Franz Dietrich

Judgment aggregation and agenda manipulation

Keywords: D70, D71 judgment aggregation, multiple issues, description-sensitivity, agenda manipulation, impossibility theorems, characterization theorems Inflation? Growth? Growth implies inflation?

When individual judgments ('yes'or 'no') on some propositions are aggregated into collective judgments, outcomes may be sensitive to the choice of propositions under consideration (the agenda). Such agenda-sensitivity opens the door to manipulation by agenda setters. I de…ne three types of agenda-insensitivity ('basic', 'full', and 'focal') and for each type axiomatically characterize the aggregation procedures satisfying it. Two axioms turn out to be central for agenda-insensitivity: the familiar independence axiom, requiring propositionwise aggregation, and the axiom of implicit consensus preservation, requiring the respect of any (possibly implicit) consensus. As the paper's second contribution, I prove a new impossibility theorem whereby these two axioms imply dictatorial aggregation for almost all agendas. JEL Class.

Introduction

Imagine that the board of a central bank has to form collective judgments ('yes'or 'no') on some propositions about the economy, such as the proposition that prices will rise. Disagreements on a proposition are resolved by taking a majority vote. The chair of the board knows that a majority believes prices won't rise. Nonetheless he wants the board to form a collective judgment that prices will rise. 2 To achieve this goal, he removes the proposition 'prices will rise'from the agenda, while putting two new propositions on the agenda: 'GDP will grow', and 'growth implies in ‡ation', i.e., 'if GDP will grow, then prices will rise'. Once it comes to voting, the two new propositions are each approved by

Majority

No Yes Yes manipulated agenda initial agenda Figure 1: An agenda manipulation reversing the collective judgment on in ‡ation 1 Paris School of Economics & CNRS. Email: fd@franzdietrich.net. Web: www.franzdietrich.net. 2 The reason might be his belief in imminent in ‡ation, or his desire for the bank to raise interest rates (which happens only if the board concludes that there is an in ‡ation risk). In the …rst case he cares about the truth of collective judgments. In the second case he cares about consequences (actions) resulting from collective judgments. This paper leaves open the motivation of agenda setters. a (di¤erent) majority. The chair is pleased, since the collective beliefs in growth and in growth implying in ‡ation logically entail a belief in in ‡ation. This agenda manipulation has successfully turned an (explicit) 'no in ‡ation'judgment into an (implicit) 'in ‡ation' judgment. Table 1 illustrates this reversal in the case of a three-member board.

This example shows that majority voting is vulnerable to agenda manipulation. Which rules (if any) are immune to agenda manipulation? This paper de…nes di¤erent types of agenda sensitivity, and characterizes the aggregation rules immune to each type. Two axioms on the aggregation rule turn out to play key roles in ensuring manipulationimmunity: independence (i.e., the analogue for judgment aggregation of Arrow's axiom of independence of irrelevant alternatives for preference aggregation), and implicit consensus preservation (i.e., the principle of respecting unanimity, in a strengthened version extended to implicit judgments). In a new impossibility theorem, I prove that these two axioms can almost never be satis…ed by an aggregation rule which is non-dictatorial (as well as having an unrestricted domain and generating rational collective judgments). This impossibility theorem is also of interest in its own right, i.e., independently of the issue of agenda manipulation. Indeed the two axioms need not be motivated by considerations of agenda manipulation. The paper therefore has two main contributions: an analysis of agenda manipulation, and the proof of a new impossibility theorem.

The present analysis of agenda sensitivity …lls a gap in the literature on judgment aggregation, in which agenda sensitivity/manipulation is often mentioned informally and was treated in a semi-formal way by Dietrich (2006). 3 Other types of manipulation have however been much studied. One type is the manipulation of the aggregation rule, more precisely of the order of priority in which a sequential aggregation rule considers the propositions in the agenda (List 2004, Dietrich and List 2007c, Nehring, Pivato and Puppe 2014). Another type of manipulation is strategic voting, in which voters do not report truthfully their judgments. Strategic voting has been studied using two di¤erent approaches. One approach focuses on opportunities to manipulate, setting aside the behavioural question of whether voters take these opportunities or vote truthfully (e.g., Dietrich and List 2007b, Dokow and Falik 2012). The other approach focuses on incentives to manipulate, i.e., on actual voting behaviour (e.g., Dietrich and List 2007b, Dokow and Falik 2012, Ahn and Oliveros 2014, Bozbay, Dietrich and Peters 2014, DeClippel and Eliaz 2015; see also Nehring and Puppe 2002). The …rst approach requires only a basic, preference-free judgment-aggregation setup, whereas the second approach requires modelling voters'preferences (and their private information, if any). The present paper studies whether an agenda setter has opportunities to manipulate via the choice of agenda. I leave open whether he is himself a voter or an external person, and whether he takes such opportunities or refrains from manipulation. The latter question depends on his preferences, which are not modelled here. Although manipulation behaviour is not addressed explicitly, it is overly clear that manipulation opportunities will lead to manipulation behaviour under many plausible preferential assumptions. 4 The paper's second contribution -a new impossibility theorem -connects to a series of impossibility results in the …eld; see for instance List and Pettit (2002), Pauly and van Hees (2006), Dietrich (2006), Dietrich and List (2007a), Mongin (2008), Nehring and Puppe (2008), Duddy and Piggins (2013), and papers in the Symposium on Judgment Aggregation in Journal of Economic Theory (C. List and B. Polak eds., 2010). Of particular interest to us is a theorem which generalizes Arrow's Theorem from preference to judgment aggregation (Dietrich and List 2007a and Dokow and Holzman 2010, both building on Nehring and Puppe 2010 and strengthening Wilson 1975). The new theorem shows that if in the generalized Arrow theorem the Pareto-type unanimity condition is extended towards implicit agreements, then, perhaps surprisingly, the dictatorship conclusion now holds for almost all agendas, not just agendas of a quite special structure.

I should mention a growing branch of the literature which constructs concrete judgment aggregation rules, and whose attention I hope to draw to agenda manipulation. Many proposals have been made. Our analysis will imply that almost all proposals are vulnerable to agenda manipulation, yet in di¤erent ways and to di¤erent degrees. 5 The paper is structured as follows. Section 2 de…nes the framework. Section 3 states and explains the impossibility theorem on propositionwise and implicit consensus preserving aggregation. Sections 4 and 5 address agenda-sensitivity, stating characterization and impossibility results. Section 6 adds concluding remarks. Appendix A de…nes an alternative framework (more typical for judgment-aggregation theory) in which all our results continue to hold. Appendix B contains all proofs.

2 The framework I now de…ne the judgment-aggregation framework (e.g., List and Pettit 2002 and Dietrich 2007, 2014). I de…ne it in a semantic version, which takes propositions to be sets of possible worlds ('events') rather than abstract or syntactic objects. The semantic way of thinking is uncommon in the …eld, but familiar elsewhere in economics, and convenient in this paper. 6 But nothing hinges on using this framework: all formal results in the main text continue to hold in a general framework which is de…ned in Appendix A.

A group of n individuals, labelled i = 1; :::; n, needs to form yes/no judgments on some interconnected propositions. We assume that n 3. 7 The agenda. Let be a …xed non-empty set of possible worlds or states. A proposition or event is a subset p ; its negation or complement is denoted p := nA. Those propositions on which judgments ('yes' or 'no') are formed make up the agenda. As usual, I assume that the agenda is a union of pairs fp; pg, the issues on the agenda. A board of a central bank might deal with the issues fgrowth, no-growthg, fin ‡ation, no-in ‡ationg, and so on. Formally:

De…nition 1 An agenda is a proposition set X 2 which is closed under negation, i.e., p 2 X , p 2 X, and which (in this paper) is …nite and contains at least one proposition p 6 = ; ?. Each pair fp; pg X is an issue of the agenda. 8 The closure under negation of a proposition set Y is denoted Y := S p2Y fp; pg. So I can conveniently write an m-issue agenda as X = fp 1 ; :::; p m g , where p i belongs to the i th issue. An individual's judgment set is the set of propositions in X he believes.

The following are standard requirements on judgment sets:

De…nition 2 Given an agenda X, a judgment set J X is consistent if

T p2J p 6 = ?,
complete if it contains a member of each issue fp; pg X, and rational if it is both consistent and complete. The set of rational judgment sets is denoted J X or just J .

As a concrete example, assume = f0; 1g 3 . In a world (j; k; l) 2 the …rst component j indicates whether it is sunny (1) or not (0), the second one k whether it is warm (1) or not (0), and the third one l whether it is windy (1) or not (0). Consider the propositions p = f(j; k; l) 2 : j = 1g (it's sunny), q = f(j; k; l) 2 : k = 1g (it's warm) and r = f(j; k; l) 2 : l = 1g (it's windy). Here are some potential agendas: X = fp; q; rg ; X = fp \ q; p \ r; q \ rg ; X = fp; p [ q; p [ rg :

(1)

The …rst of these agendas has no logical interconnections between its issues: all 2 3 = 8 judgment sets consisting of one proposition from each issue (i.e., fp; q; rg, fp; q; rg, fp; q; rg, fp; q; rg, ...) are consistent, hence in J . The other two agendas have interconnected issues. For instance, the judgment set fp \ q; p \ r; q \ rg is inconsistent, hence not in J .

The scope of the agenda. A judgment set J X typically settles many more propositions than those it explicitly contains, where 'settling a proposition'means entailing whether it is true or false. For instance, although the …rst agenda in (1) does not contain the proposition p \ q, this proposition is settled by the judgments on p and on q.

In fact, for the …rst agenda any judgment set J 2 J settles all propositions p ; I shall say that all propositions are in the agenda's scope. By contrast, for the other two agendas in (1) some propositions p may remain unsettled, i.e., are out of the agenda's scope. The following de…nitions make all this precise. De…nition 3 A proposition p (or proposition set J) entails a proposition p 0 (or proposition set J 0 ) if p (resp.

T q2J q) is a subset of p 0 (resp. T q2J 0 q).
De…nition 4 (Dietrich 2006) A proposition set J settles a proposition p if it entails p or entails p. The scope of an agenda X is the set X of propositions settled by each rational judgment set J 2 J X ; equivalently, it is the closure of X under union (or intersection) and negation, i.e., the algebra generated by X.

The scope of an agenda can be quite large. It evidently contains all agenda propositions and all unions and intersections of agenda propositions. We can continue: it contains all negations of unions of agenda propositions, all intersections of negations of unions of agenda propositions, and so on. In short, the scope contains all propositions constructible from agenda propositions. For instance the scope of the …rst agenda in (1) contains all propositions: X = 2 . Indeed, each rational judgment set uniquely determines a single world (e.g., fp; q; rg determines the world (1; 1; 0)), hence is able to settle every proposition. Note that the scope X of an agenda X is itself an agenda, where X X.

De…nition 5 Two agendas X and X 0 are equivalent if they have same scope X = X 0 .

For instance agendas X = fp; qg and X = fp \ q; p \ q; p \ q; p \ qg have the same scope and are thus equivalent. Equivalent agendas represent essentially the same decision problem, but framed di¤erently.

Note that for any agenda X the atoms of the scope X (the minimal non-empty propositions in X) are the intersections of maximally many mutually consistent propositions in X, i.e., the propositions T p2J p where J 2 J X .

Aggregation rules. An aggregation rule for an agenda X is a function F which to every pro…le of 'individual'judgment sets (J 1 ; :::; J n ) (from some domain, usually J n ) assigns a 'collective'judgment set F (J 1 ; :::; J n ). For instance, majority rule is given by F (J 1 ; :::; J n ) = fp 2 X : more than half of J 1 ; :::; J n contain pg and generates inconsistent collective judgment sets for many agendas and pro…les. We shall be concerned with aggregation rules whose individual inputs and collective output are rational. Such rules are functions F : J n ! J . Note that we exclude ties in this paper: our aggregation rules are by de…nition 'resolute'.

The example of preference aggregation. For a (…nite non-empty) set A of 'alternatives', let be the set of strict linear orders on A, where x y reads 'x is better than y'according to a given (objective) criterion. So worlds describe how the alternatives are (objectively) ranked. The group disagrees on the ranking. The preference agenda is de…ned as X = fxP y : x; y 2 A; x 6 = yg, where xP y is the proposition that x is better than y, i.e., xP y = f 2 : x yg (note that xP y = yP x). There is a one-to-one correspondence between rational judgment sets J 2 J and strict linear orders on A, given by xP y 2 J , x y. Aggregation rules F : J n ! J can thus be regarded as preference aggregation rules. Aside from this formal analogy between preference aggregation and judgment aggregation for the preference agenda, there is an interpretational di¤erence: preferences are usually viewed as attitudes of comparative desire, not judgments (beliefs) about an objective ranking.

3 The impossibility of implicit consensus preserving propositionwise aggregation

I now state two axioms on an aggregation rule F : J n ! J for a given agenda X; they will jointly lead to an impossibility result. Each axiom is interesting in itself, but also matters 'instrumentally'by helping to limit agenda manipulation, as will be shown in depth in Section 4.

The …rst axiom is the classical condition of 'independence' or 'propositionwise aggregation'. It requires the collective judgment on any given proposition in the agenda to depend solely on the individuals' judgments on this proposition -the judgmentaggregation analogue of Arrow's axiom of independence of irrelevant alternatives (to which it reduces in the case of the preference agenda).

