Methods of Proof involving  Symbolic Logic February 19, 2001

Outline Valid Steps –Logical Truth –Conjunction Elimination –Conjunction Introduction –Disjunction Introduction Proof by Cases Negation Introduction

Logical Truth If a sentence is a logical truth, it is always true, no matter what. Since it is always truth, you are allowed to assert it without any other justification. We have seen this already with Reflexivity of Identity. Because a = a is a logical truth, you can introduce it at any time.

Examples a = a (because an object is always identical with itself (def. of identity)) P   P (because of the truth table for  )  (P   P) (because of the truth table for  )

Conjunction Elimination/ Simplification If we know P  Q, we also know P (as well as Q). Thus, 1. P  Q 2. P1, Conjunction Elimination and, 1. P  Q 2. Q1, Conjunction Elimination

Conjunction Introduction The converse is true as well. If we know P and we know Q, we know P  Q. So, 1. P 2. Q 3. P  Q1, 2 Conjunction Intro.

Disjunction Introduction/ Addition Because of the truth table for , we know that a disjunction is true if one of its disjuncts is true. This allows us to introduce anything we want using a disjunction. So, given P, we can deduce P  Q P  (Q  R) P   ((M   S)  (T  (Q  R)))

Disjunction Introduction/ Addition In English, we know that if “It is President’s Day” is true, “Either it is President’s Day or frogs will fly out of my butt” is true as well. So, 1. P 2. P  Q 1 Disjunction Intro.

Proof by Cases/ Disjunction Elimination Suppose we know that (P  Q)  (P  R) In order for this disjunction to be true, one or the other sides must be true. Since P is true in both cases (by conjunction elimination), P must be true. So, since P  Q implies P and, P  R also implies P, P must be true no matter what.

Proof by Contradiction/ Reductio ad absurdum or Negation Intro. Writers and speakers who want to disprove a claim often assume the claim and then show how it leads to a contradiction. If the claim or set of claims leads to a contradiction, it cannot possibly be true. We can use this technique in more formal situations as well.

Negation Introduction Consider, 1. Either Max is a Cat or Max is an Owl 2. Sofie is an Aardvark 3. Assume Max = Sofie If Max = Sofie, then Either Sofie is a Cat or Sofie is an Owl (1, 3 Ind. Id.) Sofie can’t be either a Cat or an Owl, since Sofie is an Aardvark (2) Therefore Max  Sofie

Contradictory Premises Note that 1. P   P 2. R Since P   P cannot possibly be true, if we assert that it is true, it implies every possible state of affairs.

Homework Note You don’t have to do problems 28, 37, or 38 for the next homework set (or ever). Because this isn’t a math class, we will not have problems like this on the test.