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Open Educational Resources for School Teachers from Developing Nations

'''This course concluded on May 2nd, 2011. A new detailed OER introductory course is being planned for August 2011.'''

A new course blog has been created. This blog will provide you with step-by-step instructions to complete your assignments.

Course objective
Introductory Open Educational Resources (OER) course is focused mainly for elementary school teachers from developing nations. After this course participants will be able to:

 * recognise open and free educational resources online
 * understand the different license policies
 * learn how to reuse, create new, remix and share/distribute educational resource
 * be a member of the OER community and spread the awareness of OER

Course length and Registration
4 weeks ( 4 classes, each 2 to 3 hrs long. Total - not more than 12 Hours)

Target audience
Anyone interested to learn the basics of searching, creating, re-using and sharing license free English resources for the purpose of educating the elementary and secondary school students. This course is prepared keeping in mind that all elementary school teachers from developing nations might now have computers and internet access at their homes, but still are willing to learn by finding internet access at an educational institution or internet centres.

Participants
The course is intended to commence from April 4th, 2011. The course facilitator plans to start the course after there are at least 20 participant registrations. To view a list of registered participants, visit the participants page.

Language
This course is focused towards school teachers from developing nations teaching in English medium schools, hence this course will be in English language. The participants are encouraged and are most welcome to search, remix and re-use educational materials in other languages that are meant to be re-used and re-distributed. urdu

Course Assignments
Participants are encouraged to open their own blog or a course community discussion page (common blog space) where in they are expected to submit their assignments and share opinions and comments about the suggested readings. The participants are required to invest at least 2-3 hours of work each week for this course.

Class meetings
This class will meet semi-synchronously online. This means, the first and fourth weekly class will be synchronous online meeting (skype conference with participants at an agreed time) and the second and third will be asynchronous online meeting (online self study time).

Course Schedule
Week 1. Introduction to OER (Online synchronous meeting) Class 1- online synchronous meeting
 * What are Open_educational_resources (OER)?
 * Brief history of OER.
 * Definition of OER.
 * Introduction to Creative Commons licenses and understanding Copyright.

Assignment: You can email your answers to the course facilitator or choose to post it on your blog or the community web blog.
 * Create an individual blog or a community blog at wordpress, or at blogger.
 * Write one post blog (one or two paragraphs) introducing yourself (who you are and what you teach) and what you are looking forward to learn from this course.
 * Read the suggested readings and write one page, introducing OER to a school that you know, which does not use OER.
 * Write half a page talking about the main characteristics of OER.
 * Watch the clips/short videos on this website and discuss on your blog/common webspace what you think about them.

Suggested readings:
 * 1) http://wikieducator.org/OER_Handbook/educator
 * 2) Open_educational_resources
 * 3) http://www.col.org/SiteCollectionDocuments/Introducing_Copyright_online_edition.pdf - by Julien Hofman and Commonwealth of Learning, 2009.
 * 4) http://creativecommons.org/licenses/

Week 2. OER Authoring Tools (Asynchronous meeting) Class 2 – asynchronous meeting
 * Introduction to OER Authoring tools by Christopher Pappas.
 * Introduction to LeMill and Wikiversity.

Assignment:

You can email your answers to the course facilitator or choose to post it on your blog or the community web blog.
 * Please register yourself in LeMill and Wikiversity.
 * Find and list at least 4/5 other open educational resources websites like LeMill that are relevant/applicable to your subject area, list these on your blog or group blog.
 * Write one blog post about your favourite authoring tool and its features, why you think it is better than the others.

Suggested readings:
 * 1) http://lemill.net/content/webpages/overview-of-lemill
 * 2) http://lemill.net/content/webpages/lemill-faq
 * 3) Help:The_original_tour_for_newcomers
 * 4) Introduction

Week 3. Reusing OER and Open Content (Asynchronous meeting) Class 3 – asynchronous meeting
 * Remixing lesson plans and distributing it.
 * How to create and re-use resources in Curriki and LeMill
 * Open License compatibility issues, Creative Commons image collections of flickr for use in OER.

Assignment: You can email your answers to the course facilitator or choose to post it on your blog or the community web blog.
 * Find at least 5 educational materials that are relevant to your subject and syllabus and submit/post it them on your blog or community web space.
 * Create your own or re-mix at least 3 free and open educational resource relevant to your subject area.
 * Read the third suggested reading for this week and write a post on your blog about the different CC licenses.

Suggested readings:
 * 1) http://www.curriki.org/
 * 2) http://www.curriki.org/xwiki/bin/view/Coll_Group_OERi-OpenEducationalResourcesintegration/RemixModule3?bc=
 * 3) http://www.oerafrica.org/ResourceDownload.aspx?assetid=317&userid=-1

Week 4. Implementing OER in developing nations (Online synchronous meeting) Class 4 - online synchronous meeting
 * Implementing OER in developing nations.
 * Feedback and discussion.

Assignment:
 * Participants are requested to read the suggested readings for this week and write their opinion on blog/group blog about how OER can be implemented successfully in developing countries.
 * Participants are hereby requested to write one post on your blog/group blog about what they have learnt from this course and how they will use this knowledge in their classes.

Suggested readings:
 * 1) http://cnx.org/content/m14422/latest/ by Sunil Kumar Singh

Recognition
Participants who have submitted all the assignments and upon completion of these 4 week course will be awarded a certificate recognising your understanding of introductory course in OER, from Tallinn University, Estonia. This online course is endorsed by the Institute of Informatics, Tallinn University.

Additional useful links

 * 1) http://www.sciencenetlinks.com/index.php
 * 2) http://www.readwritethink.org/
 * 3) http://www.oercommons.org/
 * 4) http://www.merlot.org/merlot/index.htm
 * 5) http://www.wetheteachers.com/
 * 6) http://oer.issuelab.org/research
 * 7) http://www.100people.org/lesson_plans.php
 * 8) http://ck-12.com/flexbook/
 * 9) http://www.khanacademy.org/
 * 10) http://www.collegeopentextbooks.org/

Acknowledgments
I would like to thank Hans Põldoja - my thesis supervisor - and Sirje Virkus - my teacher - for their guidance. Without their ideas and help I would not have plunged this deep into the fascinating world of Open Educational Resources. Comment added by User:Nithinlaxman (talk), on or before 12:48, 17 March 2011


text:
User:DavidZip9

This is my user page.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User:DavidZip9


text:
MediaWiki:Ipboptions

3 hours:3 hours,8 hours:8 hours,1 day:1 day,5 days:5 days,1 week:1 week,2 weeks:2 weeks,1 month:1 month,3 months:3 months,6 months:6 months,1 year:1 year,indefinite:indefinite

title: MediaWiki:Ipboptions


text:
User talk:Alpha Quadrant

 Hello Alpha Quadrant, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. All users are expected to abide by our Privacy policy, Civility policy, and the Terms of Use while at Wikiversity.

