (2.5), M u ′ ∩Y is a module of σ[Y ]. We have (X ∖ {t}) ⊆ (M u ′ ∩ Y ), so M u ′ ∩ Y ≥ 2. Furthermore, we have u ∈ (Y ∖ (M u ′ ∩ Y )). Therefore, M u ′ ∩ Y is a nontrivial module of σ[Y ], which contradicts the fact that σ[Y ] is prime. It follows that X ∖ M u ′ ≠ ∅. It follows from Remark 3.16 that we obtain M u ′ ∩ X ≤ 1. Since p (σ,X) = {X σ (y), X σ (z)}, we obtain M u ′ ⊆ ({y} ∪ X σ (y)) or M u ′ ⊆ ({z} ∪ X σ (z)).
It follows that u ′ ∈ E (σ) and σu ′ is uncuttable. By Proposition 2.12, σu ′ is connected. Moreover, for t ∈ X ∖ {y, z}, we obtain {t} ∈ Υ(σu ′ ). It follows from Proposition 10.5 that u ′ ∈ W π (σ). Finally, we establish (10.30). For a contradiction, suppose that there exists Proof. Suppose that W π (σ) ≠ ∅. To begin, we show that (RT) holds when W π (σ) is finite. Indeed, suppose that W π (σ) is finite. By Proposition 10.9,

M u ′ ∈ Υ(σ -u ′ ) such that M u ′ ⊆ ({y} ∪ X σ (y)) and M u ′ ≥ 2. Since w ∈ M u ′ , we have w ←→ σ M u ′ . Since M u ′ ≥ 2,
W (σ) is equipotent to V (σ). Thus, W (σ) ∖ W π (σ) is equipotent to V (σ). Let v ∈ W π (σ). Since (σ -v) Υ(σ -v) is prime, it suffices to verify that Υ(σ -v) is equipotent to W (σ) ∖ W π (σ). Consider w ∈ W (σ) ∖ W π (σ). By Corollary 10.18, we have (v, w) ∈ A(C(σ)). It follows from Lemma 10.20 that (w, v) ∈ A(C(σ)). Hence, {w} ∈ Υ(σ -v). Therefore, we have {{w} ∶ w ∈ W (σ) ∖ W π (σ)} ⊆ Υ(σ -v). It follows that Υ(σ -v) is equipotent to W (σ)∖W π (σ), so Υ(σ -v) is equipotent to V (σ).
Consequently, (RT) holds.

In the sequel, we suppose that W π (σ) is infinite. By Corollary 10.17 and Propositions 10.24 and 10.28, we can assume that C(σ)[W π (σ)] is a linear order. Furthermore, by Lemma 10.26, C(σ)[W π (σ)] embeds neither L N nor its dual ( L N ) ⋆ . It follows from Remark 10.27 that

C(σ)[W π (σ)] is isomorphic to L N , (L N ) ⋆ , or L Z . First, suppose that C(σ)[W π (σ)] is isomorphic to L N or L Z .
In particular, observe that W π (σ) is countable. There exists a sequence (v n ) n≥0 of elements of W π (σ) such that the function

N → {v n ∶ n ≥ 0} n ≥ 0 → v n is an isomorphism from L N onto C(σ)[{v n ∶ n ≥ 0}]. For a contradiction, suppose that W ∅ (σ) ∪ W δ (σ) ≠ ∅. Consider w ∈ W ∅ (σ) ∪ W δ (σ). Let n ≥ 0.
W ∅ (σ) ∪ W δ (σ) = ∅, so W (σ) = W π (σ).
It follows from Proposition 10.9 that

V (σ) is countable. Since v 0 ∈ W π (σ), Υ(σ -v 0 ) is a modular partition of σ -v 0 and (σ -v 0 ) Υ(σ -v 0 ) is prime. Let n ≥ 1. Since (v n , v 0 ) ∈ A(C(σ)), we have {v n } ∈ Υ(σ -v 0 ). Consider X ⊆ V (σ -v 0 ) such that X ∩ M = 1 for every M ∈ Υ(σ -v 0 ). We obtain that X is a countable proper subset of V (σ) such that σ[X] is prime. Thus, (RT) holds. Second, suppose that C(σ)[W π (σ)] is isomorphic to (L N ) ⋆ . There exists a sequence (v n ) n≥0 of elements of W π (σ) such that W π (σ) = {v n ∶ n ≥ 0} and the function N → {v n ∶ n ≥ 0} n ≥ 0 → v n is an isomorphism from (L N ) ⋆ onto C(σ)[W π (σ)]. Let w ∈ W ∅ (σ) ∪ W δ (σ).
As seen previously, we obtain that {w} ∈ Υ(σv 0 ).

Thus, W ∅ (σ) ∪ W δ (σ) is subpotent to Υ(σ -v 0 ). Suppose that W ∅ (σ) ∪ W δ (σ) is infinite. Since W π (σ) is countable, we obtain that W (σ) is equipotent to W ∅ (σ) ∪ W δ (σ). It follows from Proposition 10.9 that V (σ) is equipotent to W ∅ (σ) ∪ W δ (σ). Since W ∅ (σ) ∪ W δ (σ) is subpotent to Υ(σ -v 0 ), it follows from Bernstein-Schröder theorem that Υ(σ -v 0 ) is equipotent to V (σ). Consider X ⊆ V (σ -v 0 ) such that X ∩ M = 1 for every M ∈ Υ(σ -v 0 ). We obtain that X is proper subset of V (σ) such that X is equipotent to V (σ) and σ[X]
is prime. Therefore, (RT) holds. In the sequel of the proof, we suppose that W ∅ (σ) ∪ W δ (σ) is finite. Since W π (σ) is countable, W (σ) is countable as well. It follows from Proposition 10.9 that V (σ) is countable.

To continue, we show that for each p ≥ 0,

{v m ∶ m > p} is a module of σ[W π (σ)] -v p . ( 10.31)
Indeed, consider p ≥ 0. Let q ≥ p + 2. We have (v p+1 , v p ), (v q , v p ) ∈ A(C(σ)) and (v p , v p+1 ) ∈ A(C(σ)). By Corollary 10.16, σv p admits a nontrivial module M q p containing v p+1 and v q . Since v

p ∈ W π (σ), Υ(σ -v p ) is a modular partition of σ -v p and (σ -v p ) Υ(σ -v p ) is prime. Denote by M p the unique element of Υ(σ -v p ) containing v p+1 . It follows from the maximality of M p that M q p ⊆ M p . By Assertion (M2) of Proposition 2.5, M p ∩ ({v n ∶ n ≥ 0} ∖ {v p }) is a module of σ[{v n ∶ n ≥ 0}] -v p . We obtain {v m ∶ m > p} ⊆ (M p ∩ ({v n ∶ n ≥ 0} ∖ {v p })). Let m < p. Since (v m , v p ) ∈ A(C(σ)), we have {v m } ∈ Υ(σ -v p ), so v m ∈ M p . Hence, when p ≥ 1, we have {v 0 , . . . , v p-1 } ∩ M p = ∅. (10.32) It follows that (M p ∩ (W π (σ) ∖ {v p })) = {v m ∶ m > p}. (10.33)
Hence, (10.31) holds. Consider X p ⊆ V (σv p ) such that X p ∩ N p = 1 for every N p ∈ Υ(σv p ). We have σ[X p ] is prime. Since M p ∈ Υ(σv p ), denote by y p the unique element of X p ∩ M p . We obtain (M p ∖ {y p }) ⊆ (X p ) σ (y p ) (see Notation 3.12). (10.34) Since {v m } ∈ Υ(σv p ) for each m < p, we obtain

{v m ∶ m < p} ⊆ X p . (10.35)
As previously seen, it follows from Lemma 10.20 that

(W ∅ (σ) ∪ W δ (σ)) ⊆ X p . ( 10
∈ W (σ), that is, w ∈ E (σ). Hence, V (σ)∖{w, F E (σ) (w)} is a module of σ -w (see Notation 10.8). We have X p ≥ 3 because σ[X p ] is prime. Thus, (V (σ) ∖ {w, F E (σ) (w)}) ∩ X p ≥ 2. Since σ[X p ] is prime, we obtain X p ⊆ (V (σ) ∖ {w, F E (σ) (w)}). It follows from Fact 10.6 that (V (σ) ∖ {v p , w}) ⊆ (V (σ) ∖ {w, F E (σ) (w)}). We obtain v p = F E (σ) (w).
In particular, we obtain v p ←→ σ M p (see Notation 2.1). Hence, M p is a nontrivial module of σ, which contradicts the fact that σ is prime. Consequently, we have

(V (σ -v p ) ∖ M p ) ⊆ X p . ( 10
, v p+1 ∈ M 0 . Recall that X 0 is a subset of V (σ -v 0 ) such that X 0 ∩ N 0 = 1 for every N 0 ∈ Υ(σ -v 0 ).
Furthermore, y 0 denotes the unique element of X 0 ∩ M 0 . Hence, we can assume that y 0 = v p+1 so that y 0 ≠ v p and y 0 ≠ w. It follows from (10.34) Proof. Suppose that there exists u ∈ W δ (σ) such that λ(u) = 2. Hence, there exist distinct e, f ∈ E(σ) such that λ(u) = {e, f }. By Corollary 10.18 and Lemma 10.22, we have

that v p ∈ (X 0 ) σ (v p+1 ). Since (v p , v 0 ), (w, v p ) ∈ A(C(σ)) and (v 0 , v p ) ∈ A(C(σ)), it follows σ[M ] is prime. Indeed, let N be a module of σ[M ] such that N ≥ 2. We must show that N = M . Consider m, n ∈ N such that m < n and N ∩ {p ∈ Z ∶ m ≤ p ≤ n} = {m, n}. Suppose that m -1 ∈ M . Since [m -1, m] σ ≠ (e, f
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ W (σ) = W δ (σ) and for every v ∈ W δ (σ), λ(v) = {e, f }. (10.42) Let v ∈ W δ (σ). There exist e v , f v ∈ E(σ -v) such that σ -v is not {e v , f v }- connected. Moreover, we have ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ e v = e ∩ (V (σ -v) × V (σ -v)) and f v = f ∩ (V (σ -v) × V (σ -v)). We show that C {ev,fv} (σ -v) = 2. ( 10
v , Y v , Z v } of σ -v such that [X v , Y v ∪ Z v ] σ = (e, f ) and [Y v , Z v ] σ = (e, f ). Since σ is prime, σ is {e, f }- connected. Thus, there exists x v ∈ X v and z v ∈ Z v such that [x v , v] σ ≠ (e, f ) and [v, z v ] σ ≠ (e, f ). Therefore, for every w ∈ Y v , σ -w is {e, f }-connected. It follows from (10.42) that Y v ⊆ E (σ). Consider w ∈ Y v . Since w ∈ E (σ), there exists F E (σ) (w) ∈ V (σ) ∖ {w} (see Notation 10.8) such that F E (σ) (w) ←→ σ V (σ) ∖ {w, F E (σ) (w)} (see Notation 2.1). Suppose that Y v ≥ 2. Let w ∈ Y v . Since [X v , Y v ∖ {w}] σ = (e, f ) and [Y v ∖ {w}, Z v ] σ = (e, f ), we have F E (σ) (w) ∈ Y v . For a contradiction, suppose that F E (σ) (w) ∈ Z v . Since [X v , Z v ] σ = (e, f ) and [w, Z v ] σ = (e, f ), we obtain that V (σ) ∖ {F E (σ) (w)} is a module of σ, which contradicts the fact that σ is prime. It follows that F E (σ) (w) ∈ Z v . Similarly, we have F E (σ) (w) ∈ X v . Therefore, we obtain F E (σ) (w) = v. Since F E (σ) is injective (see Notation 10.8), we have Y v = 1. Denote by w the unique element of Y v . We have F E (σ) (w) = v. Since [v, x v ] σ ≠ (f, e) and [v, z v ] σ ≠ (e, f ), there exist e ′ , f ′ ∈ E(σ) such that [v, V (σ) ∖ {v, w}] σ = (e ′ , f ′ ) and {e ′ , f ′ } ≠ {e, f }. It follows that σ -u is {e, f }- connected for every u ∈ X v ∪ Z v . By (10.42), u ∈ W (σ) for every u ∈ X v ∪ Z v . Since Y v ⊆ E (σ)
, we obtain W (σ) = {v}, which contradicts Proposition 10.9. Consequently, (10.43) holds for each v ∈ W δ (σ).

Let v ∈ W δ (σ). By (10.43), there exists a unique nontrivial modular cut

C v of σ -v such that [C v , V (σ -v) ∖ C v ] σ = (e, f ). We define a digraph L δ (σ) on V (L δ (σ)) = W δ (σ) as follows. Given distinct v, w ∈ W δ (σ), (v, w) ∈ A(L δ (σ)) if (C v ∪ {v}) ⊆ C w .
We verify that L δ (σ) is a linear order. Clearly, L δ (σ) is a partial order. Consider distinct v, w ∈ W δ (σ). We prove that

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ w ∈ C v if and only if (w, v) ∈ A(L δ (σ)) and v ∈ (V (σ -w) ∖ C w ) if and only if (w, v) ∈ A(L δ (σ)). (10.44) Clearly, if (w, v) ∈ A(L δ (σ)), then (C w ∪ {w}) ⊆ C v , so w ∈ C v . Conversely, suppose that w ∈ C v . Since σ is prime, σ is {e, f }-connected. Thus, there exists x v ∈ C v and y v ∈ V (σ -v) ∖ C v such that [x v , v] σ ≠ (e, f ) and [v, y v ] σ ≠ (e, f ).
We distinguish the following two cases.

• Suppose that there exists

t v ∈ C v ∖ {w} such that [t v , v] σ ≠ (e, f ). We obtain that (V (σ -v) ∖ C v ) ∪ {v, t v } is {e, f }-connected. Therefore, we have (V (σ -v) ∖ C v )) ∪ {v, t v } ⊆ D w , where D w = C w or V (σ -w) ∖ C w . Let u ∈ (V (σ -w) ∖ D w ). Since [u, V (σ -v) ∖ C v ] σ = (e, f ), we obtain D w = V (σ -w) ∖ C w , and hence (V (σ) ∖ C v ) ⊆ (V (σ -w) ∖ C w ). It follows that (w, v) ∈ A(L δ (σ)). • Suppose that w = x v and [C v ∖ {w}, v] σ = (e, f ). Since w ∈ E (σ), there exists z v ∈ (V (σ) ∖ C v ) such that [z v , v] σ ≠ (e, f ). Since [v, y v ] σ ≠ (e, f ), we obtain that (V (σ-v)∖C v )∪{v}, which is V (σ)∖C v , is {e, f }-connected. Therefore, we have (V (σ) ∖ C v ) ⊆ D w , where D w = C w or V (σ -w) ∖ C w . Let u ∈ V (σ -w) ∖ D w . Since [u, V (σ -v) ∖ C v ] σ = (e, f ), we obtain D w = V (σ-w)∖C w . Thus, we have (C w ∪{w}) ⊆ C v , so (w, v) ∈ A(L δ (σ)).
It follows that w ∈ C v if and only if (w, v) ∈ A(L δ (σ)). We show similarly that v ∈ (V (σw) ∖ C w ) if and only if (w, v) ∈ A(L δ (σ)). Hence, (10.44) holds. It follows that (w, v) ∈ A(L δ (σ)) or (v, w) ∈ A(L δ (σ)). Consequently, L δ (σ) is a linear order. Now, we prove that L δ (σ) does not embed L N . Otherwise, there exist a sequence (v n ) n≥0 of elements of W (σ) and v ∞ ∈ W (σ) such that the bijection

N ∪ {∞} → {v n ∶ n ≥ 0} ∪ {v ∞ } n ≥ 0 → v n , ∞ → v ∞ is an isomorphism from L N onto L δ (σ)[{v n ∶ n ≥ 0} ∪ {v ∞ }]. For each n ≥ 0, we have ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (C vn ∪ {v n }) ⊆ C vn+1 and (C vn+1 ∪ {v n+1 }) ⊆ C v∞ . (10.45)
Set C = ⋃ n≥0 C vn .
It follows from (10.45) that C is a nontrivial module of σ, which contradicts the fact that σ is prime. Consequently, L δ (σ) does not embed L N .

To pursue, we prove that L δ (σ) does not embed ( L N ) ⋆ . Otherwise, there exist a sequence (v n ) n≥0 of elements of W (σ) and v ∞ ∈ W (σ) such that the bijection

N ∪ {∞} → {v n ∶ n ≥ 0} ∪ {v ∞ } n ≥ 0 → v n , ∞ → v ∞ is an isomorphism from ( L N ) ⋆ onto L δ (σ)[{v n ∶ n ≥ 0} ∪ {v ∞ }]. For each n ≥ 0, we have ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (C vn+1 ∪ {v n+1 }) ⊆ C vn and (C v∞ ∪ {v ∞ }) ⊆ C vn . (10.46) Set C = ⋂ n≥0 C vn .
It follows from (10.46) that C is a nontrivial module of σ, which contradicts the fact that σ is prime. Consequently, L δ (σ) does not embed ( L N ) ⋆ . It follows that L δ (σ) embeds neither L N nor its dual ( L N ) ⋆ . By (10.42), we have W (σ) = W δ (σ). It follows from Proposition 10.9 that L δ (σ) is an infinite linear order. By Remark 10.27, L δ (σ) is isomorphic to L N , (L N ) ⋆ , or L Z . Given v, w ∈ W δ (σ) such that v < L δ (σ) w, we prove that For a contradiction, suppose that E (σ) ≠ ∅. As seen at the end of the proof of Proposition 10.9, {E 0 (σ), E 1 (σ), E 2 (σ)} is a partition of E (σ) and E 2 (σ) ≤ E 1 (σ) ≤ E 0 (σ) . It follows that E 0 (σ) ≠ ∅. Hence, there exists u ∈ E (σ) such that F E (σ) (u) ∈ E (σ), so F E (σ) (u) ∈ W (σ). Since L δ (σ) is isomorphic to L N , (L N ) ⋆ , or L Z , there exist v, w ∈ W (σ) such that Lastly, suppose for a contradiction that L δ (σ) is isomorphic to L N . There exists a sequence (v n ) n≥0 of elements of V (σ) such that There exists e v ∈ E(σv) such that σv is not {e v }-connected. Moreover, we have e v = e ∩ (V (σv) × V (σv)). For each w ∈ V (σ) ∖ {v}, we denote by C w v the unique element of C {ev} (σv) containing w. We prove that E (σ) = ∅. Otherwise, as observed in the proof of Proposition 10.32, there exists u ∈ E (σ) such that F E (σ) (u) ∈ W (σ). For convenience, set v = F E (σ) (u).

