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Unformatted text preview: SOLUTIONS TO HOMEWORK 3 MATH 150, FALL 09 Problem 1. Section 1.5/ Exercise 12 { , > , } is not complete. To show that, we claim that for every with one sentence symbol A , and connective symbols among , > , , either is a tautology, a contradiction, or A  = . The proof, as usual, is by induction on : (1) If is A , or > , or , then the conclusion is clear. (2) If = > , then is equivalent to . Using the induction hypoth esis for , we get the conclusion for . (3) If = , then is equivalent to , and so is a contradiction. (4) If = , then we have a the following cases: if both and are tautologies, then so is . if at least one of and is a contradiction, then so is . if A  = and either A  = or is a tautology, then A  = . if A  = and either A  = or is a tautology, then A  = ....
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 Spring '09
 SMITH
 Math

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