To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.
 * Explore Wikiversity with the links to your left.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.
 * Follow Wikiversity on twitter (http://twitter.com/Wikiversity) and identi.ca (http://identi.ca/group/wikiversity).

You don't need to be an educator to edit. You only need to be bold to contribute and to experiment with the sandbox or your userpage. See you around Wikiversity! --TeleComNasSprVen 03:46, 12 March 2011 (UTC)

title: User talk:Alpha Quadrant


text:
MediaWiki:Protect-expiry-options

8 hours:8 hours,1 day:1 day,5 days:5 days,1 week:1 week,2 weeks:2 weeks,1 month:1 month,3 months:3 months,6 months:6 months,1 year:1 year,indefinite:indefinite

title: MediaWiki:Protect-expiry-options


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User:Vivianc.

Hi ♥ My name is Vivian Chailerd. I'm 19 years old and my birthday is on February 27, 1992. I'm studying at MUIC and my major is TIM. I studied high school at British International School Phuket. “I love computer essential class and I love Brian”

Best PC Computer Specs

Processor: Core I7 980X 3.33 GHz. Intel Core i7 Processors. Number processor i7-980X. Cache memory, 12 MB L3. Maximum memory 24GB. Signals Clock Speed 3.33 GHz. Max Turbo Frequency 3.60 GHz. Intel QPI Speed 6.4 GT / s. Yes Hyper-Threading Technology. Number Core 6 Core. Number of Threads 12 Threads.

Memory: Kingston HyperX "blu" DDR3. Kingston Part Number: KHX1600C9D3B1/2G. Description: 2GB 1600MHz DDR3 Non-ECC. Features: DDR3-1600 Default = DDR3-1333. Latency Timings: CL9-9-9-27 CL9-9-9. Voltage: 1.7V-1.9V 1.5V.

Hard Drive: Seagate Barracuda 7200.9 120 GB ATA-100 Hard Drive.Capacity: 	120.0 GB Designation: Desktop Computer. Enclosure: Internal. HDD Form Factor: 3.5". Interface: IDE/EIDE. Spindle Speed: 7200.0 RPM

Optical drive: Oppo Digital BDP-83 Blu-Ray Player: Playable Disk Types: DVD Video, DVD-R, DVD-RW, DVD+R, DVD+RW, DVD Audio, SACD, CD (Audio), CD-R, CD-RW, HDCD, BD-RE, DVD-R DL, DVD+R DL, Blu-ray. Playable File Formats: MP3, JPEG, DivX, WMA. Progressive Scan: yes. Number of Discs: 1.0. TV Tuner: With TV Tuner.

AGP: ASUS GTX580. Codename: GF110. Transistor Count (Millions): 3000 million. Fabrication Process: 40 nm. Memory (Max): 1.536GB. Core Speed: 782 MHz. Shader Speed: 1564 MHz. Memory Speed: 4000 MHz. Memory Bus Width: 384-bit.

Spec of mobil

1: HTC Magic Processor: Qualcomm® MSM7200A™, 528 MHz Operating system: Android Memory: ROM=512MB RAM=288MB Device control: trackball with enter button Internal GPS antenna Bluetooth® 2.0 with Enhanced Data Rate and A2DP for wireless stereo headsets Wi-Fi®: IEEE 802.11 b/g HTC ExtUSB™ (11-pin mini-USB 2.0 and audio jack in one) Camera:3.2 megapixel color camera with auto focus Audio support formats:AAC, AAC+, AMR-NB, MP3, WMA, WAV, AAC-LC, MIDI, OGG HTC Sync™ system requirements:Microsoft Windows XP Service Pack 2 (SP2) or higher or Microsoft Windows Vista® Microsoft® Office Outlook® 2000/2002/2003/2007, Windows Address Book (Outlook Express), or Windows Contacts Microsoft® Exchange Server synchronization requirements: Microsoft Exchange Server 2003 Service Pack 2 (SP2) or Microsoft Exchange Server 2007 Battery:Rechargeable Lithium-ion battery Capacity: 1340 mAh Talk time: Up to 400 minutes for WCDMA Up to 450 minutes for GSM Standby time3: Up to 660 hours for WCDMA Up to 420 hours for GSM (The above are subject to network and phone usage.) MicroSD™ memory card (SD 2.0 compatible) AC adapter:Voltage range/frequency: 100 ~ 240V AC, 47/63 Hz DC output: 5V and 1A G-sensor Digital Compass

2:Ipad (wifi+3g) Height: 9.50 inches (241.2 mm) Width: 7.31 inches (185.7 mm) Depth: 0.34 inch (8.8 mm) Weight: 1.35 pounds (613 g) (Wi-Fi + 3G model) Weight: 1.34 pounds (607 g) (Wi-Fi + 3G for Verizon model) Storages: 16GB,32GB,64GB Wi-Fi + 3G model: UMTS/HSDPA/HSUPA (850, 900, 1900, 2100 MHz); GSM/EDGE (850, 900, 1800, 1900 MHz) Wi-Fi + 3G for Verizon model: CDMA EV-DO Rev. A (800, 1900 MHz) Data only3 Wi-Fi (802.11a/b/g/n) Bluetooth 2.1 + EDR technology 9.7-inch (diagonal) LED-backlit glossy widescreen Multi-Touch display with IPS technology 1024-by-768-pixel resolution at 132 pixels per inch (ppi) Fingerprint-resistant oleophobic coating Support for display of multiple languages and characters simultaneously Chip:1GHz dual-core Apple A5 custom-designed, high-performance, low-power system-on-a-chip Back camera: Video recording, HD (720p) up to 30 frames per second with audio; still camera with 5x digital zoom Front camera: Video recording, VGA up to 30 frames per second with audio; VGA-quality still camera Tap to control exposure for video or stills Photo and video geotagging over Wi-Fi Built-in 25-watt-hour rechargeable lithium-polymer battery Up to 10 hours of surfing the web on Wi-Fi, watching video, or listening to music Up to 9 hours of surfing the web using 3G data network Charging via power adapter or USB to computer system 30-pin dock connector port 3.5-mm stereo headphone minijack Built-in speaker Micro-SIM card tray (Wi-Fi + 3G model) Microphone Frequency response: 20Hz to 20,000Hz Audio formats supported: HE-AAC (V1 and V2), AAC (8 to 320 Kbps), Protected AAC (from iTunes Store), MP3 (8 to 320 Kbps), MP3 VBR, Audible (formats 2, 3, and 4, Audible Enhanced Audio, AAX, and AAX+), Apple Lossless, AIFF, and WAV User-configurable maximum volume limit Dolby Digital 5.1 surround sound pass-through with Apple Digital AV Adapter (sold separately) Video mirroring and video out support: Up to 1080p with Apple Digital AV Adapter or Apple VGA Adapter (cables sold separately) Video out support at 576p and 480p with Apple Component AV Cable; 576i and 480i with Apple Composite AV Cable Video formats supported: H.264 video up to 720p, 30 frames per second, Main Profile level 3.1 with AAC-LC audio up to 160 Kbps, 48kHz, stereo audio in .m4v, .mp4, and .mov file formats; MPEG-4 video, up to 2.5 Mbps, 640 by 480 pixels, 30 frames per second, Simple Profile with AAC-LC audio up to 160 Kbps per channel, 48kHz, stereo audio in .m4v, .mp4, and .mov file formats; Motion JPEG (M-JPEG) up to 35 Mbps, 1280 by 720 pixels, 30 frames per second, audio in ulaw, PCM stereo audio in .avi file format Viewable document types: .jpg, .tiff, .gif (images); .doc and .docx (Microsoft Word); .htm and .html (web pages); .key (Keynote); .numbers (Numbers); .pages (Pages); .pdf (Preview and Adobe Acrobat); .ppt and .pptx (Microsoft PowerPoint); .txt (text); .rtf (rich text format); .vcf (contact information); .xls and .xlsx (Microsoft Excel)