Independence ('propositionwise aggregation' ): For all propositions p 2 X and pro…les (J 1 ; :::; J n ); (J 0 1 ; :::

; J 0 n ) 2 J n , if p 2 J i , p 2 J 0 i for every individual i, then p 2 F (J 1 ; :::; J n ) , p 2 F (J 0 1 ; :::; J 0 n ).
This axiom is normatively no less controversial than Arrow's analogous axiom. It is known to be necessary for preventing strategic voting. We here focus on its role in preventing agenda manipulation (Dietrich and List 2007b). As shown in Section 5, it is also necessary for preventing an agenda manipulator from being able to reverse explicit collective judgments. In short, if independence is violated, then the collective judgment on a proposition p 2 X depends on other propositions in the agenda, and can thus be reversed by the agenda setter through adding or removing other propositions.

Our second axiom requires respecting consensus, in an unusually strong sense. I …rst recall the two standard consensus conditions, which pertain to judgment-set-wise resp. proposition-wise consensus:

Unanimity preservation: F (J; :::; J) = J for each unanimous pro…le (J; :::; J) 2 J n .

Unanimity principle: For all (J 1 ; :::; J n ) 2 J n and p 2 X, if each J i contains p, so does F (J 1 ; :::; J n ):

The …rst of these axioms is weaker and almost unobjectionable. The second one is analogous to the Pareto principle (and equivalent to it for the preference agenda). Our own consensus axiom resembles the latter axiom, but strengthens it by also covering 'implicit' consensus on propositions outside the agenda. The axiom can be stated in three equivalent versions.

Implicit consensus preservation (version 1):

For every proposition p in the agenda's scope X, if each judgment set in a pro…le (J 1 ; :::; J n ) 2 J n entails p, then F (J 1 ; :::

; J n ) entails p.
This axiom is demanding. It for instance implies that whenever every individual accepts at least one of some given propositions in X, i.e., implicitly endorses their disjunction (union), then so does the collective -which might con ‡ict with majority voting since each of these propositions might be rejected by a majority. In the case of the preference agenda, the axiom for instance implies that if every individual ranks a certain alternative x in 2 nd position, i.e., implicitly endorses the proposition 'x is the 2 nd best alternative', then so does the collective -although many standard preference aggregation rules (such as Borda rule) sometimes rank in 1 st position an alternative which everyone ranks in 2 nd position.

Later I give two formal arguments for this axiom, both related to the prevention of agenda manipulation. Let me anticipate them very brie ‡y. Firstly, the axiom prevents a particularly bad form of agenda sensitivity, in which unanimously supported collective judgments, explicit or implicit ones, are being reversed (see Section 4). Secondly, the axiom is e¤ectively insensitive to redescribing ('reframing') the decision problem: the set of propositions p on which consensus must be preserved stays the same if the agenda X is replaced by a new one which has the same scope and is thereby equivalent (see Section 6).9

The axiom can be reformulated using the notion of a feature of a judgment set. Examples are the feature of containing a given proposition p 2 X, and the feature of containing at most two propositions from a given set S X. We may identify each feature with the set K J of judgment sets having the feature. In its second version, our axiom requires the collective judgment set to have each feature shared by all individual judgment sets:

Implicit consensus preservation (version 2): For every K J (every feature), if each judgment set in a pro…le (J 1 ; :::; J n ) 2 J n belongs to K (has the feature), so does the collective judgment set F (J 1 ; :::; J n ).

Intuitively, the versions 1 and 2 are equivalent because a judgment set J 2 J has a given feature just in case it entails a certain proposition from the scope. For instance, J contains two given propositions q and r from X just in case it entails the proposition q \ r from X. In its third version, the axiom requires the collective judgment set to be selected from the set of individual judgment sets:

Implicit consensus preservation (version 3): For every pro…le (J 1 ; :::; J n ) 2 J n , the collective judgment set F (J 1 ; :::; J n ) belongs to fJ 1 ; :::; J n g. This axiom is far from an (undemocratic) dictatorship requirement, since the individual whose judgment set becomes the collective one may vary with the pro…le; he could for instance be the pro…le's 'median'voter in a suitably de…ned sense.

Proposition 1 The three versions of implicit consensus preservation are equivalent.

I now combine our two axioms into an impossibility result. An aggregation rule F : J n ! J is dictatorial if there is an individual i such that F (J 1 ; :::; J n ) = J i for all J 1 ; :::; J n 2 J . As usual in the theory, the structure of the agenda matters. The agenda X is called nested if it takes the very special form X = fp 1 ; p 2 ; :::; p m g where m is the number of issues and p 1 p 2 p m (whence also p m p m 1 p 1 ). For instance, the board of a bank might face such a nested agenda where p j is the proposition 'prices will grow by at most j percent'; and an academic hiring committee might face such a nested agenda where p j is the proposition 'candidate Smith will publish fewer than j papers per year'. But most relevant agendas are not nested. The agendas in (1) are not nested, and also the preference agenda de…ned in Section 2 is not nested (as long as there are more than two alternatives). Finally, the agenda X is tiny if it has at most two issues fp; pg (6 = f ; ?g), i.e., at most four propositions (6 = ; ?).

Theorem 1 Given an agenda, all independent and implicit consensus preserving aggregation rules F : J n ! J are dictatorial if and only if the agenda is non-nested and non-tiny.

To paraphrase the result, for almost all agendas our two axioms cannot be jointly satis…ed by any non-dictatorial aggregation rule. Indeed, far more agendas imply impossibility than in the Arrow-like theorem mentioned in the introduction (and formally stated later as Theorem 5). For instance all agendas in (1) fall under the impossibility of Theorem 1 (they are non-nested), but not under that of the Arrow-like theorem. The same is true of almost all example agendas used repeatedly in the literature to illustrate inconsistent majority judgments, such as agendas of type fp; q; p \ qg . Theorem 1's very wide class of 'impossibility agendas' is a result of requiring implicit consensus preservation, while standard impossibility theorems usually require one of the two weaker consensus axioms mentioned earlier.

Theorem 1's 'only if'part is established by showing that, for a nested agenda X = fp 1 ; :::; p m g (where p 1 p 2 p m ), propositionwise majority rule satis…es all requirements. 10 In short, this is because each rational judgment set in J takes the special form fp 1 ; ; :::; p k 1 ; p k ; :::; p m g for some cut o¤ point k (in f1; :::; m + 1g), and propositionwise majority rule returns the judgment set of an individual who is median in terms of the cut-o¤ point. This argument has the ‡avour of single-peakedness and other structural conditions in preference or judgment aggregation. More precisely, nested agendas have the special property that all pro…les in J n automatically satisfy several structural conditions which guarantee consistent majority judgments, i.e., all conditions introduced in List (2003) or Dietrich and List (2010). 11 Judgment-aggregation theorists will also be curious whether the notion of a non-nested agenda is related to any familiar kind of agenda. Non-nested agendas can in fact be related to non-simple agendas. 12 1 0 If n is even, then the majority is taken among individuals 1; :::; n 1 only, to avoid ties. 1 1 Consider, say, the condition of unidimensional alignment (a judgment-aggregation variant of singlecrossingness and Rothstein's 1990 order restriction, to which it reduces if X is the preference agenda). A pro…le (A1; :::; An) is unidimensionally aligned if the individuals can be linearly ordered such that, for each proposition p 2 X, the individuals i with p 2 Ai all come before or all come after those with p 6 2 Ai. The order might represent a political left-right order, with the individuals accepting a 'left-wing' proposition p located to the left of those rejecting it. For nested X, all pro…les in J n are necessarily unidimensionally aligned: just order the individuals by increasing cut-o¤ point (the ranking between two individuals with same cut-o¤ point can be chosen arbitrarily).

1 2 An agenda X is simple if it has no subset Y X with jY j > 2 that is minimal inconsistent, i.e., is inconsistent but becomes consistent if any member is removed (informally, simplicity means that there are no 'complex' interconnections involving more than two propositions). For instance, the preference agenda for a set of more than two alternatives is non-simple, since any 'cyclical' subset Y = fxP y; yP z; zP xg is minimal inconsistent. I show in Appendix B that a (non-tiny) agenda X is nested if and only if it satis…es a condition only subtly distinct from the de…nition of simplicity: X has no subset Y with jY j > 2 such that (Y nfpg) [ fpg is consistent for each p 2 Y . Adding 'inconsistent' before 'subset Y 'turns this characterization of nestedness into one of simplicity. This gives an idea of how nestedness strengthens simplicity.

4 Three types of agenda-insensitivity and their characterizations I now de…ne and characterize three forms of insensitivity of outcomes to the agenda choice (and hence, to agenda manipulation). The characterization results show that our two axioms -independence and implicit consensus preservation -play key roles in ensuring agenda-insensitivity, along with other axioms.

Think of the agenda X as being chosen by an agenda setter. This agenda setter has some room for maneuver, i.e., some degree of freedom in designing the agenda. Typically his in ‡uence has limits: he might be able to 'reframe'the decision problem, but not to alter its topic altogether. For instance he cannot remove all …nancial issues from the agenda of a bank's board. The agenda setter might also face restrictions on the agenda's complexity or size: perhaps he cannot set an agenda with more than three issues. To capture that only certain agendas X are feasible (choosable, settable), we consider a …xed set X of agendas X 2 deemed feasible/possible. It could consist of all agendas X 2 ; or of all agendas with at most six issues; or of all agendas without certain given issues (the 'too complex'issues, say); and so on. All we require from X is that it contains each single-issue agenda fp; pg S X2X X. In particular, X need not contain unions X [ X 0 of agendas X; X 0 2 X , the scope X of agendas X 2 X , or the maximal agenda X = 2 .

Can the agenda setter reverse collective judgments by changing the agenda? This question obviously depends on which aggregation rules would be used for the various feasible agendas. That is, it depends on what I call the 'aggregation system': De…nition 6 An aggregation system is a family (F X ) X2X containing an aggregation rule F X : J n X ! J X for each feasible agenda X 2 X (where F X represents the rule used if the agenda is X 2 X ). 13 I now de…ne three conditions on an aggregation system (F X ) X2X . Each one requires the outcomes to be in a speci…c sense insensitive to the agenda choice, hence, immune to agenda manipulation. The …rst condition states that the agenda setter cannot reverse any explicit collective judgment, i.e., any collective judgment on a proposition in the agenda:

De…nition 7 An aggregation system (F X ) X2X is basically agenda-insensitive -for short, agenda-insensitive -if any two feasible agendas X; X 0 2 X lead to the same collective judgment on any proposition p 2 X \ X 0 : for all J 1 ; :::

; J n 2 J X[X 0 , p 2 F X (J 1 \ X; :::; J n \ X) , p 2 F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ):
What is the rationale behind this axiom? Think of J i as individual i's judgment set under the (hypothetical) agenda X [X 0 , and think of J i \X and J i \X 0 as his submitted judgment sets under the (feasible) agendas X and X 0 , respectively. Note the implicit idea that individuals hold …xed, i.e., agenda-independent, judgments on all propositions p . In short, individuals are themselves agenda-insensitive in their judgments. A failure of individual agenda-insensitivity would of course open up additional sources of agenda manipulation, which we do not model here.

By the next theorem, agenda-insensitivity forces each rule F X to be independent. It also forces the aggregation rule F X to change coherently when the agenda setter extends the agenda X to a new agenda X 0 . What do I mean exactly? I start with two obvious de…nitions:

De…nition 8 An agenda X 0 extends another one X if X X 0 . De…nition 9 A set of propositions A is consistent with another one B if A [ B is consistent.
I can now formally de…ne what it means for the aggregation rule to change coherently as the agenda changes. For future purposes, the de…nition is formulated in full generality, i.e., for arbitrary agenda changes, not just agenda extensions:

De…nition 10 In an aggregation system (F X ) X2X , a rule F X 0 coheres with a rule F X if the outcome of F X 0 is not ruled out by that of F X : for any any J 1 ; :::; J n 2 J X , F X 0 (J 0 1 ; :::; J 0 n ) is consistent with F X (J 1 ; :::; J n ) for at least some J 0 1 ; :::; J 0 n 2 J X 0 consistent with J 1 ; :::; J n , respectively.

When do we call an entire aggregation system 'coherent'?

De…nition 11 An aggregation system (F X ) X2X is coherent if whenever an agenda X 2 X is extended to another X 0 2 X the rule F X 0 coheres with F X .

The following remark gives a clear idea of what it means for F X 0 to cohere with F X as the agenda is extended:

Remark 1 In case X 0 extends X, coherence of F X 0 with F X means that the outcome of F X 0 extends that of F X for at least some extension of the individual judgments: for any J 1 ; :::; J n 2 J X , F X 0 (J 0 1 ; :::; J 0 n ) extends F X (J 1 ; :::; J n ) for at least some J 0 1 ; :::; J 0 n 2 J X 0 extending J 1 ; :::; J n , respectively.

I can now state the characterization result about agenda-insensitivity.

Convention: For any property of aggregation rules (such as independence), an aggregation system (F X ) X2X is said to satisfy it if and only if each rule F X satis…es it.

Theorem 2 An aggregation system (F X ) X2X is agenda-insensitive if and only if it is independent and coherent.

