hmw1 - ( b ) (01111 10101) 01000 = 00101 01000 = 01101 ( c...

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Solutions to Homework 1. Math 55, Fall 2006. Prob 1.1.24. (a) Converse: If I stay home, then it will snow tonight. Contrapositive: If I do not stay at home, then it will not snow tonight. (b) Converse: Whenever I go to the beach, it is a sunny summer day. Con- trapositive: Whenever I do not go to the beach, it is not a sunny summer day. (c) Converse: If I speel until noon, then I stayed up late. Contrapositive: If I do not sleep until noon, then I did not stay up late. Prob 1.1.38. For parts (a) and (b) we have the following table: p p p ¬ p p ⊕ ¬ p T F F T F F T T For parts (c) and (D) the table is below (columns five and six): p q ¬ p ¬ q p ⊕ ¬ q ¬ p ⊕ ¬ q F F T T T F F T T F F T T F F T F T T T F F T F For parts (e) and (f) we have the following table (columns five and six). This time the column showing the negation of q is omitted. Note that the statement (e) is a tautology and the statement (f) is a contradiction. p q p q p ⊕ ¬ q ( p q ) ( p ⊕ ¬ q ) ( p q ) ( p ⊕ ¬ q ) F F F T T F F T T F T F T F T F T F T T F T T F Prob 1.1.38. ( a ) 11000 (01011 11011) = 11000 11011 = 11000
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Unformatted text preview: ( b ) (01111 10101) 01000 = 00101 01000 = 01101 ( c ) (01010 11011) 01000 = 10001 01000 = 11001 ( d ) (11011 01010) (10001 11011) = 11011 11011 = 11011 . 1 Prob 1.2.30. We only need to examine the case when the conclusion q r is false. This is possible if and only if q = F and r = F . But then p q = F in case p = F , and p r = F in the other case p = T . So, regardless of the value of p , the assumption ( p q ) ( p r ) is false if q r is false. Hence this is a tautology. Remark: there are many other ways of solving this problem. Prob 1.3.10. (a) x ( C ( x ) D ( x ) F ( x )). (b) x ( C ( x ) D ( x ) F ( x )). (c) x ( C ( x ) F ( x ) D ( x )). (d) x ( C ( x ) D ( x ) F ( x )). (e) xC ( x ) xD ( x ) xF ( x ). 2...
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hmw1 - ( b ) (01111 10101) 01000 = 00101 01000 = 01101 ( c...

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