3: Blackberry Bold 9700: 256MB onboard memory 2GB media card Camera: 3.2MP, 2X digital zoom+flash Height:109mm Width:60mm Depth: 14mm Weight:122g Video formats: MPEG4 part2 simple profile,H.263,WMV Audio formats: MP3, MIDI, AMR-NB, AAC/AAC+/eAAC+, WMA Wifi:802.11 b / g WPA / WPA2 Personal and Enterprise Cisco CCX certified Wi-Fi® access to BlackBerry® Enterprise Server Wi-Fi access to BlackBerry® Internet Bundle Support for UMA (carrier-dependent) High-resolution 480x360 pixel color display Network:Tri-band 3G HSDPA networks: 2100MHz/1900MHz/850MHz Tri-band UMTS networks: 2100/1900/850/800 MHz and 2100/1700/900 MHz Quad-band GSM/GPRS/EDGE networks: 850/900/1800/1900MHz Battery:1500 mAHR removable/rechargeable lithium-ion battery Talk time: Up to 6 hours (GSM or UMTS) Standby time: Up to 21 days / 504 hours (GSM), up to 17 days / 408 hours (UMTS) Music playback time: Up to 38 hours Audio and voice:3.5 mm stereo headset jack Built-in speakerphone Full QWERTY keyboard ESC key to right of trackpad/trackball Menu key to left of trackpad/trackball Keyboard backlighting Password protection with keyboard lock Sleep mode AES or Triple DES encryption with BlackBerry Enterprise Server Optional support for S/MIME Processor:FIPS validationRIM® wireless modem Tethered modem capability

People in a group: Phuminth Chitiparlungsri 5180348 Phanitnan Laohapisitpanich 5280520 Vivian Chailerd 5280570

title: User:Vivianc.


text:
User:WIKIADVERSITY

hello.

i'd like to learn and share knowledge.

i have two websites...

one of them is http://selfindulgence.org/

the other is

http:// en.wikademia.org/

advocate for your freedom.

"think for yourself, question authority"

title: User:WIKIADVERSITY


text: File:Snapshot 20110310.JPG

title: File:Snapshot 20110310.JPG


title: File:Fe1.s11.mtg29.djvu


text:
Elasticity/Plane stress

A state of plane stress exists when one of the three principal $$\left(\sigma_1, \sigma_2, \sigma_3 \right)\,\!$$, stresses is zero. This usually occurs where one dimension is very small compared to the other two, i.e. the element is flat or thin. In this case, the stresses are negligible with respect to the smaller dimension as they are not able to develop within the material and are small compared to the in-plane stresses. Therefore, the face of the element is not acted by loads and the structural element can be analyzed as two-dimensional, e.g. thin-walled structures such as plates subject to in-plane loading or thin cylinders subject to pressure loading. The other three non-zero components remain constant over the thickness of the plate. The stress tensor can then be approximated by:

 * $$\sigma_{ij} = \begin{bmatrix}

\sigma_{11} & \sigma_{12} & 0 \\ \sigma_{21} & \sigma_{22} & 0 \\ 0     &     0       & 0 \end{bmatrix} \equiv \begin{bmatrix} \sigma_{x} & \tau_{xy} & 0 \\ \tau_{yx} & \sigma_{y} & 0 \\ 0     &     0       & 0 \end{bmatrix}\,\!$$.

The corresponding strain tensor is:

 * $$\varepsilon_{ij} = \begin{bmatrix}

\varepsilon_{11} & \varepsilon_{12} & 0 \\ \varepsilon_{21} & \varepsilon_{22} & 0 \\ 0     &     0       & \varepsilon_{33}\end{bmatrix}\,\!$$

in which the non-zero $$\varepsilon_{33}\,\!$$ term arises from the Poisson's effect. This strain term can be temporarily removed from the stress analysis to leave only the in-plane terms, effectively reducing the analysis to two dimensions.

title: Elasticity/Plane stress


text: User:Bedroom/Name of new page

title: User:Bedroom/Name of new page


text: User:Bedroom/Homework I

title: User:Bedroom/Homework I


text:
MediaWiki:Infiniteblock

indefinitely

title: MediaWiki:Infiniteblock


text:
Talk:Concept mapping

David Ausubel

 * Overseas lecturing
 * http://www.davidausubel.org/resume4.html

 * Visiting Professor of Psychology, University of Carnpinas (Brazil), April 1976.
 * Lecturer in Psychology: Universities of Bonn, 1 Heidelberg. Tubingen, Bochum, Augsburg, Trier, Hamburg (West Germany), May-June 1977; Universities Nijmegen, Utrecht, Tilburg Free University of Amsterdam., May 1977.
 * Lecturer in Educational and Developmental Psychology, Universities of Tokyo, Osaka, Kyoto, Nagoya International Christian University, Japan, September-October 1978.
 * Visiting Lecturer in most of major universities of Canada and New Zealand (as well as the United States).
 * Visiting Lecturer in Psychology, University of Rio de Janeiro, Catholic University of Sao Paolo, Institute of Applied Linguistics, Brazil, April 1979.
 * Visiting Lecturer in Developmental and Educational Psychology: 25 universities in West Germany, Holland, Austria, Sweden, Norway, and Denmark, May-June 1979.
 * Visiting Lecturer in Educational and Clinical Psychology, University of Padua (Italy); Universities of Essen, Düsseldorf, MunIch, January 1984.
 * See also
 * http://wik.ed.uiuc.edu/index.php/Advance_organizers
 * http://wik.ed.uiuc.edu/index.php/File:Kwl.gif

-- KYPark [T] 07:00, 13 March 2011 (UTC)

title: Talk:Concept mapping


text:
Logical Proofs

This page is a learning resource for those people who want to learn how to do logical proofs. Disclaimer: The information provided here is solely from my own experiences in university logic classes. The goal of this page is to be an easy to understand and to-the-point guide for complete beginners to mathematical proofs.