Basic agenda-insensitivity only prevents the agenda setter from reversing explicit collective judgments, on proposition in the agenda. We now turn to a stronger requirement, which also excludes the reversal of implicit collective judgments, on propositions outside the agenda. For instance, if an agenda X = fp; qg leads the collective judgment set fp; qg, so that the collective implicitly accepts the proposition p \ q from the scope X, then the acceptance of p \ q cannot be reversed by using another agenda X 0 . Formally:

De…nition 12 An aggregation system (F X ) X2X is fully agenda-insensitive if any two feasible agendas X; X 0 2 X lead to the same collective judgment on any proposition p 2 X \ X 0 : for all J 1 ; :::

; J n 2 J X[X 0 , F X (J 1 \ X; :::; J n \ X) entails p , F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ) entails p:
Here, J i , J i \ X and J i \ X 0 again represents individual i's judgment set under the (hypothetical) agenda X [ X 0 , the (feasible) agenda X resp. the (feasible) agenda X 0 . While basic agenda-insensitivity requires independence and coherence, full agendainsensitivity requires stronger versions of independence and coherence. How are these stronger versions de…ned? First, an aggregation rule F for an agenda X is called independent on Y ( X) if the collective judgment on any proposition in Y only depends on the individuals'judgments on this proposition: for all propositions p 2 Y and all pro-…les (J 1 ; :::; J n ) and (J 0 1 ; :::; J 0 n ) in the domain, if for each individual i J i entails p if and only if J 0 i entails p, then F (J 1 ; :::; J n ) entails p if and only if; F (J 0 1 ; :::; J 0 n ) entails p (see Dietrich 2006). Setting Y = X yields standard independence. Full agenda-insensitivity however requires independence on the scope Y = X; this is the 'maximal'choice of Y .

Second, I strengthen the coherence condition by requiring the aggregation rule to change coherently not just when the agenda setter extends the agenda, but more generally when he 'essentially extends'the agenda, i.e., when he extends the scope of the agenda:

De…nition 13 An aggregation system (F X ) X2X is strongly coherent if whenever an agenda X 2 X is essentially extended to another X 0 2 X , i.e., X X 0 , then the rule F X 0 coheres with F X .

Note that if X X 0 (or equivalently, X X 0 ), then the judgments for X 0 subsume those for X: each J 0 2 J X 0 entails a J 2 J X . This implies a concrete characterization of coherence of F X 0 with F X in case X 0 essentially extends X:

Remark 2 In case X 0 essentially extends X, i.e., X X 0 , coherence of F X 0 with F X means that the outcome of F X 0 entails that of F X for at least some 'essential extensions' of the individual judgments: for any J 1 ; :::; J n 2 J X , F X 0 (J 0 1 ; :::; J 0 n ) entails F X (J 1 ; :::; J n ) for at least some J 0 1 ; :::; J 0 n 2 J X 0 entailing J 1 ; :::; J n , respectively. 14 I can now state the characterization of full agenda-insensitivity:

Theorem 3 An aggregation system (F X ) X2X is fully agenda-insensitive if and only if it is independent on the entire scope X and strongly coherent.

One may regard Theorems 2 and 3 as formal counterparts of claims in Dietrich (2006) about the role of independence and independence on the scope in preventing agenda manipulation, although Dietrich (2006) does not yet invoke feasible agendas, aggregation systems, and coherence or strong coherence.

Strong coherence has an interesting consequence. If two agendas X and X 0 are equivalent (i.e., have same scope X = X 0 ), any judgment set for X is equivalent to one for X 0 , and any aggregation rule for X is equivalent to one for X 0 . Formally, any J 2 J X is equivalent to the unique J 2 J X 0 such that J and J entail each other; and any aggregation rule F : J n X ! J X is equivalent to the unique rule F 0 : J n X 0 ! J X 0 de…ned as the image of F via transforming judgment sets in J X into equivalent ones in J X 0 ; formally, [F (J 1 ; :::; J n )] = F 0 (J 1 ; :::; J n ) for all J 1 ; :::; J n 2 J X .

One easily checks that strong coherence ensures equal treatment of equivalent agendas (and thus prevents Dietrich's 2006 'logical agenda manipulation'). Formally:

Remark 3 If an aggregation system (F X ) X2X is strongly coherent, then for any equivalent agendas X; X 0 2 X the corresponding rules F X and F X 0 are equivalent.

I now consider a third agenda-insensitivity condition. Rather than requiring that all collective judgments in the scope are irreversible (by a change of agenda), let us merely require irreversibility of those collective judgments which are most important or focal in the sense of being unanimously supported by all individuals. Indeed, reversing a unanimously supported collective judgment seems particularly bad, as it goes against ('overrules') all individuals. The condition that unanimously supported collective judgments cannot be reversed by agenda manipulation is formally stated as follows:

De…nition 14 An aggregation system (F X ) X2X is is focally agenda-insensitive if any two feasible agendas X; X 0 2 X lead to the same collective judgment on any unanimously accepted proposition in X \ X 0 : for all J 1 ; :::; J n 2 J X[X 0 and all propositions p 2 X \ X 0 entailed by each J i , 15 F X (J 1 \ X; :::; J n \ X) entails p , F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ) entails p:

This condition turns out to be equivalent to the requirement that each rule F X is implicit consensus preserving, assuming a mild condition of non-degeneracy (i.e., unanimity preservation):

Theorem 4 An aggregation system (F X ) X2X is focally agenda-insensitive and unanimity preserving if and only if it is implicit consensus preserving.

Agenda-insensitive aggregation: impossibility results

Our characterization results (Theorems 2-4) establish that agenda-insensitivity requires that aggregation rules satisfy certain axioms such as independence. But such axioms imply dictatorial aggregation for many agendas, by Theorem 1 and two other theorems of the literature. This turns our characterization results into impossibility results about agenda-insensitive aggregation. These impossibility results are now stated as corollaries.

I begin with basic agenda-insensitivity. By Theorem 2, this condition implies independence. However, by the well-known Arrow-like theorem in the …eld, independent aggregation rules must, for certain agendas, be degenerate, i.e., either dictatorial or not unanimity preserving. Formally:

Theorem 5 (Dietrich-List 2007a, Dokow-Holzman 2010 16 ) Given an agenda, all independent and unanimity preserving aggregation rules F : J n ! J are dictatorial if and only if the agenda is 'strongly connected'.

Which agendas count as 'strongly connected'? The most important conceptual point is that the class of these agendas is far smaller than (and included in) the class of 'impossibility agendas'in Theorem 1, i.e., the class of non-nested non-tiny agendas. For instance, an agenda of type X = fp; q; p \ qg is not strongly connected (so escapes the Arrow-like impossibility), though it is non-nested and non-tiny (so falls into Theorem 1's impossibility). Formally, strong connectedness is the conjunction of two well-known conditions, pathconnectedness (introduced by Nehring and Puppe 2002 under the label 'total blockedness') and pair-negatability, which are in turn de…ned as follows:

Pathconnectedness: Recall …rst that a proposition p 2 X conditionally entails another q 2 X -written p ` q -if fpg [ Y entails q for some set Y X which (for non-triviality) is consistent with p and with q. Agenda X is pathconnected if for any propositions p; q 2 Xnf ; ?g there are p 1 ; :::; p m 2 X (m 1) such that p = p 1 ` p 2 ` ` p m = q. Some important agendas are pathconnected, but many others are not. 17 Pair-negatability: Recall further that a set Y X is minimal inconsistent if it is inconsistent but each proper subset of Y is consistent. The agenda X is pairnegatable if it has a minimal inconsistent subset Y X which becomes consistent after negating some two members (i.e., (Y nfp; qg) [ fp; qg is consistent for some distinct p; q 2 Y ). Most concrete agendas are pair-negatable. 18 Prominently, the preference agenda is strongly connected, and is thus subject to the 1 6 Both papers build on Nehring and Puppe (2002/2010). Dietrich and List prove only the 'if'direction, and Dokow and Holzman prove both directions.

1 7 The preference agenda for more than two alternatives is pathconnected: e.g., xP y entails xP z conditionally on Y = fyP zg (for distinct options x; y; z). But an agenda of the form X = fp; q; p \ qg is not pathconnected: check that none of p; q; p \ q conditionally entails any of p; q; p \ q (assuming ?; 6 2 X). Also the agendas in (1) are not pathconnected.

1 8 The preference agenda for more than two alternatives is pair-negatable: Y = fxP y; yP z; zP xg X is minimal inconsistent and becomes consistent after replacing xP y by yP x and yP z by zY y. Also X = fp; q; p \ qg is pair-negatable (assuming ?; 6 2 X): Y = fp; q; p \ qg is minimal inconsistent but fp; q; p \ qg is consistent, and also Y = fp; p \ qg is minimal inconsistent but fp; p \ qg is consistent. (In the de…nition of pair-negatability, one can equivalently replace 'some two members' by 'some positive even number of members'.) impossibility, as is already known from Arrow's Theorem (to which Theorem 5 indeed reduces in the case of the preference agenda).

Given this Arrow-like theorem, our Theorem 2 immediately implies that agendainsensitive aggregation systems must be degenerate:

Corollary 1 If an aggregation system (F X ) X2X is agenda-insensitive and unanimity preserving, then the rule F X is dictatorial for each strongly connected agenda X 2 X (more generally, each agenda X 2 X included in a strongly connected agenda X 0 2 X ). 19 Let us now turn to the condition of full agenda-sensitivity, which by Theorem 3 forces to independence on the scope. Unfortunately, no non-degenerate aggregation rules are independent on the scope, as long as the agenda is non-monadic i.e., contains more than one issue fp; pg (6 = f ; ?g):

Theorem 6 (Dietrich 2006, Corollary 1) Given an agenda, all aggregation rules F : J n ! J which are independent on the scope are dictatorial or constant if and only if the agenda is non-monadic. 20 Given this result, Theorem 3 implies that fully agenda-insensitive aggregation systems must be degenerate:

Corollary 2 If an aggregation system (F X ) X2X is fully agenda-insensitive, then the rule F X is dictatorial or constant for each non-monadic agenda X 2 X (more generally, each agenda X 2 X included in the scope of a non-monadic agenda X 0 2 X ).

Finally, we turn to focal agenda-insensitivity, which by Theorem 4 forces to implicit consensus preserving (under the mild assumption of unanimity preservation). This by itself does not lead into an impossibility result. But if one combines focal with basic agenda-insensitivity, then one is forced to independence (by Theorem 2) as well as implicit consensus preserving (by Theorem 4), which leads us straight into the impossibility of Theorem 1. Formally:

Corollary 3 If an aggregation system (F X ) X2X is basically and focally agenda-insensitive, and unanimity preserving, then the rule F X is dictatorial for each non-nested non-tiny agenda X 2 X (more generally, each agenda X 2 X included in a non-nested non-tiny agenda X 0 2 X ).

Conclusion

I begin by summing up. I have …rstly derived a new impossibility theorem on judgment aggregation, based on the familiar independence axiom and a particularly strong consensus axiom. Subsequently, I have de…ned and axiomatically characterized three types of agenda-insensitive aggregation: basic, full and focal agenda-insensitivity. Finally, combining these characterization results with the impossibility result (and two well-known impossibility results), I have derived impossibility results about each type of agenda-insensitivity.

Let me …nish by mentioning another type agenda-insensitivity, which pertains not to particular procedures (aggregation systems), but to axioms on aggregation rules. I call an axiom description-insensitive if whenever two agendas X and X 0 are equivalent (i.e., have the same scope), then an aggregation rule for agenda X satis…es the axiom if and only if the equivalent rule for agenda X 0 (de…ned in Section 4) satis…es the axiom. 21 One might favour description-insensitive axioms on the grounds that any dependence on how the decision problem is framed is a form of arbitrariness. The standard unanimity principle is not description-invariant: respecting unanimity on propositions in X = fp; qg is considerably di¤erent from doing so for the equivalent agenda X = fp \ q; p \ q; p \ q; p \ qg . Our stronger consensus axiom -implicit consensus preservationavoids this ‡aw; it is from this perspective more canonical. The following table classi…es axioms according to whether they are description-invariant.

A.1 The common concepts

We still consider a group of individuals i = 1; :::; n with n 3. No underlying set of worlds is introduced. Instead, I de…ne agendas from scratch:

De…nition 15 An agenda is a non-empty set X (of 'propositions') which is endowed with the notions of negation and interconnections, i.e., (a) to each p 2 X corresponds a proposition denoted :p 2 X ('not p') with :p 6 = p = ::p (so X is partitioned into pairs fp; :pg, called 'issues'), (b) certain judgment sets J X containing a single member of each issue count as 'rational', the non-empty set of them being denoted J or J X , where (in this paper) X is …nite and jJ j > 1. 22 Notationally, an agenda will be denoted simply by its set of propositions 'X', suppressing the structure on X. 23 The semantic agendas of the main text are a special case: there propositions are sets of worlds, with the set-theoretic notions of negation and interconnections. 24 Syntactic agendas are another special case: here propositions are sentences of a formal logic, with the logical notions of negation and interconnections. 25 A judgment set J X is complete if it contains a member of each issue fp; :pg, and consistent if it is extendable to a rational judgment set. So the complete and consistent judgment sets are precisely the rational judgment sets. In most concrete agendas, every proposition p 2 X is contingent, i.e., neither a contradiction (for which fpg is inconsistent), nor a tautology (for which f:pg is inconsistent). A proposition p 2 X (or set S X) entails another proposition p 0 2 X (or set S 0 X) if every rational judgment set containing p (resp. including S) also contains p 0 (resp. includes S 0 ).

Aggregation rules for an agenda X are still functions F mapping any pro…le of 'individual'judgment sets (J 1 ; :::; J n ) (from some domain, usually J n ) to a 'collective' judgment set F (J 1 ; :::; J n ).

