Math+150++F09-+Hwk+3+Solutions

Math+150++F09-+Hwk+3+Solutions - SOLUTIONS TO HOMEWORK 3...

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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 α with one sentence symbol A , and connective symbols among ∧ , > , ⊥ , either α is a tautology, a contradiction, or A | = α . The proof, as usual, is by induction on α : (1) If α is A , or > , or ⊥ , then the conclusion is clear. (2) If α = β ∧> , then α is equivalent to β . Using the induction hypoth- esis for β , we get the conclusion for α . (3) If α = β ∧ ⊥ , then α is equivalent to ⊥ , and so α is a contradiction. (4) If α = β ∧ γ , then we have a the following cases: • if both β and γ are tautologies, then so is α . • if at least one of β and γ is a contradiction, then so is α . • if A | = β and either A | = γ or γ is a tautology, then A | = α . • if A | = γ and either A | = β or β is a tautology, then A | = α ....
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This note was uploaded on 11/16/2009 for the course MATH 120 taught by Professor Smith during the Spring '09 term at UC Irvine.

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Math+150++F09-+Hwk+3+Solutions - SOLUTIONS TO HOMEWORK 3...

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