Math+150++F09-+Hwk+2+Solutions

Math+150++F09-+Hwk+2+Solutions - SOLUTIONS TO HOMEWORK 2...

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Unformatted text preview: SOLUTIONS TO HOMEWORK 2 MATH 150, FALL 09 Problem 1. Section 1.2/ Exercise 9 Let be a wff. We show that ( ) is tautologically equivalent to * by induction on the complexity of . Base step: suppose that is a sentence symbol, say = A . Then * = ( A ), and so ( ) = ( A ) is tautologically equivalent to * . Inductive step: Since s connective symbols can be only among , , , there are 3 cases. (1) = ( ). Then ( ) = ( ( )), and so ( ) is tautologically equivalent to . On the other hand, * = ( * ) and by the inductive hypothesis we have that * is tautologically equivalent to ( ), so * is equivalent to . Since both ( ) and * are equivalent to , they are equivalent to each other. (2) = ( ). Then * = ( * * ), which, by the inductive hypothesis, is equivalent to (( ) ( )), which is equivalent to ( ). (3) = ( ). Then * = ( * * ), which, by the inductive hypothesis, is equivalent to (( ) ( )), which is equivalent to ( ). Problem 2. Section 1.2/ Exercise 10 Part (a): Let be a finite set of wffs. Say = { 1 , 2 ,..., n } . Define sets k , for k n + 1 recursively on k as follows: 1 = Given k , we define k +1 to be equal to k if k is independent. Otherwise, let k be such that it is implied by the remaining wffs in k . Let k +1 = k \ { } . We claim that n +1 is independent and equivalent to . Claim 1. n +1 is independent. Proof. Suppose not. Then, since for each k n we have n +1 k , it follows that each k is not independent. So by our definition, for each k , the size of k +1 is one less than the size of k . But the size of 1 equals n , so the size of n +1 must be zero. I.e. n +1 is the empty set, and so it has to be independent. Claim 2. n +1 is tautologically equivalent to ....
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Math+150++F09-+Hwk+2+Solutions - SOLUTIONS TO HOMEWORK 2...

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