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ w < L δ (σ) v < L δ (σ) F E (σ) (u)
N → {v n ∶ n ≥ 0} n → v
Let D v ∈ (C {ev} (σv) ∖ {C u v }). Since σ is {e}-connected, there exists w ∈ D v such that ≺ v, w ≻ σ ≠ {e}. Since v ←→ σ V (σ) ∖ {u, v} (see Notation 10.8), we obtain ≺v, t≻ σ ≠ {e} for each t ∈ V (σ) ∖ {u, v}. Hence, for each t ∈ V (σv) ∖ C u v , σt is {e t }-connected, where e t = e ∩ (V (σt) × V (σt)). It follows from (10.48) that t ∈ E (σ). Thus, we have W (σ) ⊆ (C u v ∪ {v}). It follows from Proposition 10.9 that W (σ) ∖ {v} is infinite. Let w ∈ (W (σ) ∖ {v}). Recall that ≺v, t≻ σ ≠ {e} for each t ∈ V (σ) ∖ {u, v}. Hence, σ -{u, w} is {e {u,w} }-connected, where e {u,w} = e ∩ (V (σ -{u, w}) × V (σ -{u, w})). Since σw is not {e w }connected, we obtain [u, V (σ -{u, w})] σ = (e, e), which contradicts the fact that w ∈ E (σ). Consequently, we have E (σ) = ∅, so V (σ) = W (σ). We conclude as follows. Consider distinct v, w ∈ V (σ). By (10.50), we have Proof of Theorem 10.1. Let σ be an infinite prime 2-structures. Clearly, if σ is not critical, then (RT) holds. Hence, suppose that σ is critical. By Proposition 10.9, V (σ) and W (σ) are equipotent. By Proposition 10.29, if W π (σ) ≠ ∅, then (RT) holds. Thus, suppose that W π (σ) = ∅.

C w v ∪ C v w = V (σ
By Observation 10.12, we have

W (σ) = W ∅ (σ) ∪ W δ (σ).
Since W ∅ (σ) ∪ W δ (σ) is infinite, it follows from Lemma 10.23 that W ∅ (σ) ⊆ W δ (σ), so W (σ) = W δ (σ).

Finally, it follows from Propositions 10.32 and 10.33 that (RT) holds.

Given v ∈ S (σ) ∖ S c (σ), if f (v) ∈ S c (σv), then σv is prime, and neither critical nor almost critical. Lastly, suppose that S (σv) = S c (σv) = {f (v)} for every v ∈ S (σ) ∖ S c (σ). and there exists an isomorphism ϕ v from (σv)f (v) onto an element τ v of R 2n satisfying (5.7).

Observation A.3. Recall that if (A.1) or (A.2) does not hold, then we can conclude as above. In the sequel, we suppose that (A.1) and (A.2) hold. We establish the new claims below in order to finally obtain a contradiction.

Claim A.4. We have f ∶ S (σ) ∖ S c (σ) → V (σ) ∖ S (σ).

Proof. Otherwise, there exists v ∈ S (σ) ∖ S c (σ) such that f (v) ∈ S (σ). Since σ-{v, f (v)} is prime, we have f (v) ∈ S (σ)∖S c (σ). It follows that (f ○f )(v) = v.

As seen in the proof of Theorem 5.13, ϕ v and ϕ f (v) are isomorphisms from P(σ -{v, f (v)}) onto P 2n . We obtain that ϕ f (v) ○ (ϕ v ) -1 is an automorphism of P 2n . We have Aut(P 2n ) = {Id {0,...,2n-1} , π 2n } (see Notation 4.21).

(A.4)

It follows that ϕ f (v) = ϕ v or π 2n ○ ϕ v .

Recall that (5.7) holds for ϕ v and ϕ f (v) . Therefore, if ϕ f (v) = ϕ v , then {v, f (v)} is a module of σ, which contradicts the fact that σ is prime. Suppose that ϕ f (v) = π 2n ○ ϕ v . Since (5.7) holds for ϕ f (v) , we have [v, (ϕ f (v) ) -1 ({2i ∶ i ∈ {0, . . . , n -1}})] σ = [(ϕ f (v) ) -1 (0), (ϕ f (v) ) -1 (2)] σ .

Since ϕ f (v) = π 2n ○ ϕ v , we obtain [v, (ϕ v ) -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ = [(ϕ v ) -1 (2n -1), (ϕ v ) -1 (2n -3)] σ .

Since τ v is critical, with P(τ v ) = P 2n , it follows from Proposition 4.15 that

[(ϕ v ) -1 (2n -1), (ϕ v ) -1 (2n -3)] σ = [(ϕ v ) -1 (2), (ϕ v ) -1 (0)] σ .
Therefore, we obtain Consequently, {v, f (v)} is a module of σ, which contradicts the fact that σ is prime.

Claim A.5. The function f is injective.

Proof. Otherwise, there exist distinct v, w ∈ S (σ) ∖ S c (σ) such that f (v) = f (w). By Claim A.4, f (v) ∈ V (σ) ∖ S (σ). Since σ -{v, f (v)} and σ -{w, f (v)} are prime, it follows from Lemma 4.4 that {v, w} is a module of σf (v). Therefore, the bijection

ψ ∶ V (σ) ∖ {w, f (v)} → V (σ) ∖ {v, f (v)} v → w x ∈ V (σ) ∖ {v, w, f (v)} → x,
is an isomorphism from (σw)f (v) onto (σv)f (v). Thus, we can choose ϕ v ○ ψ for ϕ w . As shown in the proof of Theorem 5.13, ϕ v and ϕ v ○ ψ satisfy (5.7). Since w ∈ S (σ) and f (v) ∈ V (σ) ∖ S (σ), there exists p ∈ {0, . . . , 2n -1} such that ϕ v (w) = p. Observe that π 2n ○ ϕ v (see (A.4)) is also an isomorphism from (σv)f (v) onto (τ v ) ⋆ , with (τ v ) ⋆ ∈ R 2n , satisfying (5.7). Therefore, we can assume that w = (ϕ v ) -1 (2t),

where t ∈ {0, . . . , n -1}. Since ϕ v satisfies (5.7), we obtain [f (v), w] σ = [(ϕ v ) -1 (0), (ϕ v ) -1 (2)] σ .

Since n ≥ 3 by (A.3), there exist k, l ∈ {0, . . . , n -1} such that k < l and p ∈ {2i ∶ i ∈ {0, . . . , n -1}} ∖ {2k, 2l}. Since ϕ v satisfies (5.8), we obtain

[f (v), w] σ = [(ϕ v ) -1 (2k), (ϕ v ) -1 (2l)] σ .
Since ϕ w , that is, ϕ v ○ ψ satisfies (5.7), we obtain [f (v), (ϕ v ○ ψ) -1 ({2i ∶ i ∈ {0, . . . , n -1}})] σ = [(ϕ v ○ ψ) -1 (0), (ϕ v ○ ψ) -1 (2)] σ . Furthermore, since ϕ v ○ ψ satisfies (5.8), we obtain [(ϕ v ○ ψ) -1 (0), (ϕ v ○ ψ) -1 (2)] σ = [(ϕ v ○ ψ) -1 (2k), (ϕ v ○ ψ) -1 (2l)] σ . Since p ∈ {2i ∶ i ∈ {0, . . . , n -1}} ∖ {2k, 2l}, we obtain (ϕ v ) -1 (2k), (ϕ v ) -1 (2l) ∈ V (σ) ∖ {v, f (v), w}.

Thus, ψ -1 ((ϕ v ) -1 (2k)) = (ϕ v ) -1 (2k) and ψ -1 ((ϕ v ) -1 (2l)) = (ϕ v ) -1 (2l). It follows that [f (v), (ϕ v ○ψ) -1 ({2i ∶ i ∈ {0, . . . , n-1}})] σ = [(ϕ v ) -1 (2k), (ϕ v ) -1 (2l)] σ . and [f (v), (ϕ v ) -1 (2q + 1)] σ = [(ϕ v ) -1 (2), (ϕ v ) -1 (0)] σ .

Since τ v ∈ R 2n , we have (0, 2) τv ≠ (2, 0) τv by Remark 5.12. Hence, we have [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF][START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τv ≠ [2, 0] τv . Since ϕ v is an isomorphism from (σv)f (v) onto τ v , we obtain

[(ϕ v ) -1 (0), (ϕ v ) -1 (2)] σ ≠ [(ϕ v ) -1 (2), (ϕ v ) -1 (0)] σ . (A.6)
It follows that

[f (v), (ϕ v ) -1 (2q)] σ ≠ [f (v), (ϕ v ) -1 (2q + 1)] σ ,
which contradicts (A.5).

2. Suppose that d P(σ) (w) = 2. Since f (v) ∈ N P(σ) (w), there exists u ∈ V (σ) ∖ {f (v), w} such that N P(σ) (w) = {u, f (v)}.

Since S (σv) = {f (v)} and σ -{u, w} is prime, we obtain u ≠ v. Therefore, we have u = (ϕ v ) -1 (i),

where i ∈ {0, . . . , 2n -1} ∖ {2p}. By Lemma 4.4, {f (v), (ϕ v ) -1 (i)} is a module of σw, that is, {f (v), (ϕ v ) -1 (i)} is a module of σ -(ϕ v ) -1 (2p).

(A.7) Since (5.7) is satisfied by ϕ v , we have ≺f (v), (ϕ v ) -1 (j)≻ σ =≺(ϕ v ) -1 (0), (ϕ v ) -1 (2)≻ σ for every j ∈ {0, . . . , 2n -1}. It follows that ≺(ϕ v ) -1 (i), (ϕ v ) -1 (j)≻ σ =≺(ϕ v ) -1 (0), (ϕ v ) -1 (2)≻ σ for every j ∈ {0, . . . , 2n -1} ∖ {2p, i}. Since ϕ v is an isomorphism from (σv)f (v) onto τ v , we have ≺i, j ≻ τv =≺0, 2≻ τv for every j ∈ {0, . . . , 2n -1} ∖ {2p, i}. Since τ v ∈ R 2n , it follows from (4.4) that ≺ i, 2p ≻ τv =≺ 0, 1 ≻ τv . We obtain p = 0 and i = 1. It follows from (A. Since [(ϕ v ) -1 (0), (ϕ v ) -1 (2)] σ ≠ [(ϕ v ) -1 (2), (ϕ v ) -1 (0)] σ by (A.6), we obtain f (v) ∈ ⟨X⟩ σ . Since σ is prime, it follows from (A.9) that

[v, V (σ) ∖ {v, f (v)}] σ ≠ [v, f (v)] σ . (A.11)
Since ϕ v is an isomorphism from P(σ -{v, f (v)}) onto P 2n , we obtain that σ[X] -{(ϕ v ) -1 (0), (ϕ v ) -1 (1)} is prime. Set Y = X ∖ {(ϕ v ) -1 (0), (ϕ v ) -1 (1)}.

Since v ∈ ⟨X⟩ σ , we have v ∈ ⟨Y ⟩ σ . As previously, since (5.7) is satisfied by ϕ v , it follows from (A.6) that f (v) ∈ ⟨Y ⟩ σ . Since (A.11) holds, it follows from Statements (P1) and (P2) of Lemma 3.17 that σ[Y ∪ {v, f (v)}] = σ -{(ϕ v ) -1 (0), (ϕ v ) -1 (1)} is prime. Hence, (ϕ v ) -1 (1) ∈ N P(σ) ((ϕ v ) -1 (0)).

For a contradiction, suppose that (ϕ v ) -1 (0) ∈ S (σ). Since σ -{(ϕ v ) -1 (0), (ϕ v ) -1 (1)} is prime, we obtain (ϕ v ) -1 (0) ∈ S (σ) ∖ S c (σ), which contradicts Claim A.7. It follows that (ϕ v ) -1 (0) ∈ S (σ).

Since (ϕ v ) -1 (1) ∈ N P(σ) ((ϕ v ) -1 (0)), it follows from Lemma 4.4 that

d P(σ) ((ϕ v ) -1 (0)) = 1 or 2.
For a contradiction, suppose that d P(σ) ((ϕ v ) -1 (0)) = 2. There exists w ∈ V (σ) ∖ {(ϕ v ) -1 (0), (ϕ v ) -1 (1)} such that N P(σ) ((ϕ v ) -1 (0)) = {(ϕ v ) -1 (1), w}.

Since N P(σ) (f (v)) = {v} by Claim A.6, we have (ϕ v ) -1 (0) ∈ N P(σ) (f (v)). Thus w ≠ f (v). Furthermore, since S (σv) = {f (v)}, we have (ϕ v ) -1 (0) ∈ S (σv), and hence (ϕ v ) -1 (0) ∈ N P(σ) (v). Therefore, w ≠ v. It follows that w ∈ Y . Since {(ϕ v ) -1 (1), w} is a module of σ -(ϕ v ) -1 (0) by Lemma 4.4, we obtain (ϕ v ) -1 (1) ∈ Y σ (w). But, since ϕ v is an isomorphism from P(σ -{v, f (v)}) onto P 2n , we have N P(σ-{v,f (v)}) ((ϕ v ) -1 (0)) = {(ϕ v ) -1 (1)}. Since σ -{v, f (v)} is critical, it follows from Lemma 4.4 applied to σ -{v, f (v)} that (V (σ) ∖ {v, f (v)}) ∖ {(ϕ v ) -1 (0), (ϕ v ) -1 (1)} is a module of (σ -{v, f (v)}) -(ϕ v ) -1 (0). particular, we obtain that v(σ) is even. In another vein, it follows from Corollary 5.5 (and Remark 5.6) that there exists e ∈ E(P(σ-v))∩E(P(σ)). Since e, {w, w ′ } ∈ E(P(σ-v)), it follows from Corollary 4.8 that (σ-v)-e ≃ (σv) -{w, w ′ }. Therefore, τ embeds into (σv)e, and hence τ embeds into σe. Since e ∈ E(P(σ)), σe is prime. Furthermore, since e ∈ E(P(σv)), (σe)v is prime. Thus, σe is not critical. Lastly, since v(σe) is even, it follows from Theorem 5.13 that σe is not almost critical. To conclude, it suffices to apply the induction hypothesis to σe.

2. Suppose that σv is almost critical. There exists w ∈ V (σv) such that

S (σ -v) = S c (σ -v) = {w}.
It follows from Theorem 5.13 that v(σ) = 2n + 2, where n ≥ 3, and there exists an isomorphism ϕ from (σ-v)-w onto an element ρ of R 2n satisfying (5.7).

We can conclude when w ∈ X. Indeed, suppose that X ⊆ V (σ) ∖ {v, w}. It follows from the first statement of Fact 5.18 that w ∈ Ext σ (X). In what follows, we suppose that w ∈ X.

First, suppose that x ≠ w.

Since P(ρ) = P 2n , ϕ is an isomorphism from P((σv)w) onto P 2n . Since x ≠ w, there exists y ∈ V (σ-v)∖{x, w} such that {x, y} ∈ E(P((σ-v)-w)).

As observed in Remark 5.16, we have P((σv)w) = P(σv)w. Thus {x, y} ∈ E(P(σv)), so (σ-v)-{x, y} is prime. Furthermore, since X ⊊ V (σ-v), it follows from Corollary 3.21 that there exist u, u ′ ∈ V (σ-v)∖X such that (σ-v)-{u, u ′ } is prime. Since S (σv) = {w} and w ∈ X, we obtain u ≠ u ′ . Hence {u, u ′ } ∈ E(P(σv)).

It follows from the second statement of Fact 5.18 that (σv) -{x, y} ≃ (σ -v)-{u, u ′ }. It follows that τ embeds into (σ -v)-{x, y}. Therefore, τ embeds into σ-x. To conclude, it suffices to apply the induction hypothesis to σx.

Second, suppose that x = w. Consequently, σ -e is neither critical nor almost critical. Moreover, it follows from (A.15) that τ embeds into σe. To conclude, it suffices to apply the induction hypothesis to σe.

Since ϕ(2n -2) = 2n -2 and ϕ(2n) = 2n, we obtain that {2n -1, 2n + 1}, that is, {s, t} is a module of τ , which contradicts the fact that τ is prime. It follows that d P(τ -s) (t) ≠ 1. Second, suppose that d P(τ -s) (t) = 2. Since d P(τ -s) (t) ≠ 1, t ≠ 2n -1. Hence, n ≤ t ≤ 2n -2. Similarly, by setting j = ϕ(2n + 1), we have n ≤ j ≤ 2n -2. Recall that N P(τ -(2n+1)) (t) = N P(τ -t) (2n + 1) by Corollary 5.25. It follows that {ϕ -1 (j -1), ϕ -1 (j + 1)} = {t -1, t + 1}.