The results of the main text draw on some agenda properties, whose general de…nitions are easily stated:

A agenda X is nested if it takes the form X = fp 1 ; :::; p m g , where m is the number of issues and p 1 entails p 2 , p 2 entails p 3 , and so on. An agenda X is non-tiny if it has more than two issue, i.e., more than four propositions (not counting non-contingent proposition if any, and identifying any equivalent propositions if any 26 ). An agenda is non-monadic if it has more than one issue, i.e., more than two propositions (again not counting any non-contingent proposition and identifying any equivalent propositions). Strong connectedness (the conjunction of pathconnectedness and pair-negatability) is de…ned as before, modulo replacing complements p by negations :p and replacing Xnf ; ?g by fp 2 X : p is contingentg.

A.2 Closed agendas and scope of agendas

Extending agendas is easy in the special case of semantic agendas (it su¢ ces to consider new subsets of ) or syntactic agendas (it su¢ ces to consider new sentences of the logic). But how does this work in our general framework? In principle, there are twoas we shall see, equivalent -strategies for de…ning the scope of an agenda X. They can : (a mapping p 7 ! :p satisfying :p 6 = p = ::p) and the set of issues (a partition of X into binary sets) are interde…nable objects, we could equivalently de…ne an agenda as a set X endowed with 'issues and interconnections', and de…ne the negation of p as the unique proposition :p such that fp; :pg is an issue. Algebraically, the agenda would then be the structure X (X; I; J ); where I is the set of issues.

2 4 I.e., :p is the complement p, and J consists of those sets J X (with a single member of each issue) for which T p2J p 6 = ?.

be stated informally as follows:

(a) add to X all propositions (in an extended sense) which are constructible by combining propositions in X; (b) add to X all propositions (in an extended sense) on which the judgment ('yes'or 'no') is determined by the judgments on the propositions in X.

I choose the approach (a), but also brie ‡y discuss the approach (b) (which is perhaps more in line with Dietrich's 2006 original de…nition). Starting with some natural terminology, I call an agenda closed under conjunction if for any propositions p; q 2 X there exists a proposition r (the conjunction of p and q) such that any rational judgment set contains r if and only if it contains both p and q, closed under disjunction if for any propositions p; q 2 X there exists a proposition r (the disjunction of p and q) such that any rational judgment set contains r if and only if it contains p or q (possibly both).

These two closure properties are in fact equivalent, as is seen shortly. Most relevant agendas, including the semantic ones in the main text, are redundancy-free: they contain no two distinct propositions that are equivalent, i.e., entail each other.

Proposition 2 In any redundancy-free agenda X, the conjunction resp. disjunction of two propositions p and q, if existing, is unique and denoted by p ^q resp. p _ q.

Proposition 3 An agenda X is closed under conjunction if and only if it is closed under disjunction. I then call it closed simpliciter.

To 'close'an agenda, we must extend it. First, let me spell out the obvious:

De…nition 16 An agenda X is a subagenda of another X 0 , and X 0 an extension or superagenda of X, if X X 0 , where the notions of negation and interconnections for X are those for X 0 restricted to X (i.e., the negation operator of X is that of X 0 restricted to X, and J X = fJ \ X : J 2 J X 0 g).

The following result ensures that closing an agenda is possible in a unique way:

Proposition 4 Every agenda X has a closure, i.e., a minimal closed superagenda; it is moreover unique up to relabelling 27 . De…nition 17 The scope of an agenda X is its (up to relabelling uniquely existing) minimal closed superagenda, denoted X.

The judgments within an agenda X determine those within the entire scope X. Formally: 2 7 Uniqueness up to relabelling means that between any two minimal closed superagendas X 0 and X 00 there exists an (agenda-)isomorphism that is constant on X, where an isomorphism is of course a bijection f : X 0 ! X 00 that preserves (i) the notion of negation (i.e., f (:p) = :f (p) for all p 2 X) and (ii) the notion of interconnections (i.e., J 2 JX , f (J) 2 J X 0 for all J X).

De…nition 18 (a) Given an agenda, a judgment set J settles a proposition p if J entails p or entails :p. (b) An agenda X settles a superagenda X 0 if each rational judgment set J 2 J X settles each proposition in X 0 (equivalently, is uniquely extendable to a J 0 2 J X 0 ).

Proposition 5 Every agenda settles its scope.

In fact, a stronger result can be shown: the scope of an agenda X is the (up to relabelling unique) maximal superagenda which is settled by X (and is redundancy-free outside X, i.e., contains no two equivalent propositions outside X). We could therefore have used an alternative and equivalent de…nition of the scope:

De…nition 19 (alternative statement) The scope of an agenda X is the (up to relabelling uniquely existing) maximal superagenda settled by X and redundancy-free outside X.

The following lemma gives a concrete idea of the totality of propositions in the scope. It uses conjunctions/disjunctions of any number of propositions, which are de…ned like conjunctions/disjunctions of two propositions. 28 Lemma 1 Every proposition p in the scope of a redundancy-free agenda X is a disjunction of conjunctions of propositions in X; for instance, p = W

J2J p V q2J
q, where J p := fJ 2 J X : J entails pg.

Finally, the scope carries a familiar algebraic structure:

Proposition 6 Any closed redundancy-free agenda -for instance the scope of a redundancyfree agenda -is a Boolean algebra with respect to the relation of entailment between propositions, with the meet, join, and complement given by the conjunction, disjunction, and negation, respectively. 29

Recall that Boolean algebras are de…ned as follows. First, a lattice is a partially ordered set L (L; ) such that any two elements p; q 2 L have a meet p ^q (greatest lower bound) and a join p _ q (smallest upper bound). It is distributive if p _ (q ^r) = (p _ q) ^(p _ r) and p ^(q _ r) = (p ^q) _ (p ^r) for all p; q; r 2 L. A Boolean algebra is a distributive lattice (L; ) such that L contains a greatest element | (the 'top' or 'tautology') and a bottom ? (the 'bottom'or 'contradiction'), and every element has an algebraic complement, i.e., an element whose join with p is | and whose meet with p is ?. The paradigmatic Boolean algebras are the set-theoretic ones: here there exists a set such that L 2 , = , | = , ? = ?, and the meet, join and complement are given by the set-theoretic intersection, union and complement. By Stone's representation theorem, every Boolean algebra is isomorphic to such a set-theoretic one. Another example is the Boolean algebra generated from a logic, i.e., the set of sentences modulo logical equivalence (where the logic includes classical negation and conjunction, which induce the algebraic negation, meet and join).

A.3 Feasible agendas and aggregation systems

Just as in the main text, aggregation systems are families (F X ) X2X containing an aggregation rule F X : J n X ! J X for each agenda X from a given set X of 'feasible' agendas X (see De…nition 6). However, we need to say what 'feasible agendas' are in the present framework. In the semantic framework, they were subagendas of the 'universal'agenda 2 which contains the totality of all propositions at the disposal of the agenda setter. Our present framework has no set of worlds and thus no pre-de…ned universal agenda. We therefore enrich the framework by assuming a ('universal') agenda L, which we take to be closed and redundancy-free (so L de…nes a Boolean algebra by Proposition 6). Think of L as a reservoir of propositions. It could be as large as an entire language, or as small as a set of propositions on a relevant topic (such as a given court trial). Now X is simply a set of subagendas X of L: those deemed feasible/possible. All we assume about X is, here again, that it contains at least each binary agenda fp; :pg S X2X X. Since L is closed, the scope X of an agenda X 2 X is again a subagenda of L (the smallest one that includes X). 30 Note that quite possibly X 6 2 X since X might be too rich and complex for being feasible. 31

B Proofs

All proofs are formulated for the general framework of Appendix A. The set of individuals is denoted N := f1; :::; ng. Recall that for an agenda X the set of rational judgment sets is denoted 'J 'or sometimes, to avoid ambiguity, 'J X '.

B.1 Results of Section 3 on single aggregation rules

Proof of Proposition 1. For the agenda X, consider an aggregation rule F : J n ! J . I write ICP1, ICP2 and ICP3 for the three versions of implicit consensus preservation, respectively.

'ICP1 ) ICP3 ': Assume ICP1. Consider any (J 1 ; :::; J n ) 2 J n . In the scope we can form the proposition p :=

V q2J 1 q ! W W V q2Jn q ! (i.
e., the proposition that all q in J 1 or all q in J 2 ... or all q in J n hold). Each J i entails V q2J i q, and hence, entails p. So, F (J 1 ; :::; J n ) entails p by ICP1. Let J be the unique extension of F (J 1 ; :::; J n ) to a set in J X . Since F (J 1 ; :::; J n ) entails p, J contains p. So, for some i, V q2J i q 2 J, and thus J i J. It follows that J i = J \ X = F (J 1 ; :::; J n ). QED 'ICP3 ) ICP2 ': Assume ICP3 and consider a feature K J and a pro…le (J 1 ; :::; J n ) 2 J n such that J 1 ; :::; J n 2 K. By ICP3, F (J 1 ; :::; J n ) 2 fJ 1 ; :::; J n g. So, F (J 1 ; :::; J n ) 2 K. QED 'ICP2 ) ICP1 ': Assume ICP2. Consider any p 2 X and any pro…le (J 1 ; :::; J n ) 2 J n such that each J i entails p. Since each J i belongs to the feature K := fJ 2 J : J entails pg, so does F (J 1 ; :::; J n ) by ICP2.

As part of the proof of Theorem 1, I show several lemmas. For an agenda X, an aggregation rule F on J n is called systematic if there exists a set W of ('winning') coalitions C N such that F (J 1 ; :::; J n ) = fp 2 X : fi : p 2 J i g 2 Wg for all J 1 ; :::; J n 2 J .

In this case, the set W is uniquely determined and denoted by W F .

Lemma 2 Every independent and implicit consensus preserving aggregation rule F : J n ! J is systematic if and only if the agenda X is non-nested.

Proof. Let X be an agenda. We may assume without loss of generality that all p 2 X are contingent, because each side of the claimed equivalence remains true (or false) if the non-contingent propositions are removed from the agenda.

1. In this part we assume that X is non-nested and consider an independent and implicit consensus preserving rule F : J n ! J . I show that F is systematic (drawing on Dietrich and List 2013). For any p; q 2 X, I de…ne p q to mean that there exists a …nite sequence p 1 ; :::; p k 2 X with p 1 = p and p k = q such that any neighbours p l ; p l+1 are not exclusive (i.e., fp l ; p l+1 g is consistent) and not exhaustive (i.e., f:p l ; :p l+1 g is consistent). I prove …ve claims: the …rst four gradually establish that p q for all p; q 2 X, and the last shows that F is systematic.

Claim 1 : For all p; q 2 X, p q , :p :q.

It su¢ ces to show one direction of implication, as ::p = p for all p 2 X. Let p; q 2 X with p q. Then there is a path p 1 ; :::; p k 2 X between p to q where any neighbours p j ; q j+1 are non-exclusive and non-exhaustive. To see why :p :q, note that :p 1 ; :::; :p k is a path between :p and :q where any neighbours :p j ; :p j+1 are non-exclusive (as p j ; p j+1 are non-exhaustive) and non-exhaustive (as p j ; p j+1 are nonexclusive). QED Claim 2 : If p 2 X entails q 2 X, then p q.

If p 2 X entails q 2 X, then p q in virtue of a direct connection: p; q are neither exclusive nor exhaustive (for instance, fp; qg is consistent because p is not a contradiction and entails q. QED Claim 3 : is an equivalence relation on X, and for all p; q 2 X, p q or p :q. (So each equivalence class contains at least one member of each issue fq; :qg, and it is the only equivalence class if it contains both members of some issue.) Re ‡exivity, symmetry and transitivity are all obvious (where re ‡exivity uses that every p 2 X is contingent). Now consider p; q 2 X such that p 6 q; we have to show that p :q. Since p 6 q, fp; qg or f:p; :qg is inconsistent. In either case, one of p and :q entails the other, so that p :q by Claim 2. QED Claim 4 : p q for all p; q 2 X.

Let X + be an equivalence class w.r.t. and suppose for a contradiction that X + 6 = X. Then, by Claim 3, X + must contain exactly one member of each issue fr; :rg. We show that X + is weakly ordered by the entailment relation between propositionsimplying that X is nested, a contradiction. As the entailment relation on X + is of course transitive, it remains to show that it is complete on X + . So we consider p; q 2 X + , and have to show that p entails q or q entails p. We have p 6 :q, since otherwise X + would include the entire issue fq; :qg. So fp; :qg or f:p; qg is inconsistent. Hence, p entails q or q entails p. QED Claim 5 : F is systematic.