Proofs can be thought of as 'mathematical writing' with the purpose of showing that an idea is true or false. However, before actually looking at proofs, it's necessary to develop some mathematical language. The next section on propositions will serve as the foundation for a number of key ideas moving forward.

Propositional Translation
For the first part, we are going to learn how to translate natural language into logical expressions. For this we'll use operators shown in the following table:

In the table above, A and B are arbitrary letters that each represent a proposition, which is a statement that is strictly either true or false. For example, we could have A represent the statement "The capital of New York is Albany," and B represent the statement "2 + 2 = 1029384.234." In this case, the truth value (status of being either true or false) of A would be True and would be False for B. Here are some examples of statements that are not propositions:

"It's a nice day today." - This statement is rather subjective. There isn't a universal metric to label the 'niceness' of a given day as a binary true or false.

"x + 1 = 2" - We don't know whether this statement is true or not because we don't know what x is. While one might look at this and think, x must obviously be 1, therefore the statement can be true, recall that propositions can also be false. Rather than thinking of this as an equation to be solved, look at it like more like a question, "is x plus 1 equal to 2?" which we cannot answer without knowing the value of x. However, if the statement separately told us that x is equal to 1, then we could say that x + 1 = 2 is true.

Mathematical Or
Let us begin by looking at the Or operator, denoted by the symbol $$\vee$$.

Suppose that A and B are propositions. The expression "A or B", denoted, $$(A \vee B)$$, is true when either A or B is true.

As a quick note on 'or' in the English language, there is an 'inclusive or' and an 'exclusive or', which people usually determine through context clues. For example, the 'or' used in "The meal comes with soup or salad" is an 'exclusive or' because it's implied that you can only choose one, not both. On the other hand, "Please bring a drivers license or student ID" would have an 'inclusive or', because while you can bring one or the other, there won't be any problem if you bring both. Since there can be quite a bit of ambiguity in spoken language, moving forward we'll need to make some assumptions as to what certain words really mean. One such assumption will be that in a statement, 'or' will refer to the inclusive or, and that the exclusive or (sometimes shortened to XOR) will be explicitly stated when used.

A truth table is a table that shows the outcomes of all possible truth value combinations of the components of the proposition. Since $$(A \vee B)$$ depends on only A and B, we list the truth values of A and B, followed by a column for the truth value of A or B. Notice that $$(A \vee B)$$ is only false when both A and B are false.

Lastly, notice that both A and $$(A \vee B)$$ are propositions. As shown by the truth table, the $$(A \vee B)$$ column only has results that are True or False, depending on the components A and B; recall that the definition of a proposition is a statement that is either true or false. That is to say that propositions can be as short as one letter, or can be a convoluted expression of many components. However, in the end, a proposition will always boil down to being true or false.

Example, evaluate the truth value of this statement: "Either Chicago is in Georgia or Canada is south of Detroit."

Step 1: Convert the statement from English into mathematical notation.

Select letters to represent the smallest propositions involved in the statement and identify any operators used. Recall that choice of letters is arbitrary, so instead of A and B, we could declare that G represents the statement "Chicago is in Georgia", and D represents "Canada is south of Detroit." Then, in mathematical notation, the entire statement is $$(G \vee D)$$

Step 2: Evaluate truth values

The proposition G is false, because Chicago is in Illinois. The proposition D is true, surprisingly (look at a map). With this in mind, we can substitute in the truth values and evaluate the expression: False $$\vee$$ True. Referencing the truth table for the Or operator (which shows that for $$(A \vee B)$$, as long as at least one of the propositions A and B is true, $$(A \vee B)$$ is true), we conclude that $$(G \vee D)$$ is true, and therefore the original statement is true.

Mathematical And
The And operator is denoted by $$\land$$. Suppose that A and B are two propositions. Then the proposition "A and B", denoted $$(A \land B)$$, is true whenever both A and B are true, and false if either A or B is false. See the truth table below: Similar to the Or operator, the truth value after applying the And operator in $$(A \land B)$$ depends on the truth values of the propositions A and B. For example, consider the statement "I did my homework and I cleaned my room." Suppose H and C are propositions where H represents "I did my homework" and C represents "I cleaned my room". Then the statement can be written as $$(H \land C)$$. The entire statement is only true when both H and C are true, as in, I've actually finished my homework and cleaned my room. However, if I lied about cleaning my room, then the proposition "I cleaned my room" would be false, and in turn, the statement "I did my homework and I cleaned my room" would also be false. With the And operator, having one component being false is enough to make the entire statement false. And then if it turns out I lied about cleaning my room and doing homework, then both H and C are false, and $$(H \land C)$$ is unquestionably false.

=Negation= Next is the negation operator, denoted by $$\lnot$$. The negation operator acts on one proposition, such as $$\lnot A$$, which is read as "not A". As shown in the above truth table, negation reverses the truth value of the proposition. Suppose we have the statement "I don't have a cat," and want to express this statement using mathematical notation. One option would be to declare: let D be the proposition "I don't have a cat." Then the statement would just be D. Another option would be to declare: let C be the proposition "I do have a cat". In this case, the statement would be $$\lnot C$$, as in, the negation of C. Note that propositions D and $$\lnot C$$ mean the same thing and are interchangeable (some may prefer to keep their propositions in positive language and use negations).

Specifically, two propositions are equivalent if they have the same truth values (in the future, $$\equiv$$ will be used to specifically denote equivalence).

Let's show that D is equivalent to $$\lnot C$$:

Case 1: Suppose that the speaker actually has a cat. Then the truth value of D, "I don't have a cat," would be false, and the truth value of C, "I have a cat," would be true. However, since $$\lnot C$$ means that we must negate C, and since the negation of true is false, $$\lnot C$$ is also false.

Case 2: Suppose that the speaker does not have a cat. Then the truth value of D would be true, and the truth value of C would be false. It follows that $$\lnot C$$ is true.