It follows from (B.2) that ϕ -1 (j -1) = t -1 and ϕ -1 (j + 1) = t + 1.

Since {t -2, t -1}, {t -1, t}, {t, t + 1} ∈ E(P(τ -(2n + 1))), it follows from Lemma 4.39 that (τ -(2n + 1)) -{t, t + 1} is critical, and Similarly, we obtain (τt) -{2n + 1, ϕ -1 (j -1)} is critical, and E(P((τt) -{2n + 1, ϕ -1 (j -1)})) = {{ϕ -1 (k), ϕ -1 (k + 1)} ∶ k ∈ {0, . . . , j -3} ∪ {j + 1, . . . , 2n -2}} (B.4)

∪ {{ϕ -1 (j -2), ϕ -1 (j + 1)}}.

Since ϕ -1 (j -1) = t -1, we have (τ -(2n + 1)) -{t -1, t} = (τt) -{2n + 1, ϕ -1 (j -1)}.

Set µ = τ -{t, t + 1, 2n + 1}.

It follows from (B.3) that

N P(µ) (t + 1) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ {t -2} if t = 2n -2 {t -2, t + 2} if t < 2n -2. (B.5)
Similarly, it follows from (B.4) that N P(µ) (ϕ -1 (j + 1)) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ {ϕ -1 (j -2)} if j = 2n -2 {ϕ -1 (j -2), ϕ -1 (j + 2)} if j < 2n -2. (B.6)

Since ϕ -1 (j + 1) = t + 1, we obtain t = 2n -2 if and only if j = 2n -2.

To begin, suppose that t = 2n -2 and j = 2n -2. By (B.5) and (B.6), we have ϕ -1 (2n -4) = 2n -4. By proceeding by induction, it follows from (B.3) and (B.4) that ϕ -1 (l) = l for every l ∈ {0, . . . , 2n -4}. Since ϕ -1 (j -1) = t -1 and ϕ -1 (j + 1) = t + 1, we obtain ϕ -1 (l) = l for every l ∈ {0, . . . , 2n -3} ∪ {2n -1}. Recall that τ -(2n + 1) = σ(T 2n+1 ), and ϕ is an isomorphism from τt onto σ(T 2n+1 ). Since ϕ(2n + 1) = 2n -2, we obtain ϕ -1 (2n) = 2n. It follows that {2n -2, 2n + 1} is a module of τ , which contradicts the fact that τ is prime. Now, suppose that t < 2n -2 and j < 2n -2. By (B.5) and (B.6), we have {ϕ -1 (j -2), ϕ -1 (j + 2)} = {t -2, t + 2}. It follows from (B.2) that ϕ -1 (j -2) = t -2 and ϕ -1 (j + 2) = t + 2.

It follows from (B.3) that N P(µ) (t -2) = {t -3, t + 1}.

Similarly, it follows from (B.4) that N P(µ) (ϕ -1 (j -2)) = {ϕ -1 (j -3), ϕ -1 (j + 1)}.

Since ϕ -1 (j + 1) = t + 1 and ϕ -1 (j -2) = t -2, we obtain ϕ -1 (j -3) = t -3. By proceeding by induction, we obtain ϕ -1 (jk) = tk for every k ∈ {2, . . . , min(t, j)}. For instance, suppose that t ≤ j. We obtain ϕ -1 (jt) = 0. Since d P(µ) (0) = 1, we have d P(µ) (ϕ -1 (jt)) = 1. Hence, jt = 0 or 2n -1. Since j < 2n -2, we obtain j = t. Thus, we have ϕ -1 (l) = l for every l ∈ {0, . . . , t -2}. Similarly, we obtain ϕ -1 (l) = l for every l ∈ {t + 1, . . . , 2n -1}. Since ϕ -1 (j -1) = t -1 and j = t, we obtain ϕ -1 (l) = l for every l ∈ {0, . . . , 2n -1} ∖ {t}. Recall that τ -(2n + 1) = σ(T 2n+1 ), and ϕ is an isomorphism from τt onto σ(T 2n+1 ). We obtain ϕ -1 (2n) = 2n. It follows that {t, 2n + 1} is a module of τ , which contradicts the fact that τ is prime.

Similarly, we have [p, 2n + 1] σ = [p, 2n] σ when p is odd. It follows that {2n, 2n + 1} is a module of σ, which contradicts the fact that σ is prime. It follows that (0, 1) σ = (1, 0) σ , which contradicts (B.7).

Consequently, (B.12) does not hold. Therefore, d P(σ-(2n+1)) (t) = 2. Since n ≤ t ≤ 2n, we have n ≤ t ≤ 2n -1. Set j = ϕ(2n + 1). By (B.9), N P(σ-(2n+1)) (t) = N P(σ-t) (2n + 1). Hence, n ≤ j ≤ 2n -1 and {ϕ -1 (j -1), ϕ -1 (j + 1)} = {t -1, t + 1}.

It follows that

ϕ -1 (j -1) = t -1 and ϕ -1 (j + 1) = t + 1 (B.15) or ϕ -1 (j -1) = t + 1 and ϕ -1 (j + 1) = t -1. (B.16)

Suppose that (B.15) holds. We prove that have min(t, 2nj) = 2nj. For k = 2nj, we obtain ϕ -1 (2n) = t -2n + j. Therefore, t -2n + j = 0 or 2n. Since t ≤ 2n -2 and j ≤ 2n -2, we have t + j = 2n. Since t ≥ n and j ≥ n, we obtain t = j and t = n.

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ t = j, t = n, n is
It follows that for k ∈ {2, . . . , n}, we have

⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ϕ -1 (n + k) = n -k and (similarly) ϕ -1 (n -k) = n + k. (B.22)
For a contradiction with (B.17), suppose that n is even. We obtain Clearly, ψ is another isomorphism from P(σt) onto P 2n+1 . As previously for ϕ, we obtain that for any p, q ∈ {0, . . . , 2n} such that p < q, [ψ -1 (p), ψ -1 (q)] σ = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [ψ -1 (0), ψ -1 (2)] σ if p and q are even, [ψ -1 (0), ψ -1 (1)] σ otherwise. [p, q] σ = [ψ -1 (p), ψ -1 (q)] σ for any p, q ∈ {0, . . . , 2n} such that p < q. Therefore, ψ -1 is an isomorphism from σ -(2n + 1) onto σt. If j = 2n -1, then it follows from (B.20) that N P(µ) (ϕ -1 (j + 1)) = {ϕ -1 (j -2)}.

Therefore, we have j < 2n -1.

It follows from (B.19) that

N P(µ) (ϕ -1 (j + 1)) = {ϕ -1 (j -2), ϕ -1 (j + 2)}.

Therefore, we have for each k ∈ {1, . . . , min(t, 2nj)}. Since t, j ∈ {n, . . . , 2n}, we have min(t, 2n-j) = 2n-j. Thus, for k = 2n-j, we obtain ϕ -1 (2n) = t+j -2n. It follows that t + j = 2n or 4n. Since n ≤ t < 2n -1 and n ≤ j ≤ 2n, we obtain t + j = 2n, and hence, t = n and j = n. Therefore, for each l ∈ {0, . . . , n -1}, we have ϕ -1 (2nl) = l. Symmetrically, we obtain ϕ -1 (l) = 2nl for l ∈ {0, . . . , n -1}. It follows that for each p ∈ {0, . . . , n -1} ∪ {n + 1, . . . , 2n}, we have ϕ -1 (2np) = p. (B.33)

Set ψ = π 2n+1 ○ ϕ.
Clearly, ψ is another isomorphism from P(σt) onto P 2n+1 . By Proposition 4.27, for any p, q ∈ {0, . . . , 2n} such that p < q, we have To conclude, we verify that S c (σ) = {t, 2n + 1}. As shown above, (5.24) holds. For a contradiction, suppose that there exists t ′ ∈ S c (σ) ∖ {t, 2n + 1}. By what precedes, (5.24) holds also when t is replaced by t ′ . It follows that {t, t ′ } is a module of σ -(2n + 1), which contradicts the fact that σ -(2n + 1) is prime.

Remark B.1. Let σ be a prime 2-structure with v(σ) ≥ 7. Suppose that there exist distinct s, t ∈ S c (σ). Suppose also that P(σs) ≃ P 2n+1 . By Proposition 5. Suppose that there exists t ∈ S c (σ) ∖ {s}. We verify that we can assume that (5.25) holds. Let ϕ s be an isomorphism from P(σs) onto P 2n . Since π 2n ∈ Aut(P 2n ) (see Notation 4.21), we can assume that n ≤ ϕ s (t) ≤ 2n -1.

Denote by τ s the unique 2-structure defined on {0, . . . , 2n -1} such that ϕ s is an isomorphism from σs onto τ s . Since ϕ s is an isomorphism from P(σs) onto P 2n , τ s is critical and P(τ s ) = P 2n . Here, we can assume that V (σ) = {0, . . . , 2n}, s = 2n, and ϕ s = Id {0,...,2n-1} . Thus, we have t ∈ {n, . . . , 2n -1} and P(σ -(2n)) = P 2n , so (5.25) holds. Furthermore, note that τ s = σ -(2n Moreover, for any p, q ∈ {0, . . . , 2n -1} such that p < q, we have Moreover, set ψ = π 2n ○ ϕ.

Clearly, ψ is another isomorphism from P(σt) onto P 2n . As previously for ϕ, we obtain that for any p, q ∈ {0, . . . , 2n} such that p < q, [ψ -1 (p), ψ -1 (q)] σ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [ψ -1 (0), ψ -1 (1)] σ if p is even and q is odd, [ψ -1 (p), ψ -1 (q)] σ = [p, q] σ for any p, q ∈ {0, . . . , 2n-1} such that p < q. Consequently, ψ -1 is an isomorphism from σ -(2n) onto σ -(2n -1). Hence, (5.26) holds.

To continue with situation 2, suppose that

d P(σ-(2n)) (t) = 2.
Since n ≤ t ≤ 2n -1, we have n ≤ t ≤ 2n -2. Thus, N P(σ-(2n)) (t) = {t -1, t + 1}. Set j = ϕ(2n).

Recall that n ≤ j ≤ 2n -1. By (B.40), N P(σ-t) (2n) = {t -1, t + 1}. Hence, n ≤ j ≤ 2n -2 and {ϕ -1 (j -1), ϕ -1 (j + 1)} = {t -1, t + 1}.

It follows that ϕ -1 (j -1) = t -1 and ϕ -1 (j + 1) = t + 1 (B.50) or ϕ -1 (j -1) = t + 1 and ϕ -1 (j + 1) = t -1. ∪ {{ϕ -1 t (j -2), ϕ -1 (j + 1)}}. Since (B.50) holds, we have E(P(σ -{t -1, t, 2n})) = E(P(σ -{t, ϕ -1 (j -1), 2n}).

(B.54)

We distinguish the following two cases. Both lead us to a contradiction.

1. Suppose that ϕ -1 (j -2) = t -2. By proceeding by induction, we obtain ϕ -1 (jk) = tk for k ∈ {2, . . . , min(j, t)}. It follows that j = t. Similarly, we obtain that ϕ -1 (l) = l for k ∈ {t + 2, . . . , 2n -1}. Since (B.50) holds, we obtain (ϕ t ) -1 (l) = l for l ∈ {0, . . . , 2n -1} ∖ {t}. It follows from (B.39) and (B.42) that {t, 2n} is a module of σ, which contradicts the fact that σ is prime.

2. Suppose that ϕ -1 (j -2) ≠ t -2. Since ϕ -1 (j + 1) = t + 1, it follows from (B.52), (B.53), and (B.54) that t ≤ 2n -3, j ≤ 2n -3, and ϕ -1 (j -2) = t + 2 and ϕ -1 (j + 2) = t -2.

By proceeding by induction, we obtain ϕ -1 (jk) = t + k for k ∈ {2, . . . , min(j, 2nt -1)}. Since t, j ∈ {n, . . . , 2n -3}, we have min(j, 2nt -1) = 2nt -1.

For k = 2nt -1, we obtain j + t -2n + 1 = 0 or 2n -1, which is impossible because j, t ∈ {n, . . . , 2n -3}.

Consequently, (B.50) does not hold. Therefore, (B.51) holds. Recall that E(P(σ -{t, ϕ -1 (j -1), 2n}))

= {{ϕ -1 (k), ϕ -1 (k + 1)} ∶ k ∈ {0, . . . , j -3} ∪ {j + 1, . . . , 2n -2}} (B.55)

∪ {{ϕ -1 (j -2), ϕ -1 (j + 1)}}.

Furthermore, by Lemma 4.39, we have Since ϕ -1 (j + 1) = t -1, we obtain N P(µ) (t -1) = N P(µ) (ϕ -1 (j + 1)). (B.58)

For a contradiction, suppose that ϕ -1 (j -2) = t + 2.

Since d P(µ) (ϕ -1 (j -2)) = 2, we have d P(µ) (t + 2) = 2, so t ≤ 2n -4. By proceeding by induction, we obtain ϕ -1 (jk) = t + k for k ∈ {2, . . . , min(j, 2nt -1)}. Since j ≥ n and t ≥ n, we have min(j, 2nt -1) = 2nt -1. For k = 2nt -1, we obtain ϕ -1 (j + t + 1 -2n) = 2n -1. Thus, j + t + 1 -2n = 0 or 2n -1, which is impossible because j, t ∈ {n, . . . , 2n -2}. Consequently, we have

ϕ -1 (j -2) = t -

  (x, y) ≡ σ(G) (v, w) if ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ {x,y}, {v, w} ∈ E(G) or {x, y}, {v, w} ∈ E(G).

A

  digraph D is defined by a vertex set V (D) and an arc set A(D), where an digraph vertex set arc set arc of D is an ordered pair of distinct vertices of D. Such a digraph is denoted by (V (D), A(D)). With each digraph D we associate its dual D ⋆ defined on dualV (D ⋆ ) = V (D) as follows. Given v, w ∈ V (D ⋆ ), with v ≠ w, (v, w) ∈ A(D ⋆ ) if (w, v) ∈ A(D). Given a vertex v of a digraph D, the in-neighbourhood of v is in-neighbourhood the set N - D (v) = {w ∈ V (D) ∶ (w, v) ∈ A(D)}, and its out-neighbourhood is the out-neighbourhood set N + D (v) = {w ∈ V (D) ∶ (v, w) ∈ A(D)}. A digraph D is identified with the 2-structure σ(D) defined on V (σ(D)) = V (D) as follows. Given x, y, v, w ∈ V (σ(D)), with x ≠ y and v ≠ w, (x, y) ≡ σ(D) (v, w) if ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (x, y), (v, w) ∈ A(D) or (x, y), (v, w) ∈ A(D). A digraph D is a tournament if for any v, w ∈ V (D), with v ≠ w, A(D) ∩ tournament {(v, w), (w, v)} = 1. It is a transitive digraph provided that for any u, v, w ∈ transitive digraph V (D), if (u, v) ∈ A(D) and (v, w) ∈ A(D), then (u, w) ∈ A(D). Atransitive digraph is also called a (strict) partial order. With each partial order O, we partial order associate its comparability graph Comp(O) defined on V (Comp(O)) = V (O) comparability graph as follows. For any v, w ∈ V (Comp(O)), with v ≠ w, {v, w} ∈ E(Comp(O)) if (v, w) ∈ A(O) or (w, v) ∈ A(O). A linear order is a transitive tournament. linear order

Figure 1 . 2 :

 12 Figure 1.2: The tournament T 2n+1

Remark 2 . 3 .

 23 First, consider a graph G. Set ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ e 1 = {(v, w) ∶ {v, w} ∈ E(G)} and e 0 = {(v, w) ∶ {v, w} ∈ E(G)}.

Figure 3 . 1 : 7 Claim 3 . 3 .

 31733 Figure 3.1: The graph B 7 Claim 3.3. B 2n+1 -2n is prime.

  y] σ by (3.6), we obtain [w, {x, y, v}] σ = [x, y] σ in both instances. When [w, {x, y}] σ = [y, x] σ , we obtain [w, {x, y, v}] σ = [y, x] σ by considering σ[{y, u, w}] instead of σ[{x, u, w}].

Corollary

Figure 4 . 3 :

 43 Figure 4.3: The partial order R 2m

Theorem 4 . 19 (

 419 Boudabbous and Ille 3

  [

(4. 24 )

 24 Consider e ∈ E(τ ). By Proposition 4.15, there exists B e ⊊ E(τ ) such that e = ⋃ f ∈Be f.

. 30 )

 30 Consider p ∈ {1, . . . , 2n -2}. It follows from (4.25) and (4.26) that {p -1, p + 1} is a module of τp. Therefore, p is a critical vertex of τ . Moreover, it follows from (4.29) that {p -1, p + 1} ⊆ N P(τ ) (p). Since p is a critical vertex of τ , it follows from Lemma 4.4 that N P(τ ) (p) = {p -1, p + 1}. Hence, for each p ∈ {1, . . . , 2n -2}, N P(τ ) (p) = {p -1, p + 1}. (4.31) It follows from (4.25) and (4.26) that {2, . . . , 2n} is a module of τ -0. Therefore, 0 is a critical vertex of τ . By (4.31), 1 ∈ N P(τ ) (0). Since {2, . . . , 2n} is a nontrivial module of τ -0, with {2, . . . , 2n} ≥ 3, it follows from Lemma 4.4 that d P(τ ) (0) = 1. Thus, N P(τ ) (0) = {1}. (4.32)

Theorem 4 . 24 (

 424 Boudabbous and Ille 4 [7]). Consider a 2-structure τ defined on V (τ ) = {0, . . . , 2n}, where n ≥ 2. The following two statements are equivalent • τ is critical, and P(τ

Proposition 4 . 27 .