Since F is independent, there exists a family (W p ) p2X of sets of coalitions such that F (J 1 ; :::; J n ) = fp 2 X : fi : p 2 J i g 2 W p g for all J 1 ; :::; J n 2 J :

(2)

It su¢ ces to show that W p is the same for all p 2 X. By Claim 4 and the de…nition of , it su¢ ces to show that W p = W q for all p; q 2 X which are non-exclusive and non-exhaustive. Consider such p; q 2 X. Consider any C N and let us show that C 2 W q , C 2 W q . As fp; qg and f:p; :qg are consistent, there exist J 1 ; :::; J n 2 J such that p; q 2 J i for all i 2 C and :p; :q 2 J i for all i 2 N nC. We now apply implicit consensus preservation, in any of its three variants. Using either variant 1 (and the fact that each J i entails the proposition (p ^q) _ (:p ^:q) in the scope), or variant 2 (and the fact that each J i belongs to the feature K := fJ 2 J : p 2 J , q 2 Jg), or variant 3, it follows that p 2 F (J 1 ; :::; J n ) , q 2 F (J 1 ; :::; J n ). By (2), the left side of this equivalence holds if and only if C 2 W p and the right side holds if and only if C 2 W q . So C 2 W p , C 2 W q . QED 2. Now assume that X is nested, i.e., of the form X = fp 1 ; :::; p m g where m is the number of issues and where p 1 entails p 2 , p 2 entails p 3 , etc. I consider the aggregation rule F on J n de…ned as follows: for all J 1 ; :::; J n 2 J , F (J 1 ; :::; J n ) consists of each p j contained in all J i and each :p j contained in some J i . We have to show that F (i) maps into J , (ii) is independent, (iii) is implicit consensus preserving, and (iv) is not systematic. The properties (ii) and (iv) are obvious (where (iv) uses that n > 1 and that X contains a pair of contingent propositions p; :p because jJ j > 1). It remains to prove (i) and (iii). Now (i) follows from (iii) by version 3 of implicit consensus preservation. To see why (iii) holds, note that for each J 2 J there is a cut-o¤ level t 2 f1; :::; m + 1g such that J = f:p 1 ; :::; :p t 1 ; p t ; :::; p m g, and that therefore for all J 1 ; :::; J n 2 J we have F (J 1 ; :::; J n ) = J i where i is the (or an) individual with highest cut-o¤ level.

The next lemma is the main technical step towards Theorem 1 and provides two alternative characterizations of non-nested agendas. (Compare the characterization in (b) with the de…nition of non-simple agendas mentioned in Section 3: the only di¤erence is that (b) allows Y to be consistent.) Lemma 3 For any agenda X, the following are equivalent:

(a) X is non-nested (and non-tiny). Proof. Let X be an agenda. I write p `q to mean that p (2 X) entails q (2 X), and S `q to mean that S ( X) entails q. We may assume without loss of generality. that X contains only contingent propositions, and is redundancy-free, i.e., contains no two equivalent propositions. The reason is that otherwise it su¢ ces to do the proof for any redundancy-free subagenda containing only contingent propositions, because each of the conditions (a), (b) and (c) holds for X if and only if it holds for that subagenda; to see for instance why (b) holds for X if and only if it holds for the subagenda, note that (Y nfpg) [ f:pg can only be consistent for all p 2 Y if Y contains no two equivalent propositions and no non-contingent propositions.

The equivalence between (b) and (c) is straightforward (to see why (b) implies (c), simply replace the set Y in (b) by a three-member subset of it). It is also relatively easy to see why (c) implies (a). Indeed, whenever (a) is violated, so is (c), by the following argument. First, if X is tiny, then (c) is violated since every three-element set Y

X takes the form Y = fq; :q; pg for some p; q 2 X, and thus Y nfpg [ f:pg fails to be consistent. Second, if X is nested, say X = fr; :r : r 2 Zg for some subset Z X linearly ordered by entailment, condition (c) is violated since any three-element set Y X has elements p 6 = q which both belong to Z or both belong to f:p : p 2 Zg; so that (by the linear orderedness of Z and of f:r : r 2 Zg w.r.t. entailment) p `q or q `p, which implies that (Y nfqg) [ f:qg or (Y nfpg) [ f:pg is inconsistent.

It remains to show that (a) implies (c). Let X be non-nested and non-tiny; we show (c). We distinguish between two cases.

Case 1 : no p; q 2 X are logically independent, i.e., for no p; q 2 X each of the sets fp; qg; fp; :qg; f:p; qg and f:p; :qg is consistent.

Claim 1.1. There exists a (with respect to set-inclusion) maximal nested (sub)agenda X X.

This follows from the fact that the set of nested subagendas V X is non-empty (because it contains any single-issue subagenda fp; :pg) and …nite (because X is …nite). QED Since X is nested, we may write it as X = S p2X + fp; :pg where X + is a subset of X which contains exactly one member of each issue fp; :pg X and is linearly ordered w.r.t. entailment.

Claim 1.2. There exists an s 2 XnX such that fs; pg is consistent for all p 2 X + .

Since X is nested but X is not, we have X ( X, and thus there are r; :r 2 XnX . It su¢ ces to show that at least one of r and :r is consistent with each p 2 X + . This is true because otherwise there would exist p; p 0 2 X + such that fr; pg and f:r; p 0 g are inconsistent, which (recalling that p `p0 or p 0 `p, and writing p 00 for the logically stronger one of p and p 0 ) implies that fr; p 00 g and f:r; p 00 g are inconsistent, so that fp 00 g is inconsistent, a contradiction since p 00 is contingent. QED I de…ne

Y 1 : = fp 2 X + : p `sg; Y 2 : = fp 2 X + : :p `sg: Claim 1.3. Y 1 \ Y 2 = ?, and Y 1 [ Y 2 = X + .
First, Y 1 \ Y 2 = ?, because otherwise there would be a p 2 X + such that p `s and :p `s, a contradiction as s is not a tautology. Second, suppose for a contradiction that p 2 X + n(Y 1 [ Y 2 ). I ultimately show that the agenda X [ fs; :sg is nested, a contradiction as X is a maximal nested subagenda of X. Since p and s are not logically independent (by assumption of Case 1), and since fp; sg is consistent (by Claim 1.2), fp; :sg is consistent (as p 6 2 Y 1 ) and f:p; :sg is consistent (as p 6 2 Y 2 ), it follows that f:p; sg is inconsistent, so that s `p. We next show that s entails not just in p, but also all other propositions in X + nY 1 : s `p0 for all p 0 2 X + nY 1 :

(3)

To show this, let p 0 2 X + nY 1 , and note …rst that :p 0 and :s are entailed by f:p 0 ; :p; :sg. Hence (as s `p, i.e., :p `:s), :p 0 and :s are entailed by f:p 0 ; :pg. So, since the set f:p 0 ; :pg is consistent (as either :p 0 `:p or :p `:p 0 ), the set f:p 0 ; :sg is also consistent. Since p 0 and s are not logically independent (by assumption of Case 1), and since fp 0 ; sg is consistent (by Claim 1.2), fp 0 ; :sg is consistent (as p 0 6 2 Y 1 ) and f:p 0 ; :sg is consistent (as just shown), it follows that f:p 0 ; sg is inconsistent, so that s `p0 . This proves (3).

Note that for every p 0 in X + , either p 0 `s (if p 0 2 Y 1 ) or s `p0 (if p 0 6 2 Y 1 , by ( 3)). So the augmented (sub-)agenda X [ fs; :sg is nested, a contradiction as X is a maximal nested subagenda of X. QED Claim 1.4. Y 1 ; Y 2 6 = ?.

By Claim 1.3 we may equivalently show that Y 1 ; Y 2 6 = X + . Suppose for a contradiction that Y 1 = X + or Y 2 = X + . Then X [ fs; :sg is a nested agenda, a contradiction since X was de…ned as a maximal nested subagenda of X. QED The proof of condition (c) is completed by combining Claim 1.4 with the following observation:

Claim 1.5. For all q 2 Y 1 and r 2 Y 2 , the set Y := f:q; r; sg satis…es the requirements of condition (c), i.e., jY j = 3 and (Y nfpg) [ f:pg is consistent for each p 2 Y .

Consider any q 2 Y 1 and r 2 Y 2 and let Y := f:q; r; sg. To see why jY j = 3, note that :q 6 = r since r 2 X + while :q 6 2 X + , and that s 6 = :q; r since :q; r 2 X while s 6 2 X . Further: fq; r; sg is consistent, because, …rstly, fq; sg is consistent by Claim 1.2, and, secondly, q `r, as q and r belong to the linearly ordered set X + and as r 6 `q (by the fact that q 2 Y 1 and r 6 2 Y 1 ). f:q; :r; sg is consistent, because, …rstly, :r `:q (since q `r, as just shown), and, secondly, :r `s (since r 2 Y 2 ). f:q; r; :sg is consistent, because, …rstly, :s `:q (since q `s, as q 2 Y 1 ), and, secondly, :s `r (since :r `s, as r 2 Y 2 ). QED Case 2 : p; q 2 X are logically independent, i.e., all of fp; qg; fp; :qg; f:p; qg and f:p; :qg are consistent. Consider such p; q 2 X. Since jXj > 4 there is an r 2 Xnfp; :p; q; :qg. As r is non-contradictory, it can be consistently added to at least one of the (consistent) sets fp; qg; fp; :qg; f:p; qg and f:p; :qg. We may assume without loss of generality that fp; :q; rg is consistent (otherwise, simply interchange p with :p and/or q with :q). The argument distinguishes between two subcases. Subcase 2.1 : f:p; :q; :rg and fp; q; :rg are both consistent. In this case, condition (c) holds for Y := fp; :q; :rg, since each of the sets f:p; :q; :rg, fp; q; :rg and fp; :q; rg is consistent.

Subcase 2.2 : f:p; :q; :rg or fp; q; :rg is inconsistent (perhaps both are). We assume without loss of generality that fp; q; :rg is inconsistent, i.e., fp; qg `r. (The proof is analogous in the other case.) There are three subsubcases.

Subsubcase 2.2.1 : f:p; q; :rg and fp; :q; :rg are both consistent. Here, condition (c) holds for Y := fp; q; :rg, since each of the sets f:p; q; :rg, fp; :q; :rg and fp; q; rg is consistent (the latter set being consistent because fp; qg is consistent and entails r).

Subsubcase 2.2.2 : f:p; q; :rg is inconsistent. So f:p; qg `r. As also fp; qg `r, we have q `r. We once again distinguish between cases: First assume f:p; :q; :rg is consistent. Then condition (c) holds with Y = f:p; :q; rg, because fp; :q; rg, f:p; q; rg and f:p; :q; :rg are consistent (where f:p; q; rg is consistent as f:p; qg is consistent and q `r). Second assume f:p; :q; :rg is inconsistent, i.e., f:p; :qg `r. Since also q `r, we have :r `:q; p. Condition (c) holds with Y = fp; :q; rg, because f:p; :q; rg is consistent (as f:p; :qg is consistent and entails r), fp; q; rg is consistent (as fp; qg is consistent and entails r) and fp; :q; :rg is consistent (as :r `:q; p). Subsubcase 2.2.3 : fp; :q; :rg is inconsistent. (If in the following proof for the current subsubcase we interchange p and q, then we obtain an alternative, but longer, proof for Subsubcase 2.2.2.) Since fp; :q; :rg is inconsistent, fp; :qg `r. As also fp; qg `r, it follows that p `r. We now show that (*) f:p; q; rg and f:p; :q; :rg are consistent or (**) f:p; :q; rg and f:p; q; :rg are consistent.

(4)

To show this, we assume that (*) is violated and show that (**) holds, by distinguishing between two cases:

First, let f:p; q; rg be inconsistent. It follows, on the one hand, that f:p; q; :rg is consistent (as f:p; qg is consistent), and, on the other hand, that f:p; :q; rg is consistent (as otherwise, by the inconsistency of f:p; q; rg, f:p; rg would be inconsistent, i.e., r `p, a contradiction since p `r and p 6 = r). This proves (**). Second, let f:p; q; rg be consistent. Then f:p; :q; :rg is inconsistent as (*) is violated. It follows, one the one hand, that f:p; :q; rg is consistent (as f:p; :qg is consistent), and, on the other hand, that f:p; q; :rg is consistent (as otherwise f:p; :rg would be inconsistent, i.e., :r `p, a contradiction since p `r). This proves (**).

We can now prove condition (c). In the case of (*), (c) holds with Y = f:p; :q; rg, since fp; :q; rg is consistent (as assumed without loss of generality under Case 2), f:p; q; rg is consistent (by (*)) and f:p; :q; :rg is consistent (by (*)). In the case of (**), (c) holds with Y = f:p; q; rg, since fp; q; rg is consistent (as fp; qg is consistent and p `r), f:p; :q; rg is consistent (by (**)) and f:p; q; :rg is consistent (by (**)).

Drawing on Lemma 3, I next show that for almost every agenda the set of winning coalitions of a systematic and implicit consensus preserving aggregation rule de…nes an ultra…lter (which would not be true if implicit consensus preservation were replaced by the standard unanimity condition).

Lemma 4 Consider a systematic and implicit consensus preserving aggregation rule F : J n ! J for an agenda X, and coalitions C;

C 0 N . (a) If X satis…es jJ j > 2, then [C 2 W F and C C 0 ] ) C 0 2 W F . (b) If X is non-nested and non-tiny, then C; C 0 2 W F ) C \ C 0 2 W F . (c) C 2 W F , N nC 6 2 W F .
Note that jJ j > 2 if and only if X has more than two propositions (one issue), 32 a very mild assumption, satis…ed notably by non-tiny agendas.

Proof. Let X, F , C and C 0 be as speci…ed.

(a) Suppose jJ j > 2, C 2 W F and C C 0 . We show that C 0 2 W F . As jJ j > 2, there exist contingent and pairwise non-equivalent propositions p; :p; q; :q 2 X. There must exist a member of fp; :pg which entails neither q nor :q, as can be shown using that the propositions p; :p; q; :q are contingent and pairwise non-equivalent. Without loss of generality. we assume that p entails neither q nor :q (otherwise simply interchange p and :p). So fp; qg and fp; :qg are each consistent. Note that at least one of f:p; qg and f:p; :qg is consistent, as :p is not a contradiction. Without loss of generality. we assume the latter (otherwise interchange q and :q). To summarize, each of the sets fp; qg, fp; :qg and f:p; :qg is consistent. We may therefore consider a pro…le (J 1 ; :::; J n ) 2 J n such that J i 8 < : fp; qg for all i 2 C, fp; :qg for all i 2 C 0 nC, f:p; :qg for all i 2 N nC 0 . First, since each J i contains p or :q, so does F (J 1 ; :::; J n ) by implicit consensus preservation (version 2). Second, q 2 F (J 1 ; :::; J n ) since fi 2 N : q 2 J i g = C 2 W F . These two facts imply that p 2 F (J 1 ; :::; J n ). So, as fi : p 2 J i g = C 0 , we have C 0 2 W F .

