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Unformatted text preview: SOLUTIONS TO HOMEWORK 2 MATH 150, FALL 09 Problem 1. Section 1.2/ Exercise 9 Let α be a wff. We show that ( ¬ α ) is tautologically equivalent to α * by induction on the complexity of α . Base step: suppose that α is a sentence symbol, say α = A . Then α * = ( ¬ A ), and so ( ¬ α ) = ( ¬ A ) is tautologically equivalent to α * . Inductive step: Since α ’s connective symbols can be only among ∨ , ∧ , ¬ , there are 3 cases. (1) α = ( ¬ β ). Then ( ¬ α ) = ( ¬ ( ¬ β )), and so ( ¬ α ) is tautologically equivalent to β . On the other hand, α * = ( ¬ β * ) and by the inductive hypothesis we have that β * is tautologically equivalent to ( ¬ β ), so α * is equivalent to β . Since both ( ¬ α ) and α * are equivalent to β , they are equivalent to each other. (2) α = ( β ∨ γ ). Then α * = ( β * ∧ γ * ), which, by the inductive hypothesis, is equivalent to (( ¬ β ) ∧ ( ¬ γ )), which is equivalent to ( ¬ α ). (3) α = ( β ∧ γ ). Then α * = ( β * ∨ γ * ), which, by the inductive hypothesis, is equivalent to (( ¬ β ) ∨ ( ¬ γ )), which is equivalent to ( ¬ α ). Problem 2. Section 1.2/ Exercise 10 Part (a): Let Σ be a finite set of wffs. Say Σ = { φ 1 ,φ 2 ,...,φ n } . Define sets Σ k , for k ≤ n + 1 recursively on k as follows: • Σ 1 = Σ • Given Σ k , we define Σ k +1 to be equal to Σ k if Σ k is independent. Otherwise, let φ ∈ Σ k be such that it is implied by the remaining wffs in Σ k . Let Σ k +1 = Σ k \ { φ } . We claim that Σ n +1 is independent and equivalent to Σ. Claim 1. Σ n +1 is independent. Proof. Suppose not. Then, since for each k ≤ n we have Σ n +1 ⊂ Σ k , it follows that each Σ k is not independent. So by our definition, for each k , the size of Σ k +1 is one less than the size of Σ k . But the size of Σ 1 equals n , so the size of Σ n +1 must be zero. I.e. Σ n +1 is the empty set, and so it has to be independent. Claim 2. Σ n +1 is tautologically equivalent to Σ ....
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 Spring '09
 SMITH
 Math, Logic, Mathematical Induction, Equals sign, Mathematical logic, Σk, sentence symbol

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