 427 we obtain ϕ ↾{0,...,2n-1} = π 2n . Consequently, we have ϕ = π 2n . It follows from Note 4.25 that Aut(σ(T 2n+1 )) = {Id {0,...,2n} , π 2n }. (4.34) 4.2.3 The type P 2n+1 Given n ≥ 2, consider a 2-structure τ defined on V (τ ) = {0, . . . , 2n}. The following two statements are equivalent 1. τ is critical, and P(τ

Figure 4 . 4 :

 44 Figure 4.4: The digraph D 2m+1 .

It follows from ( 4 .

 4 43), (4.44), and (4.45) that τ = σ(D 2n+1 ) ∧ σ((D 2n+1 ) ⋆ ) (see Remark 4.11 and Fact 4.14).

Figure 4 . 5 :

 45 Figure 4.5: The tournament U 2m+1 .

  It follows from (4.45), (4.46), and (4.47) that τ = σ(D 2n+1 )∧σ((D 2n+1 ) ⋆ )∧σ(U 2n+1 ) (see Remark 4.11 and Fact 4.14).

  We obtain π 2n+1 (e) = e ⋆ for each e ∈ E(τ ) (see Notation 4.21). (4.48) Consider e ∈ E(τ ). By Proposition 4.27, there exists B e ⊊ E(τ ) such that e = ⋃ f ∈Be f.

Figure 4 . 6 :

 46 Figure 4.6: The tournament W 2m+1 .

Fact 4 . 33 .

 433 Given m ≥ 1, Cay 2m+1 is prime.Proof. Let M be a module of Cay 2m+1 such that M ≥ 2. We have to show that M = Z 2m+1 . As previously noted, the permutation of Z 2m+1 , defined by p ↦ (p + 1) mod (2m + 1) for each p ∈ Z 2m+1 , is an automorphism of Cay 2m+1 . Hence, we can assume that 0 ∈ M . Moreover, the permutation of Z 2m+1 , defined by p ↦ -p mod (2m + 1) for each p ∈ Z 2m+1 , is an isomorphism from Cay 2m+1 onto (Cay 2m+1 ) ⋆ . Since Cay 2m+1 and (Cay 2m+1 ) ⋆ share the same modules, we can assume that there exists q ∈ M ∩ {1, . . . , m}. Since Cay 2m+1 [{0, . . . , m}] = L m+1 , we obtain {0, . . . , q} ⊆ M . Since (1, m + 1), (m + 1, 0) ∈ A(Cay 2m+1 ), we have m + 1 ∈ M . Hence {0, . . . , q} ∪ {m + 1} ⊆ M. (4.49) Now, we show that {0, . . . , m + 1} ⊆ M. (4.50)Clearly, (4.50) follows from (4.49) when q = m. Thus, suppose that q ≤ m -1. Let p ∈ {q + 1, . . . , m}. Since (q, p), (p, m + 1) ∈ A(Cay 2m+1 ), we have p ∈ M . It follows that {q + 1, . . . , m} ⊆ M . Since {0, . . . , q} ∪ {m + 1} ⊆ M by (4.49), we obtain {0, . . . , m + 1} ⊆ M . Consequently, (4.50) holds. If m = 1, then M = Z 2m+1 by (4.50). Lastly, suppose that m ≥ 2. Let p ∈ {m+2, . . . , 2m}. Since (m + 1, p), (p, 0) ∈ A(Cay 2m+1 ), we have p ∈ M . We obtain {m + 2, . . . , 2m} ⊆ M . It follows from (4.50) that M = Z 2m+1 . Consequently, Cay 2m+1 is prime.

Fact 4 . 34 .

 434 Given m ≥ 2, Cay 2m+1 is critical, and

Fact 4 . 35 .

 435 Given m ≥ 2, W 2m+1 is critical, and P(W 2m+1 ) = C 2m+1 . Proposition 4.36. Given n ≥ 2, consider a 2-structure τ defined on V (τ ) = {0, . . . , 2n}. The following two statements are equivalent 1. τ is critical, and P(τ ) = C 2n+1 ;

Theorem 4 .

 4 37 (Boudabbous and Ille 6 [7]). Consider a 2-structure τ defined on V (τ ) = {0, . . . , 2n}, where n ≥ 2. The following two statements are equivalent • τ is critical, and P(τ ) = C 2n+1 ; • τ = σ(W 2n+1 ) (see Figure 4.6).

Lemma 4 .

 4 40. Let τ be a critical 2-structure, with v(τ ) ≥ 7. If u, x, y are distinct vertices of τ such that {u, x}, {x, y} ∈ E(P(τ )), and d P(τ ) (y) = 1, then τ -{x, y} is critical, τ and τ -{x, y} share the same type, and E(P(τ -{x, y})) = E(P(τ )) ∖ {{u, x}, {x, y}}.

Theorem 5 . 10 (

 510 Boubabbous and Ille[START_REF] Boudabbous | Indecomposability graph and critical vertices of an indecomposable graph[END_REF] 2 ). Consider a prime 2-structure σ such that v(σ) ≥ 7. If S (σ) ≥ 2, then there exists e ∈ E(P(σ)) such that e ∩ S (σ) ≠ ∅. (In other words, if S (σ) ≥ 2, then S (σ) ∖ S c (σ) ≠ ∅.) Proof. By Theorem 5.3, E(P(σ)) ≠ ∅. Hence, P(σ) admits a component C such that v(C) ≥ 2. Since S (σ) ≥ 2, it follows from Proposition 4.5 that

  and 4.3).

Figure 5 . 1 :

 51 Figure 5.1: The digraph R 2n+1

Theorem 5 . 19 (

 519 Chudnovsky and Seymour [10] 5

Theorem 5 . 23 (

 523 Sayar 7 [31] ). For every prime 2-structure σ, with v(σ) ≥ 7, we have S c (σ) ≤ 2. Theorem 5.23 is an immediate consequence of Corollary 4.6 and of Propositions 5.26, 5.27, 5.28, and 5.29 below. The proofs of Propositions 5.26, 5.27, 5.28, and 5.29 share the same approach and have similar arguments. Moreover, they are technical and the proofs of the last three ones are long. In order to keep this section at a satisfactory length, we provide the proofs of Propositions 5.27, 5.28, and 5.29 in Appendix B.

  ], it follows from (5.15) that (τ X ) ⋆ = τ Y [{0, . . . , 2m}]. By Remark 4.30, π 2m+1 is an isomorphism from τ X onto (τ X ) ⋆ . Thus, τ X embeds into τ Y . Therefore, σ[X] embeds into σ[Y ]. Now, suppose that Z = 6. It follows from Corollary 4.7 that there exist 3 ≤ m ≤ n such that X = 2m and Y = 2n. If m = 3, then X = Z, and hence σ[X] embeds into σ[Y ]. Thus, suppose that m ≥ 4. By Corollary 4.6, P(σ[X]) ≃ P 2m and P(σ[Y ]) ≃ P 2n . We proceed as previously, using Proposition 4.15 instead of Proposition 4.27. We obtain a prime 2-structure µ = τ X [{0, . . . , 5}] such that

Notation 6 . 3 .Remark 6 . 4 .Theorem 6 . 5 (

 636465 We denote by M 1 the set of the 2-structures σ defined on V (σ) = {0, . . . , 4} and satisfying the following assertions 1. ≺0, 1≻ σ ≠≺0, 2≻ σ (see the second statement of Proposition 4.15);2. [0, 1] σ = [0, 3] σ = [2, 3] σ and [0, 2] σ = [1, 2] σ = [1, 3] σ (see(4.4) in the second statement of Proposition 4.15 2 );3. [0, 4] σ = [4, 3] σ = [0, 3] σ and [1, 4] σ = [4, 2] σ = [1, 2] σ .It is easy to verify that the elements of M 1 are prime and minimal for {4}. Cournier and Ille 3[START_REF] Cournier | Minimal indecomposable graph[END_REF]). Consider a 2-structure σ such that v(σ) = 5. Let v ∈ V (σ). The following two assertions are equivalent

. 8 )

 8 For a contradiction, suppose that e = f . SinceC {e} (σ -Z) is a modular partition of σ -Z, C(v) and C(w) are modules of σ -Z. By Proposition 2.8, (σ -Z) C {e} (σ -Z) is constant. Thus, C(v) ∪ C(w)is a module of σ -Z as well. It follows from (6.8) that C(v), C(w), and C(v) ∪ C(w) are modules of σ, which contradicts the fact that σ is prime. Consequently, we have e ≠ f.

2 .Remark 6 . 13 .

 2613 there exists an isomorphism ϕ from σ onto an element of M 2 ∪ N 2 defined on {0, . . . , n -1} such that ϕ({v, w}) = {0, n -1}. The elements of M 2 ∪ N 2 of size 5 are not the only prime 2structures that are minimal for an unordered pair. For instance, consider the reversible 2-structure σ defined on {0, . . . , 4} by E(σ) = {{(0, 1), (2, 0), (3, 0), (4, 0), (4, 1), (4, 2), (3, 4)}, {(1, 0), (0, 2), (0, 3), (0, 4), (1, 4), (2, 4), (4, 3)}, {(1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)}}.

Corollary 8 . 3 .

 83 Given an infinite critical 2-structure σ, each nontrivial component of P(σ) is isomorphic to P N or P Z .Proof. Let C be a component of P(σ) such that v(C) ≥ 2. By Lemma 4.4, C is a cycle or an infinite or finite path. It follows from Proposition 4.5 that C is infinite. Therefore, C is isomorphic to P N or P Z .8.1 The families F Z and F N Observation 8.4. Let σ be an infinite critical 2-structure. We denote by Q the partition of V (σ) constituted by the vertex sets of the components of P(σ). Using the axiom of choice, it follows from Corollary 8.3 that there exists a function ϕ ∶ V (σ) → Z satisfying• for each Y ∈ Q such that Y > 1, ϕ ↾Y is an isomorphism from the component P(σ)[Y ] of P(σ) onto P N or P Z .Denote by ρ the unique 2-structure defined onN or Z such that ϕ ↾Y is an isomorphism from σ[Y ] onto ρ. First, consider a nontrivial component C of P(σ) such that ϕ(V (C)) = Z. Let n ∈ Z. It follows from Lemma 4.4 that {n -1, n + 1} is a module of ρn.Second, consider a nontrivial component C of P(σ) such that ϕ(V (C)) = N. By Lemma 4.4, {2, 3, . . .} is a module of ρ -0. Furthermore, by Lemma 4.4, {n -1, n + 1} is a module of ρn for every n ≥ 1. Observation 8.4 leads us to introduce the following two families of 2-structures. Notation 8.5. First, we denote by F Z the family of the 2-structures τ defined on V (τ ) = Z and satisfying• for every n ∈ Z, {n -1, n + 1} is a module of τn and not of τ .

  τ = [0, 2] τ and [0, 1] τ = [1, 3] τ . By the third assertion of Lemma 8.7, we have [1, 2] τ = [1, 3] τ . It follows that (8.2) holds. By the first assertion of Lemma 8.7, we have [1, 0] τ ≠ [1, 2] τ . By (8.2), [0, 1] τ ≠ [1, 0] τ . Therefore, τ = σ(L N ). Example 8.11. Set U N = U Z [N], W N = W Z [N], and H N = H Z [N].

Lemma 8 . 19 .

 819 Let σ be an infinite, critical, and non finitely critical 2-structure. Consider the partition Q of V (σ) and a function ϕ ∶ V (σ) → Z defined as in Observation 8.4. The following two assertions hold (I3) P (see Notation 8.14) is a modular partition of σ according to Q; (I4) for each Y ∈ Q such that V (τ Y ) = N (see Notation 8.14),

Proof.

  To begin, suppose that M Q ′ possesses a unique element Y . Hence, M ⊆ Y . By Assertion (M2) of Proposition 2.5, M is a module of σ[Y ]. Thus, ϕ Y (M ) is a module of τ Y . By Assertion (I1), τ Y ∈ F N ∪ F Z . It follows from Lemmas 8.8 and 8.10 that ϕ Y (M ) is a module of L N or L Z . Therefore, ϕ Y (M ) contains even and odd integers. It follows that M ∩ V even (σ) ≠ ∅ and M ∩ V odd (σ) ≠ ∅. Since P is a modular partition of σ according to Q by Assertion (I3), {Y ∩

Example 8 . 35 .

 835 Consider the graph H defined on V (H) = Z ∪ {∞ n ∶ n ∈ Z} and saisfying• H[Z] = H Z ;• the bijection Z → {∞ n ∶ n ∈ Z}, defined by n → ∞ n for each n ∈ Z, is an isomorphism from P Z onto H[{∞ n ∶ n ∈ Z}];

Example 8 .

 8 39. We consider the tournament T N = T Z [N]. It follows from Fact 8.38 that σ(T N ) is critical, P(σ(T N ))[N] = P N , and ∞ is isolated in P(σ(T N )).

σ

  (y) (see Notation 3.7). The set {Ext σ (X)} ∪ {⟨X⟩ (e,f ) σ ∶ e, f ∈ E(σ)} ∪ {X (e,f ) σ

Figure 4 . 1 )

 41 . It follows from Proposition 9.41 that G[{0, 1} × F ′ ] is critical. Set F = {0, 1} × F ′ . We obtain that F ⊆ F and G[ F ] is critical. (9.22) It follows from (9.21) and (9.22) that Γ (σ(G)[X∪ F ], F ) is critical. Since Statement (S5) holds, it follows from Theorem 9.6 that σ(G)[X ∪ F ] is F -critical. Consequently, σ(G) is finitely (V (σ) ∖ X)-critical.

Definition 10 . 3 .

 103 Consider an infinite prime 2-structure σ.The criticality digraph C(σ) of σ is defined on V (C(σ)) = V (σ) as follows. Given distinct criticality digraph v, w ∈ V (σ), (w, v) ∈ A(C(σ)) if σv admits a nontrivial module containing w.10.1 Modular decomposition in the infinite caseNotation 10.4. We associate with each 2-structure σ the set Υ(σ) of the modules of σ that are maximal under inclusion among the proper modules of σ. (Note that Υ(σ) can be empty when σ is infinite.) Proposition 10.5. Let σ be a connected 2-structure. If Υ(σ) ≠ ∅, then Υ(σ) is a modular partition of σ and σ Υ(σ) is prime.

  For a contradiction, suppose that N = V (σ). Hence, there exists N ∈ N v such that N ∩ M ≠ ∅. By Assertion (M5) of Proposition 2.5, N ∪ M is a module of σ. Since v ∈ N ∖ M, we have M ⊊ N ∪ M. It follows from the maximality of M that N ∪ M = V (σ). Since N ∖ M ≠ ∅, it follows from Assertion (M6) of Proposition 2.5 that M ∖ N is a module of σ. Since N ∪ M = V (σ), we have M ∖ N = V (σ) ∖ N . Consequently, N is a modular cut of σ. Since v ∈ N and N ≠ V (σ), N is a nontrivial modular cut of σ, which contradicts the fact that σ is uncuttable. It follows that N ≠ V (σ).

  where y ∈ X. SetY = (X ∖ {y}) ∪ {v}. Since v ∈ X σ (y), {y, v} is a module of σ[X ∪ {v}]. It follows that σ[Y ] is isomorphic to σ[X]. Hence, σ[Y ] is prime. By Assertion (M2) of Proposition 2.5, M ∩ Y is a module of σ[Y ]. Since (X ∖ {y}) ⊆ (M ∩ Y ), we obtain M ∩ Y = Y .

Corollary 10 . 16 .

 1016 Let σ be an infinite prime 2-structure. Consider distinct u, v, w ∈ V (σ). Suppose that u, v ∈ W (σ). If (v, u), (w, v) ∈ A(C(σ)) and (u, v) ∈ A(C(σ)), then σu admits a nontrivial module containing v and w.The next result follows from Corollary 10.16.Corollary 10.17. Let σ be an infinite prime 2-structure. Consider distinct u, v, w ∈ W (σ) (seeNotation 10.11). If (v, u), (w, v) ∈ A(C(σ)) and (u, v), (v, w) ∈ A(C(σ)), then (w, u) ∈ A(C(σ)) and (u, w) ∈ A(C(σ)). Proof. Suppose that (v, u), (w, v) ∈ A(C(σ)) and (u, v), (v, w) ∈ A(C(σ)). (10.7) Since (v, u), (w, v) ∈ A(C(σ)) and (u, v) ∈ A(C(σ)), we obtain (w, u) ∈ A(C(σ)) by Corollary 10.16. For a contradiction, suppose that (u, w) ∈ A(C(σ)). Since (w, v) ∈ A(C(σ)) and (v, w) ∈ A(C(σ)), it follows from Corollary 10.16 that (u, v) ∈ A(C(σ)), which contradicts (10.7). Consequently, we have (u, w) ∈ A(C(σ)). The next result is an immediate consequence of Corollary 10.10 and Notation 10.11. Corollary 10.18. Let σ be an infinite critical 2-structure. Consider

Corollary 10 .