In both cases, D and $$\lnot C$$ have the same truth value. Since we have looked at all possible cases, we conclude that D and $$\lnot C$$ are equivalent. (Although it would also be valid- and perhaps easier- to show this point by comparing truth table columns, a method used in the section below)

=Conditional Statements= Suppose that A and B are propositions. Then $$(A \rightarrow B)$$ is read "If A, then B." This is known as a conditional statement. (Note: things will start getting a little more complicated now) Notice that $$(A \rightarrow B)$$ is only false when A is true and B is false, and is true otherwise. Consider the statement "If you come shopping with me, then I'll buy you some takoyaki." Let S and T be propositions, where S is "You came shopping with me" and T is "I bought you takoyaki." Now, let's analyze the proposition $$(S \rightarrow T)$$. As in, suppose one morning I declared that original statement: "If you come shopping with me, then I'll buy you some takoyaki." When S and T are both true, it means that at the end of the day, you came shopping with me and I bought you takoyaki. Then, the original statement $$(S \rightarrow T)$$ is true, or I honored my word. Now suppose that you don't come shopping with me; then, S is false. In that case, whether I choose to buy takoyaki for you anyway or not, the original statement still holds true. In $$(S \rightarrow T)$$, the event where you get takoyaki was dependent on whether you shopping with me: T is dependent on S. When S is false, all bets are off on T. The only situation where I'd be a liar is if you come shopping with me and I don't buy you takoyaki, which corresponds to when S is true and T is false; this is also the only case when $$(S \rightarrow T)$$ is false.

One common area of confusion with conditionals is correctly identifying the conditional relationship when reading a statement. That is to say, while the "If A, then B" format is common and perhaps the most clear way to state a conditional relationship $$(A \rightarrow B)$$, conditionals can be implied from many different phrasings, such as: "A is sufficient for B", "A only B", "B, if A", "B whenever A", and "B is necessary for A."

Conditional statements are important because there are a number of proof methods specifically for proving statements in the $$(A \rightarrow B)$$ format, which will be discussed later down the page.

Lastly, suppose we have a conditional statement, $$(A \rightarrow B)$$.

1. The converse of $$(A \rightarrow B)$$ is $$(B \rightarrow A)$$. One can think of it as, converse is the reverse of the conditional. To compare the truth values of the original conditional and the converse, consider the following truth table: Notice that the columns for $$(A \rightarrow B)$$ and $$(B \rightarrow A)$$ are not the same. A conditional statement is not equivalent to its converse (more about equivalence later as well).

2. The inverse of $$(A \rightarrow B)$$ is $$(\lnot A \rightarrow \lnot B)$$. As in, the inverse inverts the truth value of each side of the conditional.

As shown above, the columns for the original conditional statement, $$(A \rightarrow B)$$, and its inverse, $$(\lnot A \rightarrow \lnot B)$$, do not have the same truth values for each row. Therefore they are not the equivalent.

3. The contrapositive of $$(A \rightarrow B)$$ is $$(\lnot B \rightarrow \lnot A)$$. (So, while the converse flips A and B, and the inverse negates both sides, the contrapositive does both.) Unlike the converse and inverse, the contrapositive actually is equivalent to the original conditional statement; the truth table columns match for each row. Also, notice that the converse and inverse of a conditional statement have matching truth values for matching truth value combinations of A and B. This is because, suppose we take the contrapositive of the converse of $$(A \rightarrow B)$$, the contrapositive of $$(B \rightarrow A)$$ is (flip the positions of the two propositions that the conditional is acting on and then negate both sides...) $$(\lnot A \rightarrow \lnot B)$$, which is the same as the inverse of $$(A \rightarrow B)$$. This property that a conditional statement is equivalent to its contrapositive will be important moving forward, and will be a key idea behind a certain proof technique.

Lastly, note that the contrapositive of the contrapositive is just the original conditional statement. As in, the contrapositive of $$(A \rightarrow B)$$ is $$(\lnot B \rightarrow \lnot A)$$, and then the contrapositive of $$(\lnot B \rightarrow \lnot A)$$ is $$(A \rightarrow B)$$.

=Biconditional Statement= Suppose we have propositions A and B. Then $$A \leftrightarrow B$$ is a biconditional statement, read as "A if and only if B." The keywords for a biconditional are "if and only if," which is sometimes abbreviated to "iff". Other ways of implying a biconditional statement are: "A is necessary and sufficient for B", "A is equivalent to B", and "A implies B, and conversely." As shown by the truth table, the biconditional is only true when A and B have the same truth value, and is false otherwise. Before doing any examples, some definitions:

A tautology is a proposition that is true for every assignment of truth values to its components. That is, a tautology is always true. A simple example is $$A \leftrightarrow A$$. Since A can only be true or false, the proposition can only be evaluated as True $$\leftrightarrow$$ True, or False $$\leftrightarrow$$ False. In either case, the biconditional would be true.

A contradiction is a proposition that is false for every assignment of truth values to its components. As in, a contradiction is always false. Consider the example $$A \land \lnot A$$. Once again, A can only be true or false. If A is True then the proposition is True $$\land$$ False, and if A is False then the proposition is False $$\land$$ True. In either case, since the And operator is acting on a True and a False, the proposition will always be false.

Notice that since $$A \leftrightarrow B$$ is true precisely when A and B are the same, the propositions A and B are equivalent when $$A \leftrightarrow B$$ is a tautology.

Also, the biconditional statement $$A \leftrightarrow B$$ is equivalent to the expression $$(A \rightarrow B) \land (B \rightarrow A)$$, as shown by how these two propositions have matching truth values in the truth table below. What this means is that a biconditional statement $$A \leftrightarrow B$$ is the same as saying that A implies B and B implies A. This may help when determining whether a biconditional statement is true or not. For instance, let R and C be propositions, where R is "It's raining" and C is "It's cloudy." Is $$R \leftrightarrow C$$ always true? Let's first consider $$R \rightarrow C$$. Recall that a conditional statement has only one situation where it evaluates to be false: when we have True $$\rightarrow$$ False. If R is true, then it's raining, and if it's raining, does that mean it must be cloudy? For simplicity let's ignore sunshowers. Then, $$R \rightarrow C$$ must be a tautology because there are no situations where it's raining and there are no clouds (because the rain needs to come from somewhere). Next, let us consider $$C \rightarrow R$$. Is there a situation where $$C \rightarrow R$$ is false? As in, where it's cloudy and it's not raining? Yes. That's actually what I'm seeing outside my window right now. Continuing, since it is possible for $$C \rightarrow R$$ to be false, there exists a case where $$(R \rightarrow C) \land (C \rightarrow R)$$ is also false. And since $$R \leftrightarrow C$$ is equivalent to $$(R \rightarrow C) \land (C \rightarrow R)$$, it follows that it is possible for $$R \leftrightarrow C$$ to be false. Therefore, $$R \leftrightarrow C$$ is not a tautology, and R is not equivalent to C.

=Theorems Section 1= These theorems are some of the truths that logically follow from the material we've covered so far. As in, it's possible to derive them using the ideas previously discussed. In the examples section below, it will be shown precisely how to derive some of them.