(b) Suppose X is non-nested and non-tiny, and assume C; C 2 W F . We show that C \ C 2 W F . By assumption on X and Lemma 3, there is a three-element set Y = fp; q; rg X such that each of f:p; q; rg, fp; :q; rg and fp; q; :rg is consistent. This allows us to construct a pro…le (J 1 ; :::; J n ) 2 J n such that

J i 8 < : f:p; q; rg if i 2 C \ C fp; q; :rg if i 2 C nC fp; :q; rg if i 2 N nC .
First, q 2 F (J 1 ; :::; J n ) as fi : q 2 J i g = C 2 W F . Second, as C 2 W F and C C [ (N nC ), we have C [(N nC ) 2 W F by part (a); hence r 2 F (J 1 ; :::; J n ) as fi : r 2 J i g = C [ (N nC ). Third, as each J i contains :p or :q or :r, so does F (J 1 ; :::; J n ) by implicit consensus preservation (version 2). These three facts imply that :p 2 F (J 1 ; :::; J n ). Hence, as fi : :p 2

J i g = C \ C , we have C \ C 2 W F .
(c) This claim is obvious, as (by jJ j > 1) we can choose a contingent proposition p 2 X and construct a pro…le in J n in which all i 2 C accept p and all i 2 N nC accept :p.

I can now prove Theorem 1, whose 'if'part will follow from the above lemmas.

Proof of Theorem 1. 1. In this part of the proof, let the agenda X be non-nested and non-tiny, and let F : J n ! J be independent and implicit consensus preserving. I need to show that F is dictatorial. By Lemma 2, F is systematic. By Lemma 4, the set of winning coalitions W F is an ultra…lter over the set of individuals N . As is well-known, every ultra…lter over a …nite set is principal, i.e., there is an individual j 2 N such that W F = fC N : j 2 Cg. Clearly, j is a dictator.

2. Conversely, assume the agenda X is nested or tiny. I need to construct a nondictatorial rule F : J n ! J which is independent and implicit consensus preserving.

As n 3, we may choose an odd-sized subgroup M N containing at least three individuals. (For instance M = N if n is odd, or M = f1; 2; 3g.) De…ne F as the aggregation rule on J n given by majority voting among M , i.e., F (J 1 ; :::; J n ) = fp 2 X : jfi 2 M : p 2 J i gj > jM j =2g for all J 1 ; :::; J n 2 J .

I have to show that F (i) maps into J , (ii) is independent, (iii) is implicit consensus preserving, and (iv) is not dictatorial. Properties (ii) and (iv) hold obviously; regarding (ii), F is in fact even systematic, and regarding (iv) it matters that jM j > 1 and jJ j > 1. Properties (i) and (iii) both follow as soon as we have shown version 3 of implicit consensus preservation. Consider J 1 ; :::; J n 2 J . To show that F (J 1 ; :::; J n ) 2 fJ 1 ; :::; J n g, I distinguish between two cases.

Case 1 : X is nested, i.e., of the form X = fp 1 ; :::; p m g where m is the number of issues and where p 1 entails p 2 , p 2 entails p 3 , etc. Notice that for each J 2 J there is a cut-o¤ level t = t J 2 f1; :::; m + 1g such that J = f:p 1 ; :::; :p t 1 ; p t ; :::; p m g, and that F (J 1 ; :::; J n ) = J i = f:p 1 ; :::; :p t J i 1 ; p t J i ; :::; p m g; where i is the median individual in M , i.e., the (or an) individual i in M such that more than half of the individuals j in M have a cut-o¤ level t J j t J i , and more than half of the individuals j in M have a cut-o¤ level t J j t J i .

Case 2 : X is tiny. As one easily checks, we may assume without loss of generality. that X is redundancy-free and contains only contingent propositions. Then, as X is tiny, it is either a single-issue agenda or a two-issue agenda. In the …rst case, F (J 1 ; :::; J n ) is a singleton fpg, which equals J i for any individual i accepting p. In the second case, F (J 1 ; :::; J n ) is a binary set fp; qg; since the subgroups fi 2 M : p 2 J i g and fi 2 M : q 2 J i g each contain a majority of the individuals in M , these subgroups share at least one individual i, whose judgment set is therefore J i = fp; qg = F (J 1 ; :::; J n ).

B.2 Results of Section 4 on aggregation systems

Proof of Theorem 2. We consider any aggregation system (F X ) X2X .

1. First, suppose (F X ) X2X is agenda-insensitive.

Claim 1: (F X ) X2X is coherent.

Consider X; X 0 2 X with X X 0 and J 1 ; :::; J n 2 J X . Each J i is consistent, and thus extendible to a J 0 i 2 J X 0 . I show that F X (J 1 ; :::; J n ) F X 0 (J 0 1 ; :::; J 0 n ). Consider any p 2 F X (J 1 ; :::; J n ). Applying agenda-insensitivity to the agendas X and X 0 , the proposition p (2 X = X \ X 0 ) and the judgment sets J 0 i (2 J X 0 = J X[X 0 ), and noting that each J 0 i satis…es J 0 i \ X = J i and J 0 i \ X 0 = J 0 i , we obtain that p 2 F X (J 1 ; :::; J n ) , p 2 F X 0 (J 0 1 ; :::; J 0 n ):

So, as p 2 F X (J 1 ; :::; J n ) by assumption, p 2 F X 0 (J 0 1 ; :::; J 0 n ). QED Claim 2: Each F X is independent.

Consider any X 2 X , p 2 X, and (J 1 ; :::; J n ); (J 0 1 ; :::; J 0 n ) 2 J n X such that, for all i, p 2 J i , p 2 J 0 i . De…ne Z as the agenda fp; :pg 2 X . For each i, let K i be fpg if p 2 J i (or equivalently, p 2 J 0 i ), and as f:pg otherwise. Applying agenda-insensitivity to the agendas X; Z and the judgment sets J i (2 J X = J X[X 0 ), and noting that J i \ X = J i and J i \ Z = K i , we obtain p 2 F X (J 1 ; :::; J n ) , p 2 F Z (K 1 ; :::; K n ):

(5)

Applying agenda-insensitivity again, this time to the agendas X; Z and the judgment sets

J 0 i (2 J X = J X[Z )
, and noting that J 0 i \ X = J 0 i and J 0 i \ Z = K i , we obtain p 2 F X (J 0 1 ; :::

; J 0 n ) , p 2 F Z (K 1 ; :::; K n ): (6)
By ( 5) and ( 6), p 2 F X (J 1 ; :::; J n ) , p 2 F X (J 0 1 ; :::; J 0 n ). QED 2. Now suppose (F X ) X2X is coherent and independent. I prove agenda-insensitivity. Consider any X; X 0 2 X , p 2 X \ X 0 and J 1 ; :::; J n 2 J X[X 0 , and let us show that p 2 F X (J 1 \ X; :::; J n \ X) if and only if p 2 F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ). Consider the agenda Z := fp; :pg 2 X , and for each i let K i be fpg if p 2 J i and f:pg otherwise. Note that p 2 K i is equivalent to p 2 J i \ X and also to p 2 J i \ X 0 , because each of these three statements is equivalent to p 2 J i . By coherence applied to the agendas Z and X, the judgment sets K 1 ; :::; K n 2 J Z have extensions L 1 K 1 ; :::; L n K n in J X such that F Z (K 1 ; :::; K n ) F X (L 1 ; :::; L n ). It follows that p 2 F X (L 1 ; :::; L n ) , p 2 F Z (K 1 ; :::; K n ):

(7)

Further, for any i, p 2 L i is equivalent to p 2 K i (as L i K i ), which is in turn equivalent to p 2 J i \ X (as shown above). Hence, as F X is independent, p 2 F X (L 1 ; :::; L n ) , p 2 F X (J 1 \ X; :::; J n \ X). By (7) it follows that p 2 F X (J 1 \ X; :::; J n \ X) , p 2 F Z (K 1 ; :::; K n ):

By a similar argument for the agenda X 0 , p 2 F X 0 (J 1 \ X 0 ; :::

; J n \ X 0 ) , p 2 F Z (K 1 ; :::; K n ): (9)
By ( 8) and ( 9), p 2 F X (J 1 \ X; :::; J n \ X) , p 2 F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ).

Proof of Theorem 3. Let (F X ) X2X be any aggregation system.

1. First let (F X ) X2X be fully agenda-insensitive.

Claim 1: (F X ) X2X is strongly coherent.

Consider X; X 0 2 X with X X 0 and J 1 ; :::; J n in J X . For any individual i, since J i is consistent, it is extendible to a b J i 2 J X[X 0 ; we let J 0 i := b J i \ X 0 (2 J X 0 ). We have to show that (i) each J 0 i entails J i , and (ii) F X 0 (J 0 1 ; :::; J 0 n ) entails F X (J 1 ; :::; J n ). Regarding (i), for any i, as J 0 i 2 J X 0 and X X 0 , the set J 0 i entails exactly one set in J X ; so, by the consistency of J 0 i [ J i (= b J i ), J 0 i entails J i . Regarding (ii), consider any p 2 F X (J 1 ; :::; J n ) and let us show that F X 0 (J 0 1 ; :::; J 0 n ) entails p. Applying full agendainsensitivity to the agendas X; X 0 , the proposition p (2 X = X \ X 0 ) and the sets b J i (which satisfy b J i \ X = J i and b J i \ X 0 = J 0 i ), we obtain F X (J 1 ; :::; J n ) entails p , F X 0 (J 0 1 ; :::; J 0 n ) entails p:

The left-hand side holds as p 2 F X (J 1 ; :::; J n ). So, F X 0 (J 0 1 ; :::; J 0 n ) entails p. QED Claim 2: Each F X is independent on X.

Consider any X 2 X , any p 2 X, and any (J 1 ; :::; J n ); (J 0 1 ; :::; J 0 n ) 2 J X such that, for all i, J i entails p if and only if J 0 i does so. Let Z be the agenda fp; :pg 2 X . For each i, I de…ne K i as fpg if J i (or equivalently J 0 i ) entails p, and as f:pg otherwise, and I de…ne

L i := J i [ K i and L 0 i := J 0 i [ K i .
Applying full agenda-insensitivity to the agendas X; Z and the judgment sets L i (which belong to J X[Z and satisfy L i \ X = J i and L i \ Z = K i ), we obtain F X (J 1 ; :::; J n ) entails p , F Z (K 1 ; :::; K n ) entails p:

(10)

Now applying full agenda-insensitivity to the agendas X; Z and the judgment sets L 0 i (which belong to J X[Z and satisfy

L 0 i \ X = J 0 i and L 0 i \ Z = K i ), we obtain F X (J 0 1 ; :::; J 0 n ) entails p , F Z (K 1 ; :::; K n ) entails p: (11)
The relations ( 10) and ( 11) jointly imply that F X (J 1 ; :::; J n ) entails p if and only if F Z (J 0 1 ; :::; J 0 n ) entails p. QED 2. Conversely, assume that (F X ) X2X is strongly coherent and independent on X. To show full agenda-insensitivity, we consider any X; X 0 2 X , p 2 X \ X 0 and J 1 ; :::; J n 2 J X[X 0 , and show that F X (J 1 \ X; :::; J n \ X) entails p if and only if F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ) entails p. Consider the agenda Z := fp; :pg 2 X . For each i, de…ne K i as fpg if J i entails p and f:pg otherwise. By construction, J i entails K i . So, J i \ X also entails K i (as J i \ X 2 J X and p 2 X), and so J i \ X entails p if and only if p 2 K i . For analogous reasons, J i \ X 0 entails K i , and so J i \ X 0 entails p if and only if p 2 K i . By strong coherence applied to the agendas Z; X (which indeed satisfy Z X as p 2 X) and the judgment sets K i 2 J Z , there exist some L 1 ; :::; L n 2 J X such that each L i entails K i and F X (L 1 ; :::; L n ) entails F Z (K 1 ; :::; K n ). As F X (L 1 ; :::; L n ) entails F Z (K 1 ; :::; K n ) (and as F Z (K 1 ; ::

:; K n ) is fpg or f:pg), F X (L 1 ; :::; L n ) entails p , p 2 F Z (K 1 ; :::; K n ): (12)
Similarly, for any i, as L i entails K i (and as K i is fpg or f:pg), L i entails p if and only if p 2 K i , which was shown to hold if and only if J i \ X entails p. So, as F X is independent on X, F X (L 1 ; :::; L n ) entails p if and only if F X (J 1 \ X; :::; J n \ X) entails p. By (12) it follows that F X (J 1 \ X; :::; J n \ X) entails p , p 2 F Z (K 1 ; :::; K n ):

By an analogous argument for the agenda X 0 , F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ) entails p , p 2 F Z (K 1 ; :::

; K n ): (14)
The relations ( 13) and ( 14) imply that F X (J 1 \ X; :::; J n \ X) entails p if and only if F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ) entails p.