 10 18 leads us to introduce the following notation. Notation 10.19. Let σ be an infinite critical 2-structure. Given v, w ∈ W (σ), v ∼ σ w means v = w or v ≠ w and (v, w), (w, v) ∈ A(C(σ)). Clearly, ∼ σ is a symmetric and reflexive binary relation defined on V (σ) ∖ E (σ). It follows from Corollary 10.18 that v ∼ σ w for any v, w ∈ W ∅ (σ) ∪ W δ (σ).In the next lemmas, we examine the binary relation ∼ σ . Lemma 10.20. Let σ be an infinite critical 2-structure. Consider distinct v, w ∈

1 Proposition 10 . 24 .

 11024 and henceW ∅ (σ) = {v}. 10.4 Proof of Theorem 10.The next result follows from Corollary 10.10 and Fact 10.13. Given an infinite critical 2-structure σ, consider distinct v, w ∈ W (σ). If (w, v), (v, w) ∈ A(C(σ)), then (RT) holds (see Theorem 10.1).Proof. Suppose that (w, v), (v, w) ∈ A(C(σ)). It follows from Corollary 10.10 that {w} ∈ Υ(σv) and (

. 21 )

 21 We show that Υ(σw) ≥ M(10.22) for every M ∈ Υ(σv)}. This is obvious when M = 1. Hence, considerM ∈ Υ(σv)} such that M ≥ 2. Since (v, w) ∈ A(C(σ)), itfollows from Corollary 10.10 that {v} ∈ Υ(σw) and (σw) Υ(σw) is prime. Let N ∈ Υ(σw) such that M ∩ N ≠ ∅. Since {v} ∈ Υ(σw) and v ∈ M, we have v ∈ N . It follows from Fact 10.13 that M ∩ N = 1. Therefore, we obtain Υ(σw) ≥ M , so (10.22) holds. It follows from (10.21) that Υ(σw) ≥ V (σ) . Using the axiom of choice, consider Y ⊆ V (σ)∖{w} such that Y ∩O = 1 for each O ∈ Υ(σw). We obtain that Y = Υ(σ-w) and σ[Y ] is prime. Since Υ(σ-w) ≥ V (σ) , we have Y = V (σ) . Consequently, (RT) holds. The next result follows from Corollary 10.16. We use the following notation. Notation 10.25. Recall that L N denotes the usual linear order on N. We denote by L N the linear order defined on N ∪ {∞} such that L N [N] = L N and (n, ∞) ∈ A( L N ) for every n ∈ N. Lemma 10.26. Given an infinite critical 2-structure σ, C(σ) -E (σ) embeds neither L N nor its dual ( L N ) ⋆ .

  w ∈ Z, which contradicts (10.26) with W = {y ′ , v} and W ′ = {w}.

  ) and [m -1, n] σ = (e, f ), we obtain m -1 ∈ N . By proceeding by induction, we obtain(M ∩ {. . . , m -1, m}) ⊆ N.Similarly, we obtain(M ∩ {n, n + 1, . . .}) ⊆ N. Since N ∩ {p ∈ Z ∶ m ≤ p ≤ n} = {m, n}, we have N = M ∩ ({. . . , m -1, m} ∪ {n, n + 1, . . .}).For a contradiction, suppose thatn > m + 1. Since [m, m + 1] σ ≠ (e, f ) and [m -1, m + 1] σ = (e,f ), we obtain m -1 ∈ M . Similarly, we have n + 1 ∈ M . It follows that M = {p ∈ Z ∶ m ≤ p ≤ n} and N = {m, n}. Since M ≥ 5, we have n ≥ m + 4. Since [m, m + 2] σ = (e, f ), [n, m + 2] σ = (f, e), and e ≠ f , we obtain [m, m+2] σ ≠ [n, m+2] σ , which contradicts the fact that N is a module of σ[M ]. Consequently, we have n = m+1. Since N = M ∩({. . . , m-1, m}∪{n, n+1, . . .}), we have N = M . Proposition 10.32. Let σ be an infinite critical 2-structure. If there exists v ∈ W δ (σ) such that λ(v) = 2 (see Notation 10.11), then (RT) holds.

  [v, w] σ = (e, f ) if and only if there exists u ∈ W δ (σ) such thatv < L δ (σ) u < L δ (σ) w. (10.47) Consider v, w ∈ W δ (σ) such that v < L δ (σ) w.To begin, suppose that there existsu ∈ W δ (σ) such that v < L δ (σ) u < L δ (σ) w. By (10.44), we have v ∈ C u and w ∈ (V (σu) ∖ C u ). Hence, we have [v, w] σ = (e, f ). Conversely, suppose that [v, w] σ = (e, f ). By (10.44), we have w ∈ (V (σv) ∖ C v ). Hence, we have [w, C v ] σ = (f, e). Since [w, v] σ = (f, e), we obtain [w, C v ∪ {v}] σ = (f, e). Since (v, w) ∈ A(L δ (σ)), we have C v ∪ {v} ⊆ C w . Since w ∈ E (σ), C w is a nontrivial module of σ -w. Since C w is not a nontrivial module of σ, there exists u ∈ (C w ∖ (C v ∪ {v})) such that [w, u] σ ≠ (f, e). Since u ∈ (C w ∖ (C v ∪ {v})), it follows from (10.44) that v < L δ (σ) u < L δ (σ) w.

F

  and {t ∈ W (σ) ∶ v < L δ (σ) t < L δ (σ) F E (σ) (u)} = ∅. E (σ) (u) < L δ (σ) v < L δ (σ) w and {t ∈ W (σ) ∶ F E (σ) (u) < L δ (σ) t < L δ (σ) v} = ∅.In the first instance, it follows from (10.47) that [w,F E (σ) (u)] σ = (e, f ) and [v, F E (σ) (u)] σ ≠ (e, f ), which contradicts F E (σ) (u) ←→ σ V (σ) ∖ {u, F E (σ) (u)} (seeNotation 10.8). Similarly, the second instance leads to a contradiction. It follows that E (σ) = ∅, so we have V (σ) = W δ (σ).

Proposition 10 . 33 .

 1033 Let σ be an infinite critical 2-structure. If there exists v ∈ W δ (σ) such that λ(v) = 1 (see Notation 10.11), then (RT) holds.Proof. Suppose that there exists u ∈ W δ (σ) such that λ(u) = 1. Hence, there exists e ∈ E(σ) such that λ(u) = {e}. By Lemma 10.22, we have⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ W (σ) = W δ (σ) and for every v ∈ W δ (σ), λ(v) = {e}.(10.48) Let v ∈ W δ (σ).

  distinct v, w ∈ V (σ). We show thatC w v ∪ C v w = V (σ). (10.50) Let D v ∈ (C {ev} (σv) ∖ {C w v }). Since σ is {e}-connected, there exists u ∈ D v such that ≺u, v ≻ σ ≠ {e}. It follows that V (σ) ∖ C w v is {e}-connected. (10.51) Since v ∈ (V (σ) ∖ C w v ), we obtain (V (σ) ∖ C w v ) ⊆ C v w .It follows that (10.50) holds.

(A. 2 )

 2 Let v ∈ S (σ) ∖ S c (σ). It follows from Theorem 5.13 applied to σv that v(σ) = 2n + 2, where n ≥ 3, (A.3)

  [v, (ϕ v ) -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ = [(ϕ v ) -1 (2), (ϕ v ) -1 (0)] σ .Since (5.7) holds for ϕ v , we have[f (v), (ϕ v ) -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ = [(ϕ v ) -1 (2), (ϕ v ) -1 (0)] σ . It follows that [f (v), (ϕ v ) -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ = [v, (ϕ v ) -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ .Similarly, we have[f (v), (ϕ v ) -1 ({2i ∶ i ∈ {0, . . . , n -1}})] σ = [v, (ϕ v ) -1 ({2i ∶ i ∈ {0, . . . , n -1}})] σ .

  7) that {f (v), (ϕ v ) -1 (1)} is a module of σ -(ϕ v ) -1 (0). (A.8) Since τ v ∈ R 2n , τ v is critical and P(τ v ) = P 2n . It follows from Proposition 4.15 that [1, 3] τv = [0, 2] τv . Since ϕ v is an isomorphism from (σv)f (v) onto τ v , we obtain [(ϕ v ) -1 (1), (ϕ v ) -1 (3)] σ = [(ϕ v ) -1 (0), (ϕ v ) -1 (2)] σ , Since (5.7) is satisfied by ϕ v , we have [f (v), (ϕ v ) -1 (3)] σ = [(ϕ v ) -1 (2), (ϕ v ) -1 (0)] σ .We prove that[(ϕ v ) -1 (1), V (σ) ∖ {(ϕ v ) -1 (0),(ϕ v ) -1 (1)}] σ = [(ϕ v ) -1 (0), (ϕ v ) -1 (2)] σ . (A.10)Since (5.7) is satisfied by ϕ v , we have⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ [f (v), (ϕ v ) -1 ({2i ∶ i ∈ {0, . . . , n -1}})] σ = [ϕ -1 (0), ϕ -1 (2)] σ , and [f (v), (ϕ v ) -1 ({2i + 1 ∶ i ∈ {0, . . . , n -1}})] σ = [ϕ -1 (2), ϕ -1 (0)] σ .

Set Y = 2 . 1 .

 =21 V (σ) ∖ {v, w}. Since S c (σv) = {w}, σ[Y ] is critical. Furthermore, σ[Y ∪ {v}] is prime because σ[Y ∪ {v}] = σx. It follows that v ∈ S c (σ[Y ∪ {v}]). By Corollary 5.5 (and Remark 5.6), there exists e ∈ E(P(σ[Y ∪{v}]))∩E(P(σ[Y ])). Set Z = Y ∖ e. We have σ[Z] is prime. Since e ∈ E(P(σ[Y ∪ {v}]), we obtain σ[Y ∪ {v}])e = (σe)x is prime. Since (σe)x = σ[Z ∪ {v}], we have v ∈ Ext σ (Z). Furthermore, it follows from the first statement of Fact 5.18 that x ∈ Ext σ (Z). Since (σ -e)-x = σ[Z ∪{v}] and (σ -e)-v = σ[Z ∪{x}], we obtain that (σe)v and (σe)x are prime. (A.14)Since (σ-v)-e is prime, we have e ∈ E(P(σ-v)). In another vein, it follows from Corollary 3.21 applied to σv that there exist u, u′ ∈ V (σv) ∖ X such that (σ -v)-{u, u ′ } is prime. Since S (σ -v) = {w}, we obtain u ≠ u ′ . Hence, {u, u ′ } ∈ E(P(σ -v)). Therefore, we obtain e, {u, u ′ } ∈ E(P(σ -v)).By the second statement of Fact 5.18,(σv)e ≃ (σv) -{u, u ′ }. Since τ embeds into (σv) -{u, u ′ }, τ embeds into (σv)e. (A.15)Finally, we distinguish the following two subcases. Suppose that σe is decomposable. Since σe = σ[Z ∪ {v, x}], it follows from Statement (P5) of Lemma 3.17 that {v, x} is a module of σ -e. It follows that (σ -e)-x ≃ (σ -e)-v. By (A.15), τ embeds into (σe)v. Consequently, τ embeds into (σe)x. To conclude, it suffices to apply the induction hypothesis to σx because x ∈ N (σ).

2. 2 .

 2 Suppose that σe is prime. It follows from (A.14) that v, x ∈ S (σe).

  E(P((τ-(2n + 1)) -{t -1, t})) = (E(P(τ -(2n + 1))) ∖ {{k, k + 1} ∶ k ∈ {t -2, t -1, t}}) ∪ {{t -2, t + 1}} = {{k, k + 1} ∶ k ∈ {0, . . . , t -3} ∪ {t + 1, . . . , 2n -2}} (B.3) ∪ {{t -2, t + 1}}.

2 .

 2 Suppose that ϕ ↾{0,...,2n-2} = π 2n-1 . We obtainϕ(k) = 2n -2k (B.14)for each k ∈ {0, . . . , 2n -2}. Therefore, we have(0, 1) σ = (ϕ -1 (2n -2), ϕ -1 (2n -3)) σ by (B.14) = (ϕ -1 (1), ϕ -1 (0)) σ by (B.11) = (ϕ -1 (2n -1), ϕ -1 (2n -2)) σ by (B.11) = (2n -1, ϕ -1 (2n -2)) σ because ϕ -1 (2n -1) = 2n -1 = (2n -1, 0) σ by (B.14) = (1, 0) σ by (B.8).

  odd, and for each l ∈ {0, . . . , n -2} ∪ {n + 2, . . . , 2n}, ϕ(l) = 2nl.(B.17)Recall that P(σ -(2n + 1)) = P 2n+1 . By Lemma 4.39,E(P((σ -(2n + 1)) -{t -1, t})) = (E(P(σ -(2n + 1))) ∖ {{k, k + 1} ∶ k ∈ {t -2, t -1, t}}) ∪ {{t -2, t + 1}}.

( 0 ,

 0 1) σ = (ϕ -1 (2n), ϕ -1 (2n -1)) σ by (B.22) = (ϕ -1 (n + 1), ϕ -1 (n -1)) σ by (B.11) = (n + 1, n -1) σ by (B.15) (because t = j and t = n) = (1, 0) σ by (B.8), which contradicts (B.7). It follows that n is odd. It follows from (B.22) that ϕ(l) = 2nl for each l ∈ {0, . . . , n -2} ∪ {n + 2, . . . , 2n}. Hence, (B.17) holds. Set ψ = π 2n+1 ○ ϕ.

(B. 23 )

 23 Since (B.17) holds, we obtainψ(l) = l for each l ∈ {0, . . . , n -2} ∪ {n + 2, . . . , 2n}. (B.24) Since n ≥ 3, we have [0, 1] σ = [ψ -1 (0), ψ -1 (1)] σ . Furthermore, we have [0, 2] σ = [0, 2n] σ by (B.8) = [ψ -1 (0), ψ -1 (2n)] σ by (B.24) = [ψ -1 (0), ψ -1(2)] σ by (B.23). It follows from (B.8) and (B.23) that

  Finally, since t = j and t = n, it follows from (B.15) that ψ(n -1) = n + 1 and ψ(n + 1) = n -1. (B.25)

ϕ - 1

 1 (j -2) = t -2 and ϕ -1 (j + 2) = t + 2 (B.31) or ϕ -1 (j -2) = t + 2 and ϕ -1 (j + 2) = t -2. (B.32) For a contradiction, suppose that (B.32) holds. By proceeding by induction, it follows from (B.30), (B.19), and (B.26) that ϕ -1 (j + k) = tk

[ψ - 1

 1 (p), ψ -1 (q)] σ = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ [ψ -1 (0), ψ -1 (2)] σ if p and q are even, [ψ -1 (0), ψ -1 (1)] σ otherwise. (B.[START_REF] Spinrad | P4-trees and substitution decomposition[END_REF]

  [START_REF] Maffray | A translation of Tibor Gallai's Paper: Transitiv orientierbare Graphen[END_REF], S c (σ) = {s, t}.Moreover, consider an isomorphism ϕ s from P(σ -s) onto P 2n+1 . It follows from Proposition 5.28 that⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ ϕ s (t) is odd and (ϕ s (0), ϕ s (2)) σ = (ϕ s (2), ϕ s (0)) σ .Proof of Proposition 5.29. Consider s ∈ S c (σ) such that P(σs) ≃ P 2n . Therefore, v(σ) = 2n + 1. Since v(σ) ≥ 7, we obtain n ≥ 3.

1 ]

 1 σ if p is even and q is odd,[START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF][START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] σ otherwise. (B.39) It follows from Corollary 5.25 that N P(σ-(2n)) (t) = N P(σ-t) (2n). (B.40) Since v(σ -(2n)) = 2n, it follows from Corollary 4.6 that P(σt) ≃ P 2n . As above, there exists an isomorphism ϕ from σt onto τ , where τ is a critical 2-structure such that P(τ ) = P 2n . It follows from Proposition 4.15 that ≺ϕ -1 (0), ϕ -1 (1)≻ σ ≠≺ϕ -1 (0), ϕ -1 (2)≻ σ . (B.41) Similarly, we have [v, 2n-1] σ = [v, 2n] σ when v is odd. It follows that {2n-1, 2n} is a module of σ, which contradicts the fact that σ is prime. Consequently, (B.45) does not hold. By (B.44), we have(ϕ -1 ) ↾{0,...,2n-3} ) = π 2n-2 .We obtainϕ -1 (k) = 2n -3k. (B.47)for each k ∈ {0, . . . , 2n -3}. Therefore, we have(0, 2) σ = (ϕ -1 (2n -3), ϕ -1 (2n -5)) σ by (B.47) = (ϕ -1 (2n -2), ϕ -1 (2n -5)) σ by (B.42) = (2n -2, 2) σ by (B.43) and (B.47) = (2, 0) σ by (B.39).

[ψ - 1 ( 0 )

 10 , ψ -1 (2)] σ otherwise. (B.48) Since ϕ -1 (2n -1) = 2n, it follows from (B.43) and (B.47) that ψ -1 is defined by{0, . . . , 2n -1} → {0, . . . , 2n -2} ∪ {2n} 0 → 2n, 1 → 2n -2, 2 ≤ k ≤ 2n -1 → k -2. (B.49)We obtain[ψ -1 (0), ψ -1 (1)] σ = [2n, 2n -2] σ by (B.49) = [ϕ -1 (2n -1), ϕ -1 (2n -2)] σ = [ϕ -1 (2n -3), ϕ -1 (2n -4)] σ by (B.42) = [0, 1] σ by (B.47).Similarly, we have[ψ -1 (0), ψ -1 (2)] σ = [2n, 0] σ by (B.49) = [ϕ -1 (2n -1), ϕ -1 (2n -3)] σ by (B.47) = [ϕ -1 (2n -3), ϕ -1 (2n -5)] σ by (B.42) = [0, 2] σ by (B.47). It follows from (B.48) and (B.39) that

  (B.51) For a contradiction, suppose that (B.50) holds. Recall thatE(P(σ -(2n))) = {{k, k + 1} ∶ k ∈ {0, . . . , 2n -2}}.By Lemma 4.39,E(P((σ -(2n)) -{t -1, t})) = {{k, k + 1} ∶ k ∈ {0, . . . , t -3} ∪ {t + 1, . . . , 2n -2}} (B.52) ∪ {{t -2, t + 1}}.Similarly, we haveE(P(σ -{t, ϕ -1 (j -1), 2n})) = {{ϕ -1 (k), ϕ -1 (k + 1)} ∶ k ∈ {0, . . . ,j -3} ∪ {j + 1, . . . , 2n -2}} (B.53)

E

  (P((σ -(2n)) -{t, t + 1})) = {{k, k + 1} ∶ k ∈ {0, . . . , t -2} ∪ {t + 2, . . . , 2n -2}} (B.56) ∪ {{t -1, t + 2}}.Since (B.51) holds, we obtain E(P(σ -{t, t + 1, 2n})) = E(P(σ -{t, ϕ -1 (j -1), 2n}).(B.57) Set µ = σ -{t, t + 1, 2n}.