These first couple of theorems will be for showing that certain propositions are equivalent. It's possible to verify all these by drawing a truth table and comparing columns. Although, I do recommend taking some time to look at each of these and logically reason to yourself why they must be true.

A quick note on how to decide what operators to apply first in a more complicated proposition: If a symbol appears more than once, then apply them from left to right. For example, A $$\land$$ B $$\lor$$ C $$\land$$ D would be evaluated in this order: first A $$\land$$ B, then C $$\land$$ D, and lastly use the truth values of the previous two propositions to evaluate the Or operator. Of course, parenthesis will also help show you grouping/priority (the prior proposition could've been written (A $$\land$$ B) $$\lor$$ (C $$\land$$ D)), but sometimes propositions won't be written with any parenthesis and the order will be completely implied based on the application priority shown above. I usually like to use parenthesis as much as possible for the sake of clarity though.

Theorem 1.1 : For propositions A, B, and C, the following are equivalent:

 * 1) $$A \equiv \lnot ( \lnot A)$$
 * 2) $$(A \lor B) \equiv (B \or A)$$
 * 3) $$(A \land B) \equiv (B \land A)$$
 * 4) $$A \lor (B \lor C) \equiv (A \lor B) \lor C$$
 * 5) $$A \land (B \land C) \equiv (A \land B) \land C$$
 * 6) $$A \land (B \lor C) \equiv (A \land B) \lor (A \land C)$$
 * 7) $$A \lor (B \land C) \equiv (A \lor B) \land (A \lor C)$$

If you look at Theorem 1.1.2 through 1.1.5 all together, it basically just says that if you have a proposition that exclusively has Or operators (or exclusively has And operators), then it doesn't really matter what order you evaluate all the individual Or operators (or And operators). For instance, suppose you have: True $$\land$$ True $$\land$$ False $$\land$$ True. Then, you can evaluate the proposition:

(True $$\land$$ True) $$\land$$ False $$\land$$ True $$\equiv$$ True $$\land$$ False $$\land$$ True $$\equiv$$ True $$\land$$ True $$\land$$ False $$\equiv$$ (True $$\land$$ True) $$\land$$ False $$\equiv$$ True $$\land$$ False $$\equiv$$ False

Or you can just go from left to right: True $$\land$$ True $$\land$$ False $$\land$$ True $$\equiv$$ True $$\land$$ False $$\land$$ True $$\equiv$$ False $$\land$$ True $$\equiv$$ False

Note: this is only when you have a bunch of And operators (or Or operators) together! On the other hand, for example, if you have some And operators spread out over a proposition but not all together, like $$(A \land B) \rightarrow (C \land D)$$, then evaluate them from left to right as previously mentioned for default application order, unless implied otherwise by parenthesis.

As for Theorems 1.1.6 and 1.1.7, these are the Distributive Laws. The idea is somewhat similar to when you want to distribute in an expression like x * (y + z), which would be equal to xy + xz.

Theorem 1.2 : DeMorgan's Laws. For propositions A and B, the following are equivalent:

 * 1) $$\lnot (A \land B) \equiv \lnot A \lor \lnot B$$
 * 2) $$\lnot (A \lor B) \equiv \lnot A \land \lnot B$$

DeMorgan's Laws are helpful for when you have a negation acting on some expression enclosed by parentheses, and want to get rid of those parenthesis. A quick dialogue about these:

 * MathPerson: So, what's the negation of A and B? For example, if "A and B" was "I showered and I brushed my teeth"?
 * ThisOtherGirl: Hmm. Is it "I didn't shower and I didn't brush my teeth"?
 * MathPerson: Whoa! Wait just a moment there...
 * ThisOtherGirl: Ohh no I got it, since with 'AND' statements, only one of them need to be false to make the entire statement false.. It would be "I didn't shower OR I didn't brush my teeth".
 * MathPerson: That's right! When we negate an AND statement, we're saying that the statement is false. In this case, that means that one of the propositional components (being "I showered" and "I brushed my teeth") must be false. The way that there's an AND operator on one side of the equivalence side and an OR operator on the other is important! Now, what if we wanted to negate the statement "I showered or I brushed my teeth"?
 * ThisOtherGirl: An OR statement is false when both components are false... so "I didn't shower and I didn't brush my teeth".
 * MathPerson: Excellent! I'm so proud of you!

Theorem 1.3 : For propositions P, Q, and R, the following are equivalent:

Ok, I know that the statements above look way more complicated, but the only really important parts of Theorem 1.3 are 1.3.1 and 1.3.3. Although, we already went over 1.3.3 in the biconditional section above. Theorem 1.3.1 will just be really useful when you're manipulating an expression using equivalence. The rest of the equivalences can be derived using 1.3.1 and Theorems 1.1 and 1.2 (which is exactly what you can do in the Exercises section below!). As a quick example though, let's show that Theorem 1.3.5, $$\lnot (P \rightarrow Q) \equiv P \and \lnot Q$$, is true. To do this, you can start using either side of the equivalence sign, and use other equivalences in order to get an expression that matches the other side of the equivalence. As in:
 * 1) $$P \rightarrow Q \equiv \lnot P \or Q$$
 * 2) $$\lnot (P \land Q) \equiv Q \rightarrow \lnot P$$
 * 3) $$P \leftrightarrow Q \equiv (P \rightarrow Q) \land (Q \rightarrow P)$$
 * 4) $$P \rightarrow (Q \rightarrow R) \equiv (P \land Q) \rightarrow R$$
 * 5) $$\lnot (P \rightarrow Q) \equiv P \and \lnot Q$$
 * 6) $$P \rightarrow (Q \land R) \equiv (P \rightarrow Q) \land (P \rightarrow R)$$
 * 7) $$\lnot (P \land Q) \equiv P \rightarrow \lnot Q$$
 * 8) $$(P \lor Q) \rightarrow R \equiv (P \rightarrow R) \and (Q \rightarrow R)$$

$$\lnot (P \rightarrow Q) \equiv \lnot (\lnot P \lor Q) \equiv \lnot (\lnot P) \land \lnot Q \equiv P \land \lnot Q$$

We get from $$\lnot (P \rightarrow Q) $$ to $$ \lnot (\lnot P \lor Q) $$ by using Theorem 1.3.1, which allows us to replace the $$P \rightarrow Q$$ in the parenthesis with $$\lnot P \lor Q$$. Next, to get from $$ \lnot (\lnot P \lor Q) $$ to $$\lnot (\lnot P) \land \lnot Q$$, we use Theorem 1.2.2, which helps us apply the negation to the expression within the parenthesis. Since there's already a negation acting on P, the negation you get after using DeMorgan's Law just stacks onto $$(\lnot P)$$, making the double negated $$\lnot (\lnot P)$$. Of course, $$\lnot (\lnot P)$$ can be reduced to just $$P$$ by using Theorem 1.1.1. Which is how we get from $$\lnot (\lnot P) \land \lnot Q$$ to $$P \land \lnot Q$$. Since $$P \land \lnot Q$$ is the other side of the equivalence in Theorem 1.3.5, we've shown that $$\lnot (P \rightarrow Q) \equiv P \and \lnot Q$$.

