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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3) ¬ [p q ( ¬ p ¬ q r)] ¬ (p q) ¬ ( ¬ p ¬ q r) (DeMorgan's) ¬ (p q) ¬ ( ¬ (p q) r) (DeMorgan's)
This is the end of the preview. Sign up to access the rest of the document.

## This document was uploaded on 09/16/2011.

### Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online