  By Lemma 2.4, {x, y, v} is a module of σ[{x, y, v, w}], and hence w ←→ σ {x, y, v}. 2. Suppose that ≺ x, y ≻ σ = ν(σ[{x, y, v, w}]) (see Notation 2.25). Since w ←→ σ {x, y}, we have ≺ x, w ≻ σ =≺ y, w ≻ σ . For a contradiction, suppose that ≺x, w ≻ σ ≠ ν(σ[{x, y, v, w}]).

{x} is a module of σ. By Assertion (M3) of Proposition 2.5, it suffices to verify that W ∪{x, y} is a module of σ. Consider v, w ∈ V (σ)∖{x, y} such that v ←→ σ {x, y} (i.e. v ∈ W ) and w ←→ σ {x, y}. We prove that w ←→ σ {x, y, v}.

(

3

.7) By Remark 3.3, B[{x, y, u, w}] is not connected. We distinguish the following two cases. 1. Suppose that ≺ x, y ≻ σ ≠ ν(σ[{x, y, v, w}]) (see Notation 2.25). Since ≺x, v ≻ σ =≺y, v ≻ σ =≺x, y ≻ σ by (3.5), it follows from Lemma 2.4 and Proposition 2.8 that {x, y, v} ∈ C(σ[{x, y, v, w}]). By Lemma 2.4 and Proposition 2.8, {x, y, w} ∈ C(σ[{x, y, v, w}]). By Lemma 2.4, {x, y, w} is a module of B[{x, y, v, w}], which contradicts v ←→ σ {x, y}. It follows that ≺ x, w ≻ σ = ν(σ[{x, y, v, w}]). Since w ←→ σ {x, y}, [w, {x, y}]

  is critical by Corollary 4.7. To conclude, it suffices to apply the induction hypothesis to σ -{x, x ′ } with σ[X] and σ[Y ′ ].

	Lastly, we obtain the following result.
	Corollary 4.10. Let σ be a critical 2-structure σ such that

  ). By Fact 2.7, σ(T 2n-1 ) is prime. Thus, σ(T 2n+1 ) -{p, p + 1} is prime as well. Consequently, (4.28) holds.

		It follows from
	(4.28) that
	E(P 2n ) ⊆ E(P(τ )).	(4.29)

  Since {2, . . . , 2n} is a nontrivial module of τ -0, with {2, . . . , 2n} ≥ 3, it follows from Lemma 4.4 that d P(τ ) (0) = 1.

	Therefore,
	N P(τ ) (0) = {1}.	(4.42)

.41) Since (4.35) holds, {2, . . . , 2n} is a module of τ -0. Thus 0 is a critical vertex of τ . It follows from (4.41) that 1 ∈ N P(τ ) (0). Finally, since (4.35) holds, {0, . . . , 2n -2} is a module of τ -(2n). Thus 2n is a critical vertex of τ . It follows from (4.41) that 2n -1 ∈ N P(τ ) (2n) and N P(τ ) (2n)∩{1, . . . , 2n-2} = ∅. By (4.42), 0 ∈ N P(τ ) (2n). Therefore, N P(τ ) (2n) = {2n -1}. Consequently, τ is critical, and P(τ

  2m} is prime as well. Since (ψ 2m+1 (W 2m+1 )) ⋆ = Cay 2m+1 by Fact 4.32, we obtain that Cay 2m+1 ∖ ψ 2m+1 ({2m -1, 2m}) = Cay 2m+1 ∖ {m, 2m} is prime. The permutation of Z 2m+1 , defined by p ↦ (pm) mod (2m + 1) for each p ∈ Z 2m+1 , is an automorphism of Cay 2m+1 . Thus Cay 2m+1 ∖ {0, m} is prime. The permutation of Z 2m+1 , defined by p ↦ -p mod (2m + 1) for each p ∈ Z 2m+1 , is an isomorphism from Cay 2m+1 onto (Cay 2m+1 ) ⋆ . Therefore, Cay 2m+1

  which contradicts Lemma 3.13 because u e ≠ u f by (5.5). Thus, for distinct e, f ∈ E, we have u e ∈ f or u f ∈ e.

(5.6)

Let e ∈ E. Recall that the elements of E are pairwise disjoint. Since E ≥ 3, there exists f ∈ E ∖ {e} such that u e ∈ f . By (5.6), u f ∈ e. Let g ∈ E ∖ {e, f }. By (5.6) applied to f and g, u g ∈ f , because u f ∈ g. By

(5.6)

applied to e and g, u e ∈ g because u g ∈ e. Therefore, every element of E ∖ {e, f } contains u e . Consequently, E = 3.

  and suppose that σu is decomposable for every u ∈ V (σ) ∖ Y . It follows from Corollary 3.21 that there exist distinct v, w ∈ V (σ) ∖ Y such that σ -{v, w} is prime. Thus, τ embeds into σ -{v, w}. Denote by C the component of P(σ) containing v and w. For a contradiction, suppose that

  structure, it follows from Corollary 4.6 and Propositions 4.15, 4.23, 4.27, and 4.36 that P(σv) = P 2n , where n ≥ 3. Consequently, there exists y

  and σ[Z ′′ ] is prime. Consequently, suppose that Z = 5 or 6. Furthermore, σ[Z] is critical by Corollary 4.7. To begin, suppose that Z = 5. It follows from Corollary 4.7 that there exist 2 ≤ m ≤ n such that X = 2m + 1 and Y = 2n + 1. Moreover, it follows from Corollary 4.41 that σ[X], σ[Y ], and σ[Z] share the same type. By Corollary 4.6, P

  4}] is critical and P(τ Y [{0, . . . , 4}]) = P 5 . Since P(τ X [{0, . . . , 4}]) = P 5 and P(τ Y [{0, . . . , 4}]) = P 5 , ψ is an automorphism of P 5 . Therefore, we have ψ = Id {0,...,4} or π 5 (see Notation 4.21). Id {0,...,4} . We obtain τ X [{0, . . . , 4}] = τ Y [{0, . . . , 4}]. It follows from (5.13) and (5.15) that

	To conclude, we distinguish the following two cases.
	1. Suppose that ψ =

.15) Moreover, σ[Z] is isomorphic to τ Y [{0, . . . , 4}]. Consequently, there exists an isomorphism ψ from τ X [{0, . . . , 4}] onto τ Y [{0, . . . , 4}]. We obtain also that τ Y [{0, . . . ,

  .16) and consider t ∈ S c (σ) ∖ {2n + 1}. Since θ 2n+1 , π 2n+1 ∈ Aut(C 2n+1 ) by Remark 4.38, we can assume that t = 2n. C 2n+1 . Consider an isomorphism ϕ from P(σ -(2n)) onto C 2n+1 . Since θ 2n+1 , π 2n+1 ∈ Aut(C 2n+1 ), we can assume that

	Hence, N P(σ-(2n+1)) (2n) = {0, 2n -1}. By Corollary 5.25, N P(σ-(2n)) (2n + 1) =
	{0, 2n-1}. Moreover, since v(σ) ≥ 8, it follows from Corollary 5.25 that σ-(2n) ≃
	σ -(2n + 1). Therefore, P(σ -(2n)) ≃

  .13) By Assertion (I1), τ Y ∈ F N ∪ F Z . Since ϕ Y (Y ∩ M ) is a nontrivial module of τ Y ,it follows from Lemmas 8.8 and 8.10 that there exists n

  Since (8.16) holds, (8.7) holds too. Since Assertion (I5) holds, σ Q P is prime. It follows from Fact 8.22 that σ is prime.

	Lemma 8.29. Let σ be a locally critical 2-structure. Consider a partition Q
	of V (σ) and a function ϕ ∶ V (σ) → Z satisfying Assertions (I1) and (I2).
	Suppose also that Assertions (I3) and (I4) hold. Consider Y

  we obtain that σ -({x} ∪ F ′′ ) is prime. It follows from Theorem 5.8 applied to σ -({x} ∪ F ′′ ) that there exist distinct v, w ∈ ({x} ∪ F ′′ ) such that σ -{v, w} is prime.

	Remark 8.34. Observe that Corollary 8.32 does not hold if we only suppose
	that σ is prime but not finitely critical. Similarly, Corollary 8.33 does not hold if
	we only suppose that the finite subset F of V (σ) is nonempty. Indeed, consider
	the graph

  Thus, Assertion (I6) holds. Consequently, Assertions (I1),...,(I6) hold. Since Q is infinite, (8.8) holds. It follows from Theorem 8.27 that σ is critical and non finitely critical. Precisely, it follows from Assertion (J2) of Proposition 8.30 that P(σ(H))[Z] = P Z and P(σ(H))[Z] is a component of P(σ(H)). Finally, we verify that for each n ∈ Z, ∞ n is isolated in P(σ(H)). It follows from (8.24) and (8.25) that H -(∞ n ) admits a module M such that M and (

  is not a module of σ Q P . It follows from the definition of the generalized quotient (seeDefinition 8.16) that σ Q P is prime. Thus, Assertion (I5) holds. Obviously, Assertion (I4) holds.To verify that σ is prime, we utilize Fact 8.22 with Q ′ = Q as follows. Clearly, (8.7) holds. Moreover, we have P ′ = P . As previously observed, σ Q P is prime. Since Assertions (I3) and (I4) hold, it follows from Fact 8.22 that σ is prime.Since Assertions (I3) and (I4) hold, it follows from Lemma 8.29 that σn is decomposable for each n ∈ Z. Since (8.30) or (8.31) hold, {0, 1, .

  (X ∪ {u, w}) ∩ {z, w} is a module of σ[X ∪ {u, w}]. Since {u, w} ∈ E(Γ (σ,X) ), (X ∪ {u, w}) ∩ {z, w} is a trivial module of σ[X ∪ {u, w}]. Since w ∈ (X ∪ {u, w}) ∩ {z, w}, we obtain z ∈ X ∪ {u}. It follows that z = v. Therefore, w ∈ Y σ (v), that is, {v, w} is a module of σ[X ∪ {u, v, w}]

	Hence, {z, w} is a module of σ[X ∪ {u, v, w}]. By Assertion (M2) of Propo-
	sition 2.5,

X ∪ {u, v} is a module of σ[X ∪ {u, v, w}]. By Assertion (M2) of Proposition 2.5, X ∪ {u} is a module of σ[X ∪ {u, w}], which contradicts {u, w} ∈ E(Γ (σ,X) ). Consequently, w ∈ ⟨Y ⟩ σ .

It follows from Lemma 3.13 that there exists z ∈ Y such that w ∈ Y σ (z).

  By Remark 9.4, Statement (S3) holds because Statement (S5) holds. It follows from Proposition 9.29 that C is bipartite. It follows from Theorem 9.6 that C is critical. Moreover, since Statement (S5) holds, C does not embed P 5 by Corollary 9.31. In Theorem 9.38 below, we characterize the bipartite graphs Γ such that Γ does not embed P 5 and Γ is critical. We need the following three definitions (see Definitions 9.33, 9.35, and 9.36).

  .20) By Assertion (M2) of Proposition 2.5, M w ∖ {v} and M v ∖ {w} are modules of σ -{v, w}. Since w ∈ E (σ), it follows from (10.19) that M v ∖M w ≥ 2. Hence, we have (M v ∖{w})∖M w , which is (M v ∖{w})∖(M w ∖{v}), is nonempty. It follows from Assertion (M6) of Proposition 2.5 that (M w

  Let M z be a nontrivial module of σz. As seen in Remark 3.16, we haveM z ∩ Y ≤ 1 or Y ⊆ M z . Suppose that M z ∩ Y ≤ 1. It follows from (10.28) that M z ⊆ (({y} ∪ X σ (y)) ∖ {v}) or there exists t ∈ (Y ∖ {y}) such that M z = {t, v}. If t = w,then {y, w} is a module of σ[Y ], which contradicts the fact that σ[Y ] is prime. Thus, t ∈ (Y ∖ {y, w}). Consequently, (10.29) holds when M z ∩ Y ≤ 1.In the other case, we have a nontrivial module M z of σz such that Y ⊆ M z . By Fact 10.6, we have Y

	Given a nontrivial module M z of σ -z, we verify that
	M z ⊆ (({y} ∪ X σ (y)) ∖ {v})
	or	(10.29)
	there exists t ∈ (Y ∖ {y, w}) such that M z = {t, v}.
		(10.28)

y) ∖ {v} and p (σ-z,Y ) = {X σ (y) ∖ {v}, {v}}. σ (y) ⊆ M z , and hence v ←→ σ (V (σ) ∖ {z, w}).

  it follows from(10.26) that v ∈ M u ′ . Therefore, we obtain u ′ ←→ σ M u ′ by(10.25). Hence, M u ′ is a module of σ, which contradicts the fact that σ is prime. It follows that(10.30) holds. Since {y} ∪ X σ (y) is equipotent to V (σ), Υ(σu ′ ) is equipotent to V (σ) as well. Since u ′ ∈ W π (σ), (σu ′ ) Υ(σu ′ )is prime. Consequently, (RT) holds. The next result follows from Corollary 10.16, Remark 10.27, and Proposition 10.28. Proposition 10.29. Let σ be an infinite critical 2-structure. If W π (σ) ≠ ∅, then (RT) holds.

  By Corollary 10.18, we have (v n , w) ∈ A(C(σ)). It follows from Lemma 10.20 that (w, v n ) ∈ A(C(σ)). Consequently, we obtain C(σ)[{v n ∶ n ≥ 0} ∪ {w}] ≃ L N , which contradicts Lemma 10.26. It follows that

  .36) Let w ∈ V (σ -v p )∖X p . There exists N p ∈ Υ(σ -v p ) such that w ∈ N p . We obtain w ∈ (X p ) σ (z), where z denotes the unique element of X p ∩ N p . It follows from Lemma 3.13 that V (σ -v p )∩⟨X p ⟩ σ = ∅. For a contradiction, suppose that there exists N p ∈ (Υ(σv p ) ∖ {M p }) such that N p ≥ 2. Let w ∈ N p ∖ X p . It follows from (10.33),(10.35), and (10.36) that w

  Since v p ←→ σ {v p+1 , v p+2 }, we have v p ←→ σ {v m ∶ m > p}. It follows from(10.33) that v p ←→ σ (M p ∩ (W π (σ) ∖ {v p })). Since σ is prime, M p is not a module of σ. Therefore, v p ←→ σ M p . Thus, there exists w ∈ M p such that v p ←→ σ {v p+1 , w}. Since v p ←→ σ {v m ∶ m > p}, we have w ∈ {v m ∶ m > p}. It follows from (10.35) that w ∈ {v m ∶ m < p}. We obtain w ∈ W π (σ). Furthermore, it follows from (10.36) that w ∈ W ∅ (σ) ∪ W δ (σ). We obtain w ∈ E (σ). Hence, V (σ) ∖ {w, F E (σ) (w)} is a module of σw (seeNotation 10.8). As previously, it follows from Fact 10.6 that (V (σ) ∖ {v p , w}) ⊆ (V (σ) ∖ {w, F E (σ) (w)}) and we obtain v p = F E (σ) (w). By (10.33), we have v p

.37) Finally, we show that for each p ≥ 1, v p ←→ σ {v p+1 , v p+2 } (see Notation 2.1). (10.38) Otherwise, there exists p ≥ 1 such that v p ←→ σ {v p+1 , v p+2 }. Recall that M p denotes the unique element of Υ(σv p ) containing v p+1 . By (10.31), {v m ∶ m > p + 1} is a module of σ[W π (σ)]v p+1 . We obtain v p ←→ σ {v m ∶ m > p + 1}.

  n is an isomorphism from L N onto L δ (σ). Let n ≥ 0. It follows from(10.47) that [v n , v n+1 ] σ ≠ (e, f ) and [v n , v n+p ] σ = (e, f ) for every p ≥ 2. Consequently, there exists a directed path τ on Z (seeDefinition 10.30) such that σ is isomorphic to τ [N]. By Remark 10.31, σv 0 is prime, which contradicts the fact that σ is critical. Similarly, if L δ (σ) is isomorphic to (L N ) ⋆ , then σ is not critical. It follows that L δ (σ) is isomorphic to L Z . Hence, there exists a sequence (v n ) n∈Z of elements of V (σ) such that ϕ ∶ Z → {v n ∶ n ≥ 0} n → v n is an isomorphism from L Z onto L δ (σ).It follows from (10.47) that ϕ is an isomorphism from σ onto a directed path on Z. By Remark 10.31, σ[{v n ∶ n ≥ 0}] is prime, so (RT) holds.