Analogously, for any p 2 X, the disjunction p _ q is the smallest upper bound of p and q. QED Claim 2 : The lattice (X; `) is distributive.

Let p; q; r 2 X. Since p `p_q and p `p_r, we have (*) p `(p_q)^(p_r). Since q^r entails q (which entails p_q) and entails r (which entails p_r), (**) q^r `(p_q)^(p_r). By (*) and (**), p _ (q ^r) `(p _ q) ^(p _ r):

We next show the converse implication,

(p _ q) ^(p _ r) `p _ (q ^r): (16)
Consider any J 2 J containing (p _ q) ^(p _ r), and let us show that p _ (q ^r) 2 J.

As (p _ q) ^(p _ r) entails p _ q and also p _ r, we have p _ q; p _ r 2 J. So, J contains p or q (or both), and contains p or r (or both). So, J contains p or contains both q and r; in the latter case, q ^r 2 J. Since, as we have shown, p 2 J or q ^r 2 J, we have p _ (q ^r) 2 J, as desired. By ( 15) and ( 16), and by the asymmetry of `, p _ (q ^r) = (p _ q) ^(p _ r). By analogous arguments, p ^(q _ r) = (p ^q) _ (p ^r). QED Claim 3 : X has a smallest element ? and a greatest element |, namely the contradiction V p2X p and the tautology W p2X p, respectively.

It is obvious that V p2X p entails each q 2 X and that each q 2 X entails W p2X p. QED Claim 4 : For each p 2 X, p ^:p = ? and p _ :p = | (i.e., :p is the algebraic complement of p).

Let p 2 X. Since fp; :pg is inconsistent, p ^:p = ?. Since every J 2 J contains p or :p, every J 2 J contains p _ :p, whence p _ :p = |.

Lemma 5 The notions of consistency, entailment, conjunction and disjunction are preserved by any extension of the agenda (and thus can be used without referring explicitly to an agenda). Formally, for any agenda X and any superagenda X 0 (e.g., the scope of X), (a) a set S X is consistent w.r.t. X if and only if it is so w.r.t. X 0 , (b) a proposition p 2 X (or set S X) entails a proposition p 0 2 X (or set S 0 X) w.r.t. X if and only if it does so w.r.t. X 0 , (c) a proposition r 2 X is the (or a) conjunction/disjunction of certain propositions in X w.r.t. X if and only if it is so w.r.t. X 0 .

Proof. Part (b) follows from part (a), since the entailment notion is reducible to the consistency notion (e.g., p entails p 0 if and only if fp; :p 0 g is inconsistent). Further, part (c) follows from part (b), since the notions of conjunction and disjunction are reducible to the entailment notion: r is a conjunction of a set of propositions S if and only if r and S entail each other, and r is a disjunction of the set of propositions S if and only if :r and f:p : p 2 Sg entail each other. To prove part (a), recall that (*) J X = fJ 0 \ X : J 0 2 J X 0 g. Consider any S X. First, let S be consistent w.r.t. X. Then there is a J 2 J X such that S J. By (*), we may write J = J 0 \ X for some J 0 2 J X 0 . Clearly, S J 0 , whence S is consistent w.r.t. X 0 . Conversely, assume S is

To show this, we consider any agenda V and construct a semantic agenda X of the given sort to which V is isomorphic. Let the set of worlds be := J V . To each p 2 V corresponds a set of worlds, the 'extension'of p, given by [p] := f! 2 : p 2 !g. Note that the assignment p 7 ! [p] de…nes a bijection from V to the set X := f[p] : p 2 V g. I de…ne an agenda by the set X, endowed with issues de…ned as the sets f[p]; [:p]g (which indeed partition X into pairs, since the sets fp; :pg partition V into pairs and since p 7 ! [p] maps V bijectively to X), rational judgment sets de…ned as the sets J X containing exactly one member of each issue and satisfying T

A2J

A 6 = ?.

This agenda X satis…es (i) since [:p] = [p] for all p 2 V , and satis…es (ii) immediately by de…nition. To show that it satis…es (iii), we …rst show that for each J 2 J X the intersection T

A2J

A (6 = ?) is indeed a singleton. Assume for a contradiction that it contains distinct !; ! 0 2 . Since ! 6 = ! 0 , there is a p 2 V such that p 2 ! 0 n! and :p 2 !n! A, a contradiction. Second, one has to check injectivity and surjectivity of the mapping from J X to ff!g : ! 2 g; we leave this to the reader.

Finally, to show that V and X are isomorphic (as agendas), it su¢ ces to show that p 7 ! [p] de…nes an (agenda) isomorphism. This is so because the assignment p 7 ! [p] is bijective, and bijectively maps the issues fp; :pg of V to those of X, and the rational judgment sets of V to those of X (the latter can be shown by verifying that the assignment J 7 ! f[p] : p 2 Jg de…nes a bijection from J V to J X ).

2. From now on we assume that X takes the semantic form de…ned in part 1. In the current part, we show the existence claim. As one can check, X is a subagenda of the agenda X 0 := 2 whose issues are the pairs fA; Ag (A ) and whose rational judgment sets are the sets of the form fA : ! 2 Ag (! 2 ). It su¢ ces to show that X 0 is a minimal closed extension of X. First, X 0 is closed, where the conjunction is given by the intersection, and the disjunction by the union. Second, we have to show minimality. Consider any superagenda X 00 of X which is a strict subagenda of X 0 . We have to show that X 00 is not closed. As X 00 is a subagenda of X 0 , it inherits its issues from X 0 , and thus X 00 is closed under complement: A 2 X 00 ) A 2 X 00 . Since X 0 (= 2 ) is the only subset of 2 which includes X and is closed under intersection and complement, and since X 00 is closed under complement, X 00 cannot be closed under intersection. It follows that X 00 is not closed (i.e., not closed under conjunction), by the following argument. Choose any A; B 2 X 00 such that A \ B 6 2 X 00 . Suppose for a contradiction that X 00 contains a C which (relative to agenda X 00 ) is the conjunction of A and B, i.e., is equivalent to fA; Bg. Sine A \ B 6 2 X 00 , C 6 = A \ B. So, since also C

A and C B (as C entails A and B relative to the agenda X 00 ), we have C ( A \ B. Choose any ! 2 (A \ B)nC. Note that J 00 := fD 2 X 00 : ! 2 Dg belongs to J X 00 , and contains A and B but not C. So (still relative to agenda X 00 ) fA; Bg does not entail C, a contradiction since C is the conjunction of A and B.

3. Finally, we show the uniqueness claim. Since the agenda X 0 de…ned in part 2 is a minimal closed extension of X, it su¢ ces to show that any other such extension of X is equal to X 0 up to relabelling. Let Z be an arbitrary minimal closed superagenda of X. We need to de…ne an agenda isomorphism f : X 0 ! Z which is constant on X. For all ! 2 and all Y X 0 (= 2 ), let Y ! := fA 2 Y : ! 2 Ag, and for all B 2 X 0 (= 2 ) let p B := _ !2B

^X! ( 2 Z):

Here and in what follows, let '_', '^'and ':'refer to the disjunction, conjunction and negation operators of Z (rather than of X or X 0 ). By Proposition 6, '_', '^' and ':' can alternatively be viewed as the algebraic operations of join, meet and complement in the Boolean algebra Z. So, standard algebraic rules apply, such as associativity, commutativity and distributivity of _ and ^. Also, let | and ? be the greatest and smallest elements of the Boolean algebra Z, respectively; clearly, | is the (only) tautology and ? the (only) contradiction of the agenda Z. where the last expression uses notation introduced in Claim 1. This expression is a disjunction of terms (disjuncts) of two types: any V Y A ! with ! 2 A (type 1) and any V Y A ! with ! 2 A (type 2). The result is not a¤ected by adding the following new disjuncts: any V Y A ! with ! 2 A (type 3) and any V Y A ! with ! 2 A (type 4). Indeed, by Claim 1 each new disjunct of type 3 is either ? or coincides with a disjunct of type 1, and any new disjunct of type 4 is either ? or coincides with a disjunct of type 2.

After adding these new disjuncts and re-grouping, the expression becomes " _ !2

^Y A ! # _ " _ !2 ^Y A ! # :
Noting that each Y A ! equals fAg [ Y ! and each Y A ! equals fAg [ Y ! , and then using distributivity twice, the last expression reduces to "

A ^ _ !2 ^Y! !# _ " A ^ _ !2 ^Y! !# = h A _ A i ^ _ !2 ^Y! ! = _ !2
^Y! :

This proves (18). By an analogous argument, one can show that (unless Y = ?), we have Moreover, the agenda S includes X by Claim 3, hence is a superagenda of X. Since Z is by de…nition a minimal closed superagenda of X, it follows that S = Z. QED Claim 6 : For all A; B 2 X 0 , A B if and only if p A entails p B .

W !2 V Y ! = W !2 V Y 0 ! for
For each ! 2 we have p f!g 6 = ?; this is because the set X ! is consistent with respect to agenda X, and hence p f!g = V X ! 6 = ? by Lemma 6. Now consider any A; B 2 X 0 . First, if A B, then p A clearly entails p B since p B is a disjunction of at least those terms of which p A is a disjunction. Conversely, now assume that p A entails p B . As AnB B, p AnB entails p B ; and so, as p B = :p B by Claim 2, p AnB entails :p B . Also, as AnB A, p AnB entails p A ; and so, as p A entails p B , p AnB entails p B . Since, as we have shown, p AnB entails both :p B and p B , it entails :p B ^pB = ?. Hence, p AnB = ?. It follows that AnB = ?, i.e., A B, since if there were an ! 2 AnB, then p f!g would entail p AnB , whence p f!g = ?, in contradiction with what was shown at the start of the proof of the claim. QED Claim 7 : The assignment B 7 ! p B de…nes an agenda isomorphism between X 0 and Z which is constant on X. (This completes the proof.)

This assignment -call it f -is constant on X by Claim 3, and surjective by Claim 4. To show injectivity, consider distinct A; B 2 X 0 . We may assume without loss of generality. that A 6 B (since otherwise the roles of A and B can be interchanged). By Claim 6, p A does not entail p B , and so p A 6 = p B .

It remains to show that f preserves the agenda structure: the issues (resp. negation operator) and the interconnections. This could be deduced from Claims 2, 5 and 6 since, …rstly, by these claims the (bijective) function f is a Boolean-algebra isomorphism, and, secondly, for a closed agenda, the agenda structure and the Boolean-algebra structure are interde…nable, as can be veri…ed; see Proposition 6. 33 But let me give a direct proof. First, f preserves the issues structure, since for each A 2 X 0 we have :p A = p A (by Claim 2) and A is the X 0 -relative negation of A. Second, consider a set S X 0 ; we show that S is consistent (in the sense of X 0 ) if and only if its image fp B : B 2 Sg is consistent (in the sense of Z). This holds for the following reasons. S is consistent if and only if \S 6 = ?, which is in turn equivalent to p \S 6 = p ? , i.e., to p \S 6 = ?. By Claim 4, the latter is equivalent to V B2S p B 6 = ?, which is in turn equivalent to the consistency of fp B : B 2 Sg by Lemma 6.

Proof of Lemma 1. This lemma follows from the proof of Proposition 4.

Proof of Proposition 5. Let X be an agenda. It su¢ ces to show that for each J 2 J X and p 2 X, J entails p or entails :p, or equivalently, V q2J q entails p or entails :p. This follows from the fact that, by Lemma 1, V q2J q is an atom of X, i.e., a logically strongest element of Xnf?g.

Acknowledgments: I thank two dedicated referees and an editor for very helpful comments. I gratefully acknowledge support by the French Agence Nationale de la Recherche (ANR-12-INEG-0006-01).

  ., Puppe, C. (2010) Abstract Arrovian Aggregation, Journal of Economic Theory 145(2): 467-494 Nehring, K., Pivato, M., Puppe, C. (2014) The Condorcet set: Majority voting over interconnected propositions, Journal of Economic Theory 151: 268-303 Pauly, M., van Hees, M. (2006) Logical Constraints on Judgment Aggregation, Journal of Philosophical Logic 35(6): 569-585 Pigozzi, G. (2006) Belief merging and the discursive dilemma: an argument-based account to paradoxes of judgment aggregation, Synthese 152(2): 285-298 Pivato, M. (2009) Geometric models of consistent judgement aggregation, Social Choice and Welfare 33(4): 559-574 Rothstein, P. (1990) Order Restricted Preferences and Majority Rule, Social Choice and Welfare 7(4): 331-342 Saporiti, A., Tohmé, F. (2006) Single-crossing, strategic voting and the median choice rule, Social Choice and Welfare 26(2): 363-383 Wilson, R. (1975) On the Theory of Aggregation, Journal of Economic Theory 10: 89-99 A A more standard judgment-aggregation framework for the results and concepts of the paper All displayed results of the main text (the 'theorems', 'propositions', 'corollaries' and 'remarks') and all de…nitions of properties of aggregation rules/systems (such as 'independence'and 'basic/full/focal agenda-insensitivity') continue to apply as stated under a more standard, non-semantic judgment-aggregation framework. Section A.1 de…nes the usual notions of this framework (following List and Pettit 2002 and more precisely Dietrich 2007/2014). Sections A.2 and A.3 add the notions of 'scope'and 'aggregation system', whose de…nitions are less obvious than in a semantic or syntactic framework.

  Examples of syntactic agendas are X = fa; b; a ^bg and X = fa; a ! b; bg . Here and in what follows, Y still stands for S p2Y fp; :pg.

