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COT3100Hmk01Sol

# COT3100Hmk01Sol - COT 3100 Fall 2002 Homework#1 Solutions...

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COT 3100 Fall 2002 Homework #1 Solutions Section 2.1 8e) p q p q p (p q) [p (p q)] q 0 0 1 0 1 0 1 1 0 1 1 0 0 0 1 1 1 1 1 1 8h) p q r p q q r [(p q) (q r)] p r [(p q) (q r)] ( p r) 0 0 0 1 1 1 1 1 0 0 1 1 1 1 1 1 0 1 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 1 0 0 0 1 1 1 1 1 1 1 1 1 12a) In order for an implication to be false, the first part must be true while the second false. This means that p q r is true while s t is false. The first clause could only be true if all three of p, q, and r are true, while the second clause can only be false if both s and t are false. So, there is one truth assignment that makes the entire statement false. It's p=q=r=true, s=t=false Section 2.2 4) [[[(p q) r] [(p r) ¬ r]] ¬ q] s ¬ [[[(p q) r] [(p r) ¬ r]] ¬ q] s (Definition of Implication) ¬ [[[(p q) r] [p (r ¬ r)]] ¬ q] s (Associative Law) ¬ [[[(p q) r] [p F]] ¬ q] s (Inverse Law) ¬ [[[(p q) r] F] ¬ q] s (Domination Law) ¬ [[p (q r)] ¬ q] s (Identity Law) [ ¬ [p (q r)] ¬¬ q]] s (DeMorgan's Law) [ ¬ [p (q r)] q]] s (Double Negation) [[ ¬ p ¬ (q r)] q] s (DeMorgan's Law) [( ¬ p ¬ q ¬ r) q] s (DeMorgan's Law) ( ¬ p q ) ( ¬ q q) ( ¬ r q) s (Distributive Law) ( ¬ p q ) F ( ¬ r q) s

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COT3100Hmk01Sol - COT 3100 Fall 2002 Homework#1 Solutions...

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