  ). By exchanging v and w if necessary, we can assume thatC w v is equipotent to V (σ). We verify that σ[C w v ] is prime. For a contradiction, suppose that M is a nontrivial module of σ[C w v ]. Clearly, M is a module of σ-v. Since σ is prime, there exists t ∈ M such that ≺t, v ≻ σ ≠ {e}. Let s ∈ M ∖{t}. Since C w v is {e}-connected and M is a module of σ[C w v ], σ[C w v ]s is {e}-connected as well. Since ≺t, v ≻ σ ≠ {e} and t ∈ (C w v ∖ {s}), we obtain that (C w v ∖ {s}) ∪ {v} is {e}-connected. Moreover, by (10.51), V (σ) ∖ C w v is {e}-connected. Since v ∈ (V (σ) ∖ C w v ) ∩ ((C w v ∖ {s}) ∪ {v}),

	(V (σ) ∖ C w v ) ∪ ((C w

v ∖ {s}) ∪ {v}), which is V (σ) ∖ {s}, is {e}-connected,

which contradicts (10.48) and (10.49). Consequently, σ[C w v ] is prime. It follows that (RT) holds. Theorem 10.1 follows easily:

•

  Suppose that t < 2n -1. Recall that P(σ -(2n + 1)) = P 2n+1 . It follows from Lemma 4.39 applied to (σ -(2n + 1)) -{t, t + 1} thatE(P(µ)) ={{k, k + 1} ∶ k ∈ {0, . . . , t -2} ∪ {t + 2, . . . , 2n -1}} (B.30) ∪ {{t -1, t + 2}}.We obtainN P(µ) (t -1) = {t -2, t + 2}.Since (B.16) holds, we have ϕ -1 (j + 1) = t -1.

  ). It follows from Proposition 4.15 that ≺0, 1≻ σ ≠≺0, 2≻ σ (see Notation 1.1).

(B.38)

  ←→ σ W ′ , W ←→ σ W ′ , W ⋆ , 1 W δ (σ), 174 W ∅ (σ), 174 W Z , 107 W π (σ), 174 W 2m+1 , 56 X σ (y), 27 X

			INDEX
	neighbourhood, 3 M 2 , 87	according to, 110 σ[W ], 1
	outside graph, 137 N 2 , 90	module, 7 σ ∨ τ , 38
	primality graph, 34 self-complementary, 3 E (σ), 171 F N , 104	strong module, 13 trivial, 13 σ ∧ τ , 38 σ ⋆ , 1
	the bull, 21 F Z , 104	trivial, 8 ≈ (e,f ) , 5
	Index 2-structure, 1 W -critical, 137 ν(σ)-connected, 18 {e, f }-components, 6 {e, f }-connected, 6 e-constant, 1 e-linear, 4 almost critical, 71 asymmetric, 2 component, 18 connected, 6 constant, 1 critical, 33 critical subset, 33 critical vertex, 33 cuttable, 11 decomposable, 8 directed path, 192 dual, 1 extreme vertex, 171 finitely critical, 103 indecomposable, 8 linear, 4 locally critical, 108 partially critical, 137 prime, 8 minimal, 85 primitive, 13 N σ P N , 103 P Z , 103 P ∧ Q, 18 P ∨ Q, 18 (v), 24 (e,f ) N G (v), 3 L m , 4 N + N -D (x), 3 D (v), 3 modular partition, 9 D ⋆ , 3 D 2m+1 , 53 E(G), 2 E(σ), 1 F E (σ) , 172 G ⊕ H, 3 H N , 107 H Z , 107 H 2n , 42 K S , 2 L S , 4 σ Q P , 110 vertex set, 2 identity function, 2 isomorphic, 2 isomorphism, 2 join, 17, 38 P n (σ), 21 P (σ,X) , 137 Symbol Index R n (σ), 21 S (σ), 33 S c (σ), 65 ν(σ), 18 G, 2 (W, W ′ ) σ , 5 π n , 47 lattice, 18 length, 2 linear order, 4 meet, 18, 38 modular cut, 11 trivial, 11 (W, v) σ , 5 ≺v, w ≻ σ , 1 σ(D), 3 (v, W ) σ , 5 (v, w) σ , 1 A(D), 3 B 2n+1 , 21 C n , 3 σ(G), 3 σ -W , 1 σ -w, 1 σ P , 9	vertex-transitive, 2 2-substructure, 1 automorphism, 2 automorphism group, 2 coatoms, 18 degree, 3 digraph, 3 arc set, 3 criticality digraph, 169 dual, 3 in-neighbourhood, 3 out-neighbourhood, 3 transitive, 4 vertex set, 3 graph, 2 comparability graph, 4 complement, 2 complete, 2 component, 3 connected, 3 disjoint union, 3 ⟨X⟩ σ , (e,f ) ⟨X⟩ σ , 27 ∪Q, 10 Υ(σ), 169 Π(σ), 15 Γ (σ,X) , 137 bipartite, 2 ∆(S), 38 [v, w] σ , 1 [v, W ] σ , 5 embedding, 2 up-directed, 99 σ [W, W ′ ] σ , 5 (y), 141 W (e,f ) cycle, 3 outside partition, 28 π 2n , 50 d G (x), 3 p (σ,X) , 27 partial order, 4 path, 2 quotient, 9 generalized quotient, 110 q (σ,X) , 141 q a (σ,X) , 150 q s (σ,X) , 150 v ←→ σ W , 5 U N , 107 v ←→ σ W , 5 U Z , 106 v ∼ σ w, 177 support, 33 critical support, 65 tournament, 4 Cayley tournament, 57 U 2m+1 , 54 Aut(σ), 2 Cay(Γ, Ω), 57 V (D), 3 Cay 2m+1 , 57 V (G), 2 Comp(O), 4 V (σ), 1 W P , 10 Ext
	reversible, 1 P n , 2	edge set, 2 C(σ), 169
	rigid, 2 Q 2m , 43	empty, 2 P(σ), 34
	symmetric, 2 R 2m , 44	half-graph, 42 C(Γ), 151
	uncuttable, 11 T N , 131	isolated vertex, 3 C(σ), 18
	vertex set, 1 T Z , 129	multipartite, 2 C {e,f } (σ), 6
	vertex subset, 1 T 2n+1 , 4	2. neighbour, 3 M 1 , 85	(B.59)
	237 243

σ (X), 27 Id V (σ) , 2

In general, a reversible

2-structure is neither symmetric nor asymmetric.

Ehrenfeucht et al. [13,[START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF] use primitive instead of indecomposable.

Also called cut in[START_REF] Boussaïri | The C 3 -structure of the tournaments[END_REF] for digraphs.

Ehrenfeucht, Harju and Rozenberg [14] say that a 2-structure has the 2-block property if it is cuttable.

Recall that we consider finite 2-structures.

Cournier and Ille [12] established this proposition for digraphs. Ille[START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF] proved this proposition for binary structures, that is, labeled

2-structures[START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF].

Cournier and Ille [12] proved this corollary for digraphs. Ille[START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF] proved this proposition for binary structures, that is, labeled 2-structures[START_REF] Ehrenfeucht | The Theory of 2-Structures, A Framework for Decomposition and Transformation of Graphs[END_REF]

Sumner [35] demonstrated this theorem for graphs.

Cournier and Ille [12] proved this theorem for digraphs, and Ille[START_REF] Ille | La décomposition intervallaire des structures binaires[END_REF] for binary structures by using the same proof.

The same approach is adopted in[START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] to characterize the critical infinite digraphs.

Boudabbous and Ille [7] proved this theorem for digraphs.

Boudabbous and Ille [7] proved this theorem for digraphs.

Boudabbous and Ille [7] proved this theorem for digraphs.

Boudabbous and Ille [7] proved this theorem for digraphs.

Schmerl and Trotter [33] proved this theorem for binary relational structures. The cases of partially ordered sets, graphs, and tournaments are specified.

Chudnovsky and Seymour [10] proved this theorem for graphs.

Liu [27] proved this theorem for tournaments.

Sayar [31] proved this theorem for digraphs.

Alzohairi and Boudabbous [1] characterized the minimal prime graphs for a vertex subset of size 3 that do not contain K {0,1,2} as an induced subgraph.

Cournier and Ille [12] proved this theorem for digraphs.

It is not difficult to verify that Proposition

3.8, Corollary 3.9, and hence Theorem 3.10 hold for infinite prime 2-structures as well.

Boubabbous and Ille[START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] proved this theorem (see[START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] Theorem 12]) for digraphs.

Boubabbous and Ille[START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] proved this theorem (see[START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] Theorem 13]) for digraphs.

Boubabbous and Ille[START_REF] Boudabbous | Critical and infinite directed graphs[END_REF] proved this theorem for digraphs.

Breiner et al. [5] proved this lemma for (finite) graphs (see[START_REF] Breiner | Partially critical indecomposable graphs[END_REF] Lemma

2.7]).

Breiner et al. [5] proved this lemma for (finite) graphs (see[START_REF] Breiner | Partially critical indecomposable graphs[END_REF] Lemma 4.3]).

Breiner et al. [5] proved this lemma for (finite) graphs (see[START_REF] Breiner | Partially critical indecomposable graphs[END_REF] Lemma

4.4]).

Breiner et al. [5] proved this lemma for (finite) graphs (see[START_REF] Breiner | Partially critical indecomposable graphs[END_REF] Corollary 4.5]).

Rigollet and Thomassé [29] proved this theorem for infinite digraphs

We use the axiom of choice to prove Theorem 10.1. We also use the axiom of choice to prove some of the preliminary results that follow, and we mention its use in their proofs only.

APPENDIX B. PROOFS OF PROPOSITIONS 5.27, 5.28, AND 5.29 Bibliography

Furthermore, since {m, m + 2} is a module of τ -(m + 1), we have

Therefore, we obtain

By proceeding by induction, we obtain that the second assertion holds. Conversely, suppose that both assertions above hold. Since the second assertion holds, we obtain the following. Given m, n, p, q ∈ Z such that m < n and p < q, if m ≡ p mod 2 and n ≡ q mod 2, then [m, n] τ = [p, q] τ .

It follows that for every n ∈ Z, {n -1, n + 1} is a module of τn. To conclude, we have to verify that [n, n -1] τ ≠ [n, n + 1] τ for every n ∈ Z. Let n ∈ Z. For instance, suppose that n is even. We obtain

Moreover, we have

Since [1,[START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF] τ ≠ [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ , we obtain [n, n -1] τ ≠ [n, n + 1] τ . The case n odd is handled similarly.

Lemma 8.7. Given a 2-structure τ such that V (τ ) = N, τ ∈ F N if and only if the following three assertions hold

• [1,[START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF] τ ≠ [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ ;

• for m, n ∈ N such that m < n, we have [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ , and [2m + 1, 2n + 1] τ = [1,3] τ ;

• [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ = [1,3] τ .

Proof. To begin, suppose that τ ∈ F N . As in the proof of Lemma 8.6, we obtain that the first two assertions hold. Since τ ∈ F N , {2, 3, . . .} is a module of τ -0.

Hence, [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ = [1,3] τ . Conversely, suppose that the three assertions above hold. Let n ≥ 1. As in the proof of Lemma 8.6, we obtain that {n-1, n+1} is a module of τ -n and not of τ . Moreover, it follows from the last two assertions that {2, 3, . . .} is a module of τ -0. Lastly, {0} ∪ {2, 3, . . .} is not a module of τ because [1,[START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF] τ ≠ [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ . Lemma 8.8. Given τ ∈ F Z , the following four assertions hold

• for each n ∈ Z, {n, n + 1, . . .} is a module of τ if and only if [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF]1] τ = [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF][START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ and [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ = [1,3] τ ;

• for each n ∈ Z, {. . . , n -1, n} is a module of τ if and only if [1,[START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ = [START_REF]Since {0, 1} ∈ E(P(σ -(2n))), (σ -(2n)) -{0, 1} is prime. Set X = V (σ) ∖ {0[END_REF][START_REF] Belkhechine | The prime tournaments T with W 5 (T ) = T -2[END_REF] τ and [0, 1] τ = [1,3] τ ;

Theorem 9.6 (Belkhechine et al. [3]). Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. Suppose that Statement (S5) holds. The following three assertions are equivalent 1. σ is (V (σ) ∖ X)-critical;

2. for each component C of Γ (σ,X) , σ[X ∪ V (C)] is V (C)-critical;

3. for each component C of Γ (σ,X) , v(C) = 2 or v(C) ≥ 4 and C is critical.

Remark 9.7. As seen at the beginning of Chapter 8, σ(P Z ) is finitely critical. Set X = {z ∈ Z ∶ z ≤ 0}.

As for σ(P Z ), it follows from Theorem 7.1 that σ(P Z )[X] is prime. Similarly, σ(P Z )[X ∪ {1, . . . , k}] is prime for every k ≥ 1. Consequently, for each k ≥ 1, Statement (Sk) does not hold. Moreover, {1, 2} is the only edge of Γ (σ(P Z ),X) . Hence, for every z ≥ 3, z is an isolated vertex of Γ (σ(P Z ),X) . It follows that Theorem 9.5 does not hold when Statement (S3) is not satisfied. Similarly, Theorem 9.6 does not hold when Statement (S5) is not satisfied.

We introduce a weakening of the partial criticality in the following way. We obtain the next result by using Theorem 7.4 several times. Corollary 9.8. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. The following two assertions are equivalent 1. σ is prime; 2. for each finite subset F of V (σ) ∖ X, there exists F ′ ∈ P (σ,X) such that F ′ is finite and F ⊆ F ′ .

Corollary 9.8 and the fact that Statement (S5) is supposed to be satisfied in Theorem 9.6 lead us to introduce the next definition. The next definition is a weakening of partial criticality (see Theorem 9.10). Definition 9.9. Given a 2-structure σ, consider X ⊊ V (σ) such that σ[X] is prime. We say that σ is finitely (V (σ) ∖ X)-critical if for each finite subset F of V (σ) ∖ X, there exists

The following two assertions are equivalent 1. Statement (S5) holds and σ is prime;

Theorem 9.10 is discussed in Remark 9.57. Precisely, in Remark 9.57, we provide a prime 2-structure showing that we do not have a compactness theorem with partial criticality. We prove Theorem 9.10 at the end of Section 9.5. The last main result ends this section. It shows that Theorem 5.8 is satisfied in the infinite case when the 2-structure σ is also supposed to be (V (σ) ∖ X)-critical.

Therefore, the linear order L is unique.

Furthermore, denote by ϕ(L) the unique linear order defined on Y such that ϕ is an isomorphism from L onto ϕ(L). We obtain

Consequently, Γ is also a half graph by considering the linear order ϕ(L) ⋆ defined on Y (see Section 1.3), and the bijection ϕ -1 ∶ Y → X. Definition 9.35. A linear order L is discrete [START_REF] Rosenstein | Linear orderings[END_REF] if the following two conditions are satisfied 1. for every v ∈ V (L), if v is not the least element of L, then v admits an immediate predecessor;

2. for every v ∈ V (L), if v is not the greatest element of L, then v admits an immediate successor.

Definition 9.36. A half graph is discrete if the linear order L in Definition 9.33 is discrete.

In the next observation, we explain how to decompose suitably a discrete linear order (see Definition 9.35).

Observation 9.37. Given an infinite linear order λ, λ is discrete if and only if λ admits a modular partition P satisfying the following conditions.

Proof. As in the proof of Lemma 10.20, consider the set

Let X ∈ X . It follows from (10.8) and (10.10) (see the proof of Lemma 10.20) that (V (σ) ∖ X) ⊆ ⋃ y∈X X σ (y) (see Notation 3.12). (10.11) In particular, there exist y, z ∈ X such that v ∈ X σ (y) and w ∈ X σ (z). For each t ∈ X ∖ {y}, we have

(10.12) Indeed, consider u ∈ X σ (t). Since t ≠ y, we have u ≠ v. Thus, there exists

It follows from the maximality of the elements of Υ(σv) that ({t} ∪ X σ (t)) ∈ Υ(σv). Hence, (10.12) holds. Similarly, we have

By Lemma 10.20, we have w ∈ W π (σ). Hence, Υ(σ-w) is a modular partition of σw and (σw) Υ(σw) is prime. Now, we establish the analogues of (10.12) and (10.13) for Υ(σw). We verify that for each M w ∈ Υ(σw), we have (10.9) in the proof of Lemma 10.20). (10.14) We can assume that M w ≥ 2, so M w is a nontrivial module of σw. As seen in Remark 3.16, we have M w ∩ X ≤ 1 or X ⊆ M w . For a contradiction, suppose that X ⊆ M w . Let u ∈ V (σw) ∖ M w . Since X ⊆ M w , we have u ∈ ⟨X⟩ σ (see Notation 3.12). By (10.11), u ∈ X σ (t), where t ∈ X. We obtain u ∈ X σ (t) ∩ ⟨X⟩ σ , which contradicts Lemma 3.13. It follows that M w ∩ X ≤ 1. Hence, (10.14) holds. Consequently, there exists Y ⊆ V (σ) ∖ {w} such that for each M w ∈ Υ(σw), the following three assertions hold

Suppose for a contradiction that y = z. Since {y} ∪ X σ (y) is a module of σw by (10.16) and {z} ∪ X σ (z) is a module of σv by (10.12), {y} ∪ X σ (y) ∪ X σ (z) is a module of σ, which contradicts the fact that σ is prime. It follows that y ≠ z.

