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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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- Spring '08