Note that since equivalences are a two way street, if $$\lnot (P \rightarrow Q) \equiv P \and \lnot Q$$ is true then so is $$ P \and \lnot Q \equiv \lnot (P \rightarrow Q) $$.

=Exercises Section 1= Ready to get started on some exercises? Then go ahead and try some of these problems out! (There are answers and full explanations for each problem below the questions.) Oh, but I do recommend reading the answer to Exercise 1 before taking a stab at the harder proving-equivalence problems. (More information about these types of biconditional proofs later!)

Exercise 1 : Above, we proved Theorem 1.3.5 by showing there's a chain of equivalences that let you get from $$\lnot (P \rightarrow Q) $$ to $$P \land \lnot Q$$. However, another valid approach would be to show that there's a way to get from $$P \land \lnot Q$$ to $$\lnot (P \rightarrow Q) $$ instead. Why not try to do precisely that?

Exercise 2 : Prove that Theorem 1.1.4, $$A \lor (B \lor C) \equiv (A \lor B) \lor C$$, is true using a truth table.

Exercise 3 : Prove Theorem 1.3.4, $$P \rightarrow (Q \rightarrow R) \equiv (P \land Q) \rightarrow R$$, is true using equivalences.

Exercise 4 : Prove that Theorem 1.3.2, $$\lnot (P \land Q) \equiv Q \rightarrow \lnot P$$, is true using a truth table.

Exercise 5 : Convert "Grapes grow on vines and pistachios grow on trees" into a propositional form. (Optional: evaluate the truth value of the proposition)

Exercise 6 : Explain using words why Theorem 1.3.1, $$P \rightarrow Q \equiv \lnot P \or Q$$, must be true. (Hint, look at the truth table for $$P \rightarrow Q$$)

Exercise 7 : Prove that Theorem 1.3.8, $$(P \lor Q) \rightarrow R \equiv (P \rightarrow R) \and (Q \rightarrow R)$$, is true using equivalences.

Exercise 8 : Cindy is a senior in high school. One morning she walked into her first period accounting class fashionably late while holding a cup of brand coffee. Unfortunately the accounting teacher is in a bad mood and says, "I will not write you up if and only if you have a note from the office and you have been late less than three times this year, or if you completed today's homework and your parents called the office ahead of time." Cindy does have a note from the office, this is the tenth time she's late this year, she did complete today's homework, and her parents did not call. Assuming that the accounting teacher keeps their word, what is the outcome of this situation?

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=Answers Section 1= Exercise 1 :

One approach is to just do what we did earlier except in reverse. That is:

$$P \land \lnot Q \equiv \lnot (\lnot P) \land \lnot Q \equiv \lnot (\lnot P \lor Q) \equiv \lnot (P \rightarrow Q)$$

The reasons for manipulating the proposition in each step would still correspond to the respective theorems used earlier, although the process is less intuitive compared to the process starting with $$\lnot (P \rightarrow Q) $$. We can also look at the problem starting from scratch. In general, when I'm not sure how to proceed with a proof problem, the first thing I do is reread all of the theorems, and look for anything that might be useful. At a glance, Theorem 1.1.1 could be useful, since the expression we're working with has a negation in it, but the rest of Theorem 1.1 doesn't seem super relevant because we're starting with only one Or operator and $$P \land \lnot Q \equiv \lnot Q \land P$$ won't really move anything forward. Theorem 1.2.2 might be helpful for putting in parenthesis, but it won't get us to the desired end result. It could possibly serve as an intermediary step though, so let's remember it for now. Lastly, looking at Theorem 1.3, the equivalences that stand out are the ones where one side of the equivalence looks like $$P \land \lnot Q$$, or looks like something we can get to from $$P \land \lnot Q$$ (but not including Theorem 1.3.5, because using the theorem you're trying to prove in the proof for that theorem will make a circular proof!! it's invalid to assume something is true before you've proved it's true!!). Anyway, Theorem 1.3 has a couple theorems where one side of the equivalence is of the form $$\lnot (P \land Q)$$, but we can't use these because recall from DeMorgan's laws, putting in parenthesis means we'll have to switch the And operator in $$P \land \lnot Q$$to an Or operator. As in, using Theorem 1.2.2 ($$\lnot A \land \lnot B \equiv \lnot (A \lor B)$$),

$$P \land \lnot Q \equiv \lnot (\lnot P \lor \lnot (\lnot Q)) \equiv \lnot (\lnot P \lor Q)$$.

Fortunately, the expression inside the parenthesis, $$\lnot P \lor Q$$, appears as the right hand side of Theorem 1.3.1 ($$P \rightarrow Q \equiv \lnot P \or Q$$). Therefore, we can do some substitution and get $$\lnot (\lnot P \lor Q) \equiv \lnot (P \rightarrow Q)$$. Since $$\lnot (P \rightarrow Q) $$ is exactly the proposition we want, we're done!

Exercise 2 : Ok actually, this problem might've been trickier than it looked because it actually has three components (A, B, and C), whereas previous examples had at most two. In the truth table, it's necessary to have a row for each possible assignment of True/False for each of the three components. The shortcut for finding how many rows you need is calculating two to the power of the number of components. Here, it would be 2^3, which is equal to 8, hence the 8 rows. Although honestly, you shouldn't have to draw a truth table for more than three components, so you don't really need to memorize the calculation shortcut. (If you're interested in the reasoning behind why it's two to the power of the number of components, it's based on the idea of combinations from probability.)

Also, you can also organize the columns however you think makes the most sense. In this case, the components are all at the left, the columns we want to compare are at the right, and the columns in the middle are parts that we need to evaluate to get to the columns we want to compare. This is just to make it so that it's less likely for mistakes to happen (when filling in a new column, you only need to apply one operation and you can reference the previous columns). You can also omit some columns if you're lazy, as long as you have a column for each of the smallest components (in this case A, B, and C), as well as for the two expressions you're trying to prove are equivalent.