Since {y} ∪ X σ (y) is a module of σw, we obtain {u, u ′ } ∈ E(Γ (σ,X) ) (see Definition 9.3) for any u ∈ X σ (y) and u ′ ∈ X σ (z) ∖ {w}. Similarly, since {z} ∪ X σ (z) is a module of σ -v, we obtain {u, u ′ } ∈ E(Γ (σ,X) ) for any u ∈ X σ (y)∖{v} and u ′ ∈ X σ (z). By Theorem 3.19, Γ (σ,X) is nonempty. Therefore, we have

is a nontrivial module of σ, which contradicts the fact that σ is prime. Consequently, we have

Second, suppose for a contradiction that λ(v) ≠ λ(w). There exist a nontrivial modular cut C w of σw and e ′ , f

from Corollary 10.16 that σv 0 admits a nontrivial module N 0 containing v p and w. It follows from the maximality of M 0 that N 0 ⊆ M 0 . In particular, we have w ∈ M 0 . Since V (σ) ∖ {w, v p } is a module of σw, we obtain v p ∈ ⟨X 0 ⟩ σ . Therefore, we have v p ∈ (X 0 ) σ (v p+1 ) ∩ ⟨X 0 ⟩ σ , which contradicts Lemma 3.13. Consequently, (10.38) holds.

To conclude, we verify that σ

For a contradiction, suppose that p ≥ 2. It follows from (10.31) and (10.38) that v p-1 ←→ σ {v p , v q }. Thus, we have p = 1. Proceeding by induction, it follows from (10.40) that v r ∈ M for every r ≥ q + 1. By (10.38), v q-1 ←→ σ {v q , v q+1 }. Since v q , v q+1 ∈ M , we have v q-1 ∈ M . It follows from the minimality of q that q -1 = 1. Therefore, we obtain

We use the following definition and remark in the proof of the next proposition. We need the next two results to prove Theorem 5.21. The next lemma has to be compared with Corollary 5.14.

Lemma A.1. Let σ be a 2-structure such that v(σ) ≥ 7. If σ is prime, and neither critical nor almost critical, then there exists

Proof. Since σ is not critical, we have S (σ) ≠ ∅. If S (σ) ≥ 2, then it suffices to apply Theorem 5.10. Now, suppose that S (σ) admits a unique element denoted by x. Since σ is not almost critical, we have S c (σ) = ∅. It follows that σx is prime and noncritical.

Proposition A.2. Let σ be a 2-structure such that v(σ) ≥ 7. If σ is prime, and neither critical nor almost critical, then there exists v ∈ V (σ) such that σv is prime, and neither critical nor almost critical, as well.

The proof of Proposition A.2 is long and technical. We decompose it into several claims.

The beginning of the proof of Proposition A.2. By Lemma A.1,

If there exists v ∈ S (σ) ∖ S c (σ) such that S (σv) ≥ 2, then it follows from Theorem 5.10 applied to σ -v that σ -v is prime, and neither critical nor almost critical.

To continue, suppose that

, {v, w} is a module of σ, which contradicts the fact σ is prime. It follows that f is injective.

, and hence

For a contradiction, suppose that w ∈ S (σ). Since σ -{f (v), w} is prime, we obtain w ∈ S (σ) ∖ S c (σ), and f (w) = f (v), which contradicts Claim A.5. It follows that w ∈ S (σ).

Observe that π 2n ○ ϕ v (see (A.4)) is also an isomorphism from (σv)f (v) onto (τ v ) ⋆ , with (τ v ) ⋆ ∈ R 2n , satisfying (5.7). Therefore, we can assume that

where p ∈ {0, . . . , n -1}. Since w ∈ S (σ), we have d P(σ) (w) ≤ 2 by Lemma 4.4. Since σ -{f (v), w} is prime, we obtain

We distinguish the following two cases. Each of them leads to a contradiction.

1. Suppose that d P(σ) (w) = 1. Since σ -{f (v), w} is prime, we have

It follows from Lemma 4.4 that

which contradicts the fact that ϕ v is an isomorphism satisfying (5.7). Indeed, since n ≥ 3 by (A.3), there exists q ∈ {0, . . . , n -1} ∖ {p}. Since (5.7) is satisfied by ϕ v , we have

which contradicts (A.8).

Claim A.7. We have S (σ) ∖ S c (σ) = 1.

Proof. Otherwise, consider distinct v, w ∈ S (σ)∖S c (σ). Since f is injective, we have f (v) ≠ f (w). Furthermore, it follows from Claim A.6 that N P(σ) (f (v)) = {v} and N P(σ) (f (w)) = {w}. As previously noted, by considering ϕ v ○ π 2n (see (A.4)) instead of ϕ v , we can assume that

where p ∈ {0, . . . , n -1}. Since

Since w = (ϕ v ) -1 (2p), we obtain p = n -1 and ϕ v (f (w)) = 2n -1. Therefore, we have

As observed in Remark 5. [START_REF] Fraïssé | L'intervalle en théorie des relations, ses généralisations, filtre intervallaire et clôture d'une relation[END_REF],

Consequently, {(ϕ v ) -1 (2n -3), (ϕ v ) -1 (2n -1)} is a module of σw, which contradicts the fact that w ∈ S (σ).

The end of the proof of Proposition A.2. We conclude as follows. By Claim A.7, S (σ)∖S c (σ) admits a unique element denoted by v. By Claim A.6,

Moreover, we have f (v) ∈ S (σ) by Claim A.4. Since N P(σ) (f (v)) = {v} by Claim A.6, it follows from Lemma 4.4 that

We obtain ϕ -1 (1) ∈ ⟨Y ⟩ σ . Consequently, ϕ -1 (1) ∈ Y σ (w) ∩ ⟨Y ⟩ σ , which contradicts Lemma 3.13. It follows that

Since (ϕ v ) -1 (0) ∈ S (σ), it follows from Lemma 4.4 that

It follows that (A.10) holds. In particular, we have

Similarly, we have

In particular, we obtain

By (A.6), we have

Therefore, it follows from (A.12) that

which contradicts (A.9).

Proof of Theorem 5.21. We proceed by induction on v(σ

For convenience, we denote by N (σ) the set of v ∈ V (σ) such that σv is prime, and neither critical nor almost critical. By Proposition A.2,

To begin 1 , we prove that there exists X ⊊ V (σ) such that

1 From here until (⋆) (see page 207), the proof is similar to that of Theorem 5.19.

Consider Y ⊆ V (σ) such that σ[Y ] ≃ τ , and suppose that σu is decomposable for every u ∈ V (σ) ∖ Y . It follows from Corollary 3.21 that there exist distinct v, w ∈ V (σ) ∖ Y such that σ -{v, w} is prime. Thus, τ embeds into σ -{v, w}. Denote by C the component of P(σ) containing v and w. For a contradiction, suppose that V (C) ⊆ V (σ) ∖ S (σ). By Proposition 4.5, V (σ) ∖ V (C) ≤ 1, so S (σ) ≤ 1. Since σ is not critical, we have S (σ) = 1. By Theorem 5.13, C is the unique component of P(σ) such that v(C) ≥ 2. If V (C) ∩ S (σ) = ∅, then it follows from Theorem 5.13 that σ is almost critical. Consequently, we have V (C) ∩ S (σ) ≠ ∅. Therefore, there exist distinct vertices c 0 , . . . , c p of C satisfying

• p ≥ 2, {c 0 , . . . , c p-1 } ⊆ V (σ) ∖ S (σ), and c p ∈ S (σ);

Since τ embeds into σ -{v, w}, τ embeds into σ -{c p-1 , c p } as well. Since c p ∈ S (σ), (A.13) holds. Now, we consider X ⊆ V (σ) such that (A.13) holds. There exists

If there exists w ∈ (V (σ)∖X)∩N (σ) ≠ ∅, then it suffices to apply the induction hypothesis to σw. Hence, suppose that

Since σv is prime, σv is critical or almost critical. We distinguish the following two cases.

Appendix B

Proofs of Propositions 5.27, 5.28, and 5.29

Proof of Proposition 5.27. Consider s ∈ S c (τ ) such that P(τs)

Up to isomorphism, we can assume that

The following observation is useful in what follows. Let x, y ∈ {0, . . . , 2n} such that x < y < 2n.

If {ϕ -1 (x), ϕ -1 (y)} ∩ {2n, 2n + 1} = ∅, then ϕ -1 (x) < ϕ -1 (y).

(B.2) Indeed, we have

Since x < y < 2n, we obtain

Since T 2n+1 -(2n) = L 2n and x < y, we have ϕ -1 (x) < ϕ -1 (y). First, suppose that d P(τ -( 2n+1)) (t) = 1, and denote by u the unique element of N P(τ -(2n+1)) (t). Since n ≤ t ≤ 2n-1, we obtain t = 2n-1, and hence u = 2n-2. Since N P(τ -(2n+1)) (t) = N P(τ -t) (2n + 1), we have N P(τ -t) (2n + 1) = {u}, that is, N P(τ -t) (2n + 1) = {2n -2}. Since n ≤ ϕ(2n + 1) ≤ 2n -1, we obtain ϕ(2n + 1) = 2n -1 and ϕ(2n -2) = 2n -2. Furthermore, since {t, u} ∈ E(P(τ -(2n + 1))), it follows from Lemma 4.40 that (τ -(2n + 1)) -{t, u} is critical, and

Similarly, (τt) -{2n + 1, u} is critical, and

Thus, 2n is the unique isolated vertex of P((τ -(2n + 1)) -{t, u}), that is, P(τ -{t, u, 2n + 1}). Analogously, ϕ -1 (2n) is the unique isolated vertex of P(τ -{t, u, 2n + 1}). Therefore, ϕ(2n) = 2n.

Recall that τ -(2n + 1) = σ(T 2n+1 ), and ϕ is an isomorphism from τt onto σ(T 2n+1 ). Consequently, ϕ ↾{0,...,2n-3} is an automorphism of σ(T 2n+1 ) -{2n -

) is rigid. Hence, ϕ ↾{0,...,2n-3} = Id {0,...,2n-3} .

Proof of Proposition 5.28. Consider s ∈ S c (σ) such that P(σs) ≃ P 2n+1 . Therefore, v(σ) = 2n + 2. Since v(σ) ≥ 7, we obtain

Up to isomorphism, we can assume that

Since σ -(2n + 1) is critical and P(σ -(2n + 1)) = P 2n+1 , it follows from Proposition 4.27 that

Furthermore, for any p, q ∈ {0, . . . , 2n} such that p < q, we have

As above, since P(σt) ≃ P 2n+1 , there exists an isomorphism ϕ from σt onto τ , where τ is a critical 2-structure such that P(τ ) = P 2n+1 . By Proposition 4.27,

Furthermore, for any p, q ∈ {0, . . . , 2n} such that p < q, we have

Similarly, we can assume that n ≤ ϕ(2n + 1) ≤ 2n. For a contradiction, suppose that

Since n ≤ t ≤ 2n, we have t = 2n. Hence

It follows from (B.9) that ϕ(2n + 1) = 2n, N P(σ-(2n)) (2n + 1) = {ϕ -1 (2n -1)}, and ϕ(2n -1) = 2n -1.

It follows from Lemma 4.40 that (σ -(2n + 1)) -{2n -1, 2n} is critical and

Thus, we obtain P(σ -{2n -1, 2n, 2n + 1}) = P 2n-1 . Similarly, τ -{2n -1, 2n} is critical and

Observe that (ϕ -1 ) ↾{0,...,2n-2} is an isomorphism from P(τ -{2n-1, 2n})), which is P 2n-1 , onto P(σ -{2n, 2n + 1, ϕ -1 (2n -1)}). Since ϕ(2n -1) = 2n -1, we have

It follows that ϕ ↾{0,...,2n-2} ∈ Aut(P 2n-1 ).

Therefore, we obtain ϕ ↾{0,...,2n-2} = Id {0,...,2n-2} or π 2n-1 .

We distinguish the following two cases. In each of them, we obtain a contradiction.

1. Suppose that ϕ ↾{0,...,2n-2} = Id {0,...,2n-2} . Since ϕ(2n

for each k ∈ {0, . . . , 2n -1}. We verify that {2n, 2n + 1} is a module of σ. Let p ∈ {0, . . . , 2n -1}. For instance, assume that p is even. We obtain

It follows that

Similarly, we have

(Note that if j = 2n -1, then

Since (B.15) holds, we have

We show that ϕ -1 (j -2) ≠ t -2. Otherwise, we have ϕ -1 (j -2) = t -2. By proceeding by induction, it follows from (B.18), (B. [START_REF] Harju | Decomposition of infinite labeled 2-structures, Results and trends in theoretical computer science[END_REF], and (B.21) that ϕ -1 (jk) = tk for k ∈ {1, . . . , min(t, j)}. It follows that t = j. Hence, ϕ(l) = l for l ∈ {0, . . . , t -1}. Analogously, by proceeding by induction, we obtain ϕ(l) = l for l ∈ {t + 1, . . . , 2n}. Thus, {t, 2n + 1} is a module of σ, which contradicts the fact that σ is prime. It follows that

Since ϕ -1 (j + 1) = t + 1 by (B.15), it follows from (B.18) and (B. [START_REF] Harju | Decomposition of infinite labeled 2-structures, Results and trends in theoretical computer science[END_REF]) that

By proceeding by induction, it follows from (B.18), (B. [START_REF] Harju | Decomposition of infinite labeled 2-structures, Results and trends in theoretical computer science[END_REF], and (B.21) that ϕ -1 (j + k) = tk for k ∈ {2, . . . , min(t, 2nj)}. Since t, j ∈ {n, . . . , 2n -2}, we It follows from (B.24) and (B.25) that ψ -1 satisfies (5.24). Lastly, suppose that (B.16) holds. Since (B.16) holds, we have σ -{t, t + 1, 2n + 1} = σ -{t, ϕ -1 (j -1), 2n + 1}. Thus, we have

To conclude, we distinguish the following two cases.

• Suppose that t = 2n -1. It follows from Lemma 4.40 applied to (σ -(2n + 1)) -{2n -1, 2n} that Since n ≥ 3, we obtain

It follows from (B.8) and (B.11) that [p, q] σ = (ϕ -1 (p), ϕ -1 (q)] σ for any p, q ∈ {0, . . . , 2n} such that p < q. Therefore, ϕ -1 is an isomorphism from σ-(2n+1) onto σ-t. Since t = 2n-

Furthermore, we have

It follows from (B.8) and (B.11) that

for any p, q ∈ {0, . . . , 2n} such that p < q. Therefore, ϕ -1 is an isomorphism from σs onto σt. By (B.16), we have ϕ -1 (t + 1) = t -1 and ϕ -1 (t -1) = t + 1. It follows from (B.36) that ϕ -1 satisfies (5.24).

Furthermore, for any p, q ∈ {0, . . . , 2n -1} such that p < q, we have

Similarly, we can assume that n ≤ ϕ(2n) ≤ 2n -1.

To begin with situation 1, suppose that

Since n ≤ t ≤ 2n -1, we have t = 2n -1.

By (B.40), d P(σ-t) (2n) = 1. Similarly, we have

Moreover, it follows from (B.40) that

It follows from Lemma 4.40 that (σ -(2n)) -{2n -2, 2n -1} is critical and

Observe that P(σ -{2n -2, 2n -1, 2n}) = P 2n-2 . Clearly, (ϕ -1 ) ↾{0,...,2n-3} is an isomorphism from P 2n-2 onto P(σ -{ϕ -1 (2n -2), 2n -1, 2n} for each k ∈ {0, . . . , 2n -2}. We verify that {2n -1, 2n} is a module of σ. Let v ∈ {0, . . . , 2n -2}. For instance, assume that v is even. We obtain By proceeding by induction, we obtain ϕ -1 (j -k) = t-k for k ∈ {2, . . . , min(j, t)}.

It follows that j = t. We obtain ϕ -1 (l) = l for l ∈ {0, . . . , t -2}. (B.60) Since t ≥ n and n ≥ 3, we obtain

By proceeding by induction, we obtain

We have Therefore, we have

It follows from (B.39) and (B.42) that [ϕ -1 (p), ϕ -1 (q)] σ = [p, q] σ for any p, q ∈ {0, . . . , 2n-1} such that p < q. Consequently, ϕ -1 is an isomorphism from σ -(2n) onto σt. Moreover, ϕ -1 is defined by

→ v by (B.60) and (B.62).

Consequently, (5.27) holds. We conclude as follows. For a contradiction, suppose that there exists u ∈ S c (σ) ∖ {t, 2n}. We distinguish the following two cases.

It follows from (5.26) that 2n ∈ Ext σ (X). Hence σ -{0, 1} is prime, which contradicts 0 ∈ S c (σ).

Second, suppose that d P(σ-(2n)) (t) = 2. We have t, u ∈ {1, . . . , 2n -2}. For instance, assume that t < u. We obtain that (5.27) holds, but also (5.27) holds after replacing t by u. Precisely, the function

is an isomorphism from σ -(2n) onto σt. Similarly, the function

is an isomorphism from σ -(2n) onto σu. We distinguish the following cases.

• Suppose that t ≤ u -3. Since N P(σ-(2n)) (u) = {u -1, u + 1}, it follows from Lemma 4.4 that {u -1, u + 1} is a module of (σ -(2n))u. In particular, we have Since {u -1, u + 1} is a module of (σ -(2n))u, {u -1, u + 1} is a module of σ -(2n), which contradicts 2n ∈ S (σ).

• Suppose that t = u -2. Since N P(σ-(2n)) (t + 1) = {t, t + 2}, it follows from Lemma 4.4 that {t, t + 2} is a module of (σ -(2n)) -(t + 1).

is an isomorphism from σ -(2n) onto σu. We distinguish the following cases.

• Suppose that u ≤ 2n-4. Since N P(σ-(2n)) (u) = {u-1, u+1}, it follows from Lemma 4.4 that {u -1, u + 1} is a module of (σ -(2n))u. Furthermore, we have Therefore, {u -1, u + 1} is a module of σu, which contradicts u ∈ S (σ).

• Suppose that u = 2n -3. We obtain • Suppose that u = 2n -2. We obtain Both cases above lead us to a contradiction. Consequently, S c (σ) = {t, 2n}.

Author Index

The