Hmk1Sol - COT 3100H Spring 2008 Homework #1 Solutions 1) a)...

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COT 3100H Spring 2008 Homework #1 Solutions 1) a) 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 b) 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 c) p q r ¬p r q (¬p r) ¬(q (¬p r)) p ¬(q (¬p r)) 0 0 0 1 0 1 0 0 0 1 1 0 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 0 0 0 0 1 1 1 0 1 1 0 1 1 1 1 0 0 0 1 1 1 1 1 1 1 0 0 2) 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
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3) ¬ [p q ( ¬ p ¬ q r)] ¬ (p q) ¬ ( ¬ p ¬ q r) (DeMorgan's) ¬ (p q) ¬ ( ¬ (p q) r) (DeMorgan's)
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Hmk1Sol - COT 3100H Spring 2008 Homework #1 Solutions 1) a)...

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