  (b) X has a subset Y such that jY j 3 and (Y nfpg)[f:pg is consistent for all p 2 Y . (c) X has a subset Y such that jY j = 3 and (Y nfpg)[f:pg is consistent for all p 2 Y .

Claim 1 :

 1 For all Y X, A 2 XnY and ! 2 we write Y A ! := Y ! [ fAg. For every subagenda Y of X, A 2 XnY , and! 2 A, either V Y A ! = ? or Y A ! = Y A ! 0 for some ! 0 2 A.Consider any subagenda Y of X, A 2 XnY , and! 2 A. First assume Y A ! is inconsistent w.r.t. agenda X. Then V Y A ! = ? by Lemma 6. Now assume Y A ! is consistent w.r.t. agenda X. So there is an ! 0 2 \Y A ! . In particular, ! 0 2 \Y ! . So, for each B 2 Y , ! 2 B ) ! 0 2 B.In fact, the `)' can be replaced by ',', since ! and ! 0 belong to the same number of sets B in Y (i.e., to half the these sets, asB 2 Y , B 2 Y ). So, Y ! = Y ! 0 , and hence, Y A ! = Y A ! 0 . QED Claim 2 : For all B 2 X 0 , :p B = p B .Let B 2 X 0 . Since : coincides with the algebraic complement operation in Z, it su¢ ces to show that p B _ p B = | and p B ^pB = ?.We …rst prove that p B _ p B = |. Since p B _ p B = any set of the form XnfA; Ag with A 2 X. Note that _

  a set Y 0 of the form Y nfA; Ag with A 2 Y ; which together with (18) yields that W |, as desired.We …nally have to prove that p B V p B = ?. Using distributivity twice, for each ! 2 B the set X ! contains B. Second, we show that B entails p B , or equivalently, that :p B entails :B. This follows from the previous argument applied to B rather than B, because :p B = p B by Claim 2 and because :B = B (as Z is a superagenda of X, so that B's Z-relative negation :p B coincides with B's X-relative negation B). QED Claim 4 : For all A; B 2 X 0 , p A[B = p A _ p B and p A\B = p A ^pB . Let A; B 2 X 0 . The …rst identity holds immediately by de…nition of p A and p B . As for the second identity, using de Morgan's Law (valid in Boolean algebras) and then Claim 2, p A ^pB = :(:p A _ :p B ) = :(p A _ p B ). Now using the …rst identity, it follows that p A^B = :p A[B , which reduces to p A\B by A [ B = A \ B and Claim 2. QED Claim 5 : Z = fp B : B 2 X 0 g. The set S := fp B : B 2 X 0 g ( Z) is closed under negation by Claim 2, hence de…nes a subagenda of Z. The agenda S is closed because for any B; C 2 X 0 the disjunction of p B and p C (relative to the agenda Z) equals p B[C by Claim 4, hence belongs to the agenda S (relative to which it of course still de…nes the disjunction of p B and p C ).

The limited overlap of the present paper withDietrich (2006) is explained in Section

 4 One such assumption is that the agenda setter holds preferences over outcomes that are totally independent of votes and voters'information, as in our introductory example where the agenda setter simply wants a collective judgment of rising prices.

The proposals include premise-and conclusion-based rules (e.g.,[START_REF] Kornhauser | Unpacking the Court[END_REF][START_REF] List | Aggregating sets of judgments: an impossibility result[END_REF], Dietrich 2006), sequential rules (e.g.,List 2004, Dietrich and List 2007b), distance-based

The …niteness restriction could be dropped in many results, e.g., those of Section 4.

I thank Marcus Pivato for bringing this fact to my attention.

3 An aggregation system could be viewed as a single 'extended aggregation rule'with an additional argument, the agenda. Note that each rule FX must have domain J n X and co-domain JX (this restriction might be lifted by a more general de…nition of 'aggregation system').

4 Strong coherence is equivalent to ordinary coherence if the scope of any feasible agenda is a feasible agenda, i.e., if X

X ) X 2 X .

5 Note that Ji entails p if and only if Ji \ X entails p (since p

X), and if and only if Ji \ X 0 entails p (since p 2 X 0 ). So, the requirement that each Ji entails p means that p emerges as unanimously accepted, whether agenda X or agenda X 0 is used.

9 The generalization mentioned in brackets holds because FX inherits dictatorship or constancy from F X 0 by coherence. Analogous remarks apply to Corollary

(in which the generalization draws on strong coherence) and Corollary

2 0Dietrich (2006) only states the 'if'direction; but the 'only if'direction holds trivially.

1 To be entirely precise, one can identify an axiom with the set of all aggregation rules satisfying it, i.e., the set A = fF : for some agenda X 2 , F is a rule J n X ! JX satisfying the axiomg. An axiom is thus given by a set A of rules F : J n X ! JX for agendas X 2 . The axiom is description-insensitive if the set A is closed under equivalence of rules.

2 Most results do not require the …niteness restriction. The condition that jJ j > 1 excludes trivial agendas.2

In more explicit algebraic terms, the agenda is the triple (X; :; J ) containing X and the structure on X, i.e., the negation operator : and the set of rational judgment sets J . Since the negation operator

5 Negation is given by the negation symbol (in fact, to ensure that double-negations cancel out, ':p'is de…ned from p by adding or deleting an initial negation symbol, depending on whether p already starts with a negation symbol). J contains the logically consistent sets J X containing a single member of each issue.2 6 In practice, agendas of course contain no non-contingent propositions and no two equivalent propositions. Propositions are equivalent if they entail one another.

8 Generalizing the earlier de…nition, I call a proposition r the conjunction (resp. disjunction) of a set of propositions S if any rational judgment set contains r if and only if it contains all (resp. some) p 2 S.2 9 Without assuming redundancy-freeness, the agenda is a Boolean algebra modulo equivalence between propositions.

0 Or so we may assume without loss of generality. Recall that the scope is unique up to relabelling propositions.3 1 All our results about aggregation systems remain true if we allow L and any X 2 X to be in…nite, i.e., to be agendas in a generalized sense without …niteness restriction.

2 counting only contingent propositions and counting equivalent propositions (if any) only once

3 For instance, a subset A is consistent in the agenda sense if and only if its algebraic meet is not ?, by Lemma 6.

7 References Ahn, D. S., Oliveros, S. (2014) The Condorcet Jur(ies) Theorem, Journal of Economic Theory 150: 841-51 Bozbay, I., Dietrich, F., Peters, H. (2014) Judgment aggregation in search for the truth, Games and Economic Behavior 87: 571-590 DeClippel, G., Eliaz, K. (2015) Premise-based versus outcome-based information aggregation, Games and Economic Behavior 89: 34-42 Dietrich, F. (2006) Judgment Aggregation: (Im)Possibility Theorems, Journal of Economic Theory 126(1): 286-298 Dietrich, F. (2007) A generalised model of judgment aggregation, Social Choice and Welfare 28(4): 529-565 Dietrich, F. (2014) Scoring rules for judgment aggregation, Social Choice and Welfare 42(4): 873-911 Dietrich, F. (2015) Aggregation theory and the relevance of some issues to others, Journal of Economic Theory 160: 463-493 Dietrich, F., List, C. (2007a) Arrow's theorem in judgment aggregation, Social Choice and Welfare 29(1): 19-33

Proof of Theorem 4. Consider an aggregation system (F X ) X2X .

1. First let this system be focally agenda-insensitive and unanimity preserving. Fix any X 2 X . Let p 2 X be entailed by each of J 1 ; :::; J n 2 J X . We have to show that F X (J 1 ; :::; J n ) entails p. Consider the agenda X 0 := fp; :pg 2 X . Applying focal agenda-insensitivity to the judgment sets J 0 i := J i [ fpg 2 J X[X 0 (each of which entails p 2 X \ X 0 ) and noting that each J 0 i satis…es J 0 i \ X = J i and J 0 i \ X 0 = fpg, we obtain that F X (J 1 ; :::; J n ) entails p , F X 0 (fpg; :::; fpg) entails p:

The right-hand side (in which 'entails'can be replaced by 'contains') holds since F X 0 is unanimity preserving. So the left-hand side holds, as desired.

2. Conversely, assume each F X is implicit consensus preserving. It then obviously is unanimity preserving. To show focal agenda-insensitivity, we consider any X; X 0 2 X and p 2 X \ X 0 , and any J 1 ; :::; J n 2 J X[X 0 each of which entails p. We show that F X (J 1 \ X; :::; J n \ X) entails p; for analogous reasons also F X 0 (J 1 \ X 0 ; :::; J n \ X 0 ) entails p, and since both entailments are therefore true they are automatically equivalent, completing the proof. Fix an individual i. Since J i is consistent and entails p, J i is consistent with p. So J i \ X is also consistent with p, and therefore cannot entail :p. Now, as p 2 X, every judgment set in J X (such as J i \ X) entails either p or :p. So J i \ X entails p. As this is true for all i and as F X is implicit consensus preserving, F X (J 1 \ X; :::; J n \ X) entails p.

B.3 Results of Appendix A.2 on the scope of agendas

I prove these results in a slightly di¤erent order, and draw on additional lemmas.

Proof of Proposition 2. The conjunction (disjunction) of elements p; q of a redundancyfree agenda X is unique because any two conjunctions (disjunctions) of p and q entail each other, hence coincide as X is redundancy-free.

Proof of Proposition 3. Suppose an agenda X is closed under conjunction. Let p; q 2 X. Let r 2 X be the (possibly not unique) conjunction of :p and :q. Then :r is the (possibly not unique) disjunction of p and q. Indeed, any J 2 J contains p or q if and only if it is not the case that :p; :q 2 J; which is equivalent to r 6 2 J, i.e., to :r 2 J. Analogously, one can show that if X is closed under disjunction then any p; q 2 X have a conjunction in X, namely the proposition :r where r is a disjunction of :p and :q.

Proof of Proposition 6. Let X be a closed redundancy-free agenda and `the relation of entailment between propositions. The proof proceeds in four claims.

Claim 1 : (X; `) is a lattice whose meet and join are given by the operations of conjunction ^and disjunction _, respectively. First, `is a partial order: it is clearly re ‡exive and transitive, and it is also antisymmetric as X is redundancy-free. Next, for any p; q 2 X, the conjunction p ^q is the greatest lower bound of p and q because, …rstly, it is a lower bound (i.e., p ^q `p; q), and, secondly, if r is also a lower bound, then r `p ^q, as r `p; q and fp; qg `p ^q. consistent w.r.t. X 0 . Then we may choose a J 0 2 J X 0 such that S J 0 . By (*), J X contains J := J 0 \ X. Note that S J. So S is consistent w.r.t. X.

Lemma 6 For any agenda X and any closed (redundancy-free) superagenda X 0 -possible X itself or the scope of X -a set A X is consistent if and only if, in X 0 , V p2A p 6 = ?.

Proof. Let X and X 0 be as speci…ed. By Lemma 5, we need not distinguish between consistency w.r.t. X and w.r.t. X 0 . We proceed by showing three claims.

Claim 1 : ? is the only element of X 0 which is not contained in any rational judgment set J 2 J X 0 . This follows from four facts (some of which draw on Proposition 6): (i) ? is the only element of X 0 which entails its own algebraic complement (a basic fact about Boolean algebras); (ii) the algebraic complement of an element p is its (agenda-theoretic) negation :p; (iii) an element p entails another q if and only if no J 2 J X 0 contains both p and :q; (iv) every J 2 J X 0 contains exactly one of member of each pair p; :p 2 X. QED Claim 2 : For any J 2 J X 0 and any A J, we have

Let J 2 J X 0 and A J. By Proposition 6 we can think of '^' alternatively as the conjunction operator (de…ned agenda-theoretically) or the meet (de…ned Booleanalgebraically). The claim holds by induction on the size of A. If A = ?, the claims holds because then V p2A p = | and | 2 J (as | = : ?, where ? 6 2 J by Claim 1). Now assume A has size m 1 and suppose the claim holds for any smaller size. We may write A = A 0 [ fqg with q 6 2 A 0 . By induction hypothesis, V p2A 0 p 2 J. Since J contains both V p2A 0 p and q, J contains their conjunction ( V

First, let A X 0 be consistent. Then it has an extension J 2 J X 0 , which by Claim 2 contains V Proof of Proposition 4. Let X be an agenda. As one easily checks, we may assume without loss of generality. that X is redundancy-free.

1. In this part we show that we may assume without loss of generality. that X is a 'semantic' agenda given as follows: there exists a …nite set of 'worlds' 6 = ? such that X 2 , where (i) each issue takes the form fA; Ag (I write A for the complement nA of any set A ), (ii) the set J X of rational judgment sets consists of those sets J X which contain exactly one member of each issue and satisfy T

A2J

A 6 = ?, and (iii) rational judgment sets in J X correspond to worlds in , in the sense that the assignment J 7 ! T

A2J

A de…nes a bijection from J X to ff!g : ! 2 g. i 1

A :

It thus su¢ ces to show that for all ! 2 B and ! 0 2 B we have

As A V A = ? (since A and A are complements in the algebra Z), it follows that [ V X ! ] V [ V X ! 0 ] entails ?, hence, equals ?. QED Claim 3 : p B = B for all B 2 X.

Consider any B 2 X. We regard B as an element of the extended agenda Z X. Since Z is redundancy-free, it su¢ ces to show that p B and B entail each other. We …rst show that p B entails B. Since p B is the least upper bound of all V X ! with ! 2 B, it su¢ ces to show that B is an upper bound, i.e., that each of these V X ! entails B. This