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Unformatted text preview: SOLUTIONS TO HOMEWORK 1 MATH 150, FALL 09 Problem 1. Section 1.1/ Exercise 2 If α is a formula, then either it is a sentence symbol or it is build by one of the formula building operations applied to other formulas. In the first case its length is 1, and in the second case its length is strictly greater than 3. So, there are no formulas of length 2 and 3. To show that there is no formula of length 6, suppose for contradiction that the length of α is 6. • if α = ( ¬ β ), then β must have length 3, which we showed is impos sible. • if α = ( β γ ), where ∈ {∨ , ∧ , → , ↔} , then length( β )+ length( γ ) = 3. So one of them must have length 2, which is impossible. So, there are no formulas of length 6. To show that every other length is possible, note that: • the length of ( ¬ A ) is 4, • the length of ( A ∨ B ) is 5 • the length of (( A ∨ B ) ∨ A ) is 9, • for every formula of length n , we can build a formula of length n +3, by taking its negation Using this we can build formulas of length 7, 8 and any length greater than 9. Problem 2. Section 1.1/ Exercise 5 Part (a): We use induction on the complexity of α ....
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 Spring '09
 SMITH
 Math, Logic, Formulas, Inductive Reasoning, Logical connective, β

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