This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MACM 101 — Discrete Mathematics I Exercises on Propositional Logic. Due: Tuesday, Septem ber 29th (at the beginning of the class) SOLUTIONS 1. Construct a truth table for the following compound proposition: ( p ↔ q ) → ( p ∨ q ) Solution p q ( p ↔ q ) → ( p ∨ q ) 1 