Exercise 3 :

So we want to prove $$P \rightarrow (Q \rightarrow R) \equiv (P \land Q) \rightarrow R$$. I'll start with the left hand side of the equivalence, but if you started from the right or used a different approach, no problem! In proofs there's often multiple ways to go about things. Anyway, my first thoughts looking at the left hand side there is to try doing out both of those conditionals (using Theorem 1.3.1) and see where that gets us. Reasoning being that our options are super limited with $$P \rightarrow (Q \rightarrow R)$$, and getting rid of those conditionals might open some new possibilities (and also because the expression on the right of the equivalence has an And operator, hmm).

$$P \rightarrow (Q \rightarrow R) \equiv P \rightarrow (\lnot Q \lor R) \equiv \lnot P \lor (\lnot Q \lor R)$$

From here, the path forward is a lot clearer. Let's use Theorem 1.1.4 to move those parenthesis: $$\lnot P \lor (\lnot Q \lor R) \equiv (\lnot P \lor \lnot Q) \lor R$$

This is relevant because now we can use DeMorgan's Laws, specifically Theorem 1.2.1, in order to pull out the negations get the And operator we want between P and Q: $$(\lnot P \lor \lnot Q) \lor R \equiv \lnot (P \land Q) \lor R$$

We're really close now. For the last step, we'll use Theorem 1.3.1 one more time, and then we're done: $$\lnot (P \land Q) \lor R \equiv (P \land Q) \rightarrow R$$

As a quick note about that last step, note that even though in the theorems we use singular letters to represent propositions, those letters could actually be standing in for longer propositions. Recall that $$A, (A \lor B), (A \land B \land C \rightarrow D)$$, could all be considered propositions; a proposition is anything that is ultimately true or false. So even though Theorem 1.3.1 is $$P \rightarrow Q \equiv \lnot P \or Q$$, we could just as well say that A and B are propositions where A is $$(P \land Q)$$ and B is $$R$$. Then say:

$$\lnot (P \land Q) \lor R \equiv \lnot A \lor B \equiv A \rightarrow B \equiv (P \land Q) \rightarrow R$$

(by substituting in and out the propositions for A and B). It's not necessary of course, this is just to show that the P, Q, R, or whatever letters that appear in theorems, can represent longer expressions.

Lastly, this is how you could write the full answer:

$$P \rightarrow (Q \rightarrow R) \equiv P \rightarrow (\lnot Q \lor R) \equiv \lnot P \lor (\lnot Q \lor R) \equiv (\lnot P \lor \lnot Q) \lor R \equiv \lnot (P \land Q) \lor R \equiv (P \land Q) \rightarrow R$$

Therefore, it follows that $$P \rightarrow (Q \rightarrow R) \equiv (P \land Q) \rightarrow R$$.

Exercise 4 :

No catch this time, just a 4 row truth table:

Exercise 5 :

Let G and P be propositions, where G is "Grapes grow on vines" and P is "Pistachios grow on trees". Then the statement can be expressed as $$G \land P$$. Both G and P are true, so by the definition of the And operator, $$G \land P$$ is also true.

Exercise 6 :

I'll just put a copy of the truth table for $$P \rightarrow Q$$ here:

So, let's say we wanted to create an expression that also accepts two components, P and Q, and is true when $$P \rightarrow Q$$ is also true. Recall that when P is false, $$P \rightarrow Q$$ will always be true, regardless of Q (as shown by the last two rows of the truth table). However, if we had our expression just be $$\lnot P$$, then we wouldn't be accounting for the situation when $$P \rightarrow Q$$ is true because both P and Q are true. You might've also noticed that $$P \rightarrow Q$$ is always true whenever Q is true. Hence, $$\lnot P \lor Q$$. The only situation when $$P \rightarrow Q$$ is false is when P is true and Q is false, which would also correspond to $$(\lnot True) \lor False$$, or just $$False \lor False$$, which is also false.

Exercise 7 :

Ok now we're proving $$(P \lor Q) \rightarrow R \equiv (P \rightarrow R) \and (Q \rightarrow R)$$. This time I think I'd opt to start with the right hand side (but again, no problem if you went about this in a different way!). Let's try getting rid of those conditionals again.

$$(P \rightarrow R) \and (Q \rightarrow R) \equiv (\lnot P \lor R) \land (Q \rightarrow R) \equiv (\lnot P \lor R) \land (\lnot Q \lor R)$$

Hmm. Now we've got to an expression that reminds me of the distributive laws in Theorem 1.1. Specifically Theorem 1.1.7, $$A \lor (B \land C) \equiv (A \lor B) \land (A \lor C)$$, although the formatting looks a little different. Fortunately we can just move around the order a little bit using Theorem 1.1.2.

$$(\lnot P \lor R) \land (\lnot Q \lor R) \equiv (R \lor \lnot P) \land (R \lor \lnot Q) \equiv R \lor (\lnot P \land \lnot Q) \equiv (\lnot P \land \lnot Q) \lor R$$

Now we've got to a point that should look a little familiar. Similar to Exercise 3, let's use DeMorgan's Laws (although Theorem 1.2.2 this time) to pull out the negation, and then finish with Theorem 1.3.1.

$$(\lnot P \land \lnot Q) \lor R \equiv \lnot (P \lor Q) \lor R \equiv (P \lor Q) \rightarrow R$$

Therefore, $$(P \lor Q) \rightarrow R \equiv (P \rightarrow R) \and (Q \rightarrow R)$$.

Exercise 8 :

First, let's convert the accounting teacher's words into propositional form (there are different ways of doing this, particularly with handling the negation).

Let W, N, T, H, and P be propositions, where W is "Accounting teacher writes Cindy up", N is "Cindy has a note", T is "Cindy has been late less than three times", H is "Cindy did today's homework", and P is "Cindy's parents called the office."

Then, the teachers statement is: $$\lnot W \leftrightarrow N \land T \lor H \land P$$

You can also include parenthesis to make it $$\lnot W \leftrightarrow (N \land T) \lor (H \land P)$$, which is the same as above, but maybe less confusing to look at.

Anyway, using what we know, we can assign truth values to all the propositional components on the right hand side of the biconditional: N - True, T - False, H - True, P - False. Then, evaluating the right hand side, $$(N \land T) \lor (H \land P)$$, we have:

$$(True \land False) \lor (True \land False) \equiv (False) \lor (False) \equiv False$$

Therefore, the biconditional is $$\lnot W \leftrightarrow False$$.

Since we're assuming that the accounting teacher keeps their word (which implies that the biconditional is a tautology) and the right side of the biconditional is false, the left hand side, $$\lnot W$$, must also be false. And, if $$\lnot W$$ is false, then $$\lnot (\lnot W) $$ must be true. Given that $$\lnot (\lnot W) \equiv W $$, it follows that W is true. Recall that W is "Accounting teacher writes Cindy up".

Sooooooo, in the end, it looks like Cindy ran into a spot of bad luck today.

Anyway, congratulations on making it to the end of the first exercises and answers section!

=Direct Proofs= At this point we're finally ready to get started on writing proofs.

title: Logical Proofs