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# notes_7_2x2 - Branden Fitelson Philosophy 12A Notes 1...

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Branden Fitelson Philosophy 12A Notes 1 Overview of Today’s Lecture Today’s Music: Led Zeppelin The mid-term is on Thursday, 6/10 (in class) . I’ve posted (and discussed) a sample mid-term. It has the same structure and complexity as the actual mid-term (good study guide). I have posted HW #3, which is due on Thursday @ 4pm in the drop box. It’s all chapter 3 problems — truth-table methods for validity-testing. I posted revised versions of lecture #6 and my “short method” handout. MacLogic — a useful computer program for natural deduction. You might want to download MacLogic at this point . . . We’ll be using it very soon (and for the rest of the term) . . . See http://fitelson.org/maclogic.htm Today: Chapter 3, Finalé and Chapter 4 Intro. UCB Philosophy Chapter 3 (Finalé) & Chapter 4 (Intro) 06/08/10 Branden Fitelson Philosophy 12A Notes 2 Expressive Completeness: Rewind, and More Extra-Credit Q . How can we define in terms of | ? A . If you naïvely apply the schemes I described last time, then you get a 187 symbol monster : p q A | A , where A is given by the following 93 symbol expression: (((p | (q | q)) | (p | (q | q))) | ((p | (q | q)) | (p | (q | q)))) | (((q | (p | p)) | (q | (p | p))) | ((q | (p | p)) | (q | (p | p)))) There are simpler definitions of using | . E.g. , this 43 symbol answer: p q ((p | (q | q)) | (q | (p | p))) | ((p | (q | q)) | (q | (p | p))) I offered E.C. for the shortest solution. Some students have come up with it (the shortest solution is < 25 symbols, counting parens). More E.C. Find the shortest possible definitions of (1) p q , (2) p q , and (3) p & q in terms of p , q , and the NAND operator | . If you submit EC, please prove the correctness of your solution, using a truth-table method. You may submit these E.C. solutions to your GSI. UCB Philosophy Chapter 3 (Finalé) & Chapter 4 (Intro) 06/08/10 Branden Fitelson Philosophy 12A Notes 3 Presenting Your “Short-Cut” Truth-Table Tests In any application of the “short” method, there are two possibilities: 1. You find an interpretation ( i.e. , a row of the truth-table) on which all the premises p 1 , . . . , p n of an argument are true and the conclusion q is false. All you need to do here is ( i ) write down the relevant row of the truth-table, and ( ii ) say “Here is an interpretation on which p 1 , . . . , p n are all true and q is false. So, p 1 , . . . , p n q is in valid.” 2. You discover that there is no possible way of making p 1 , . . . , p n true and q false. Here, you need to explain all of your reasoning (as I do in lecture, or as Forbes does, or as I do in my handout). It must be clear that you have exhausted all possible cases , before concluding that p 1 , . . . , p n q is valid . This can be rather involved, and should be spelled out in a step-by-step fashion. Each salient case has to be examined. Consult my handout and lecture notes for model answers of both kinds.

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notes_7_2x2 - Branden Fitelson Philosophy 12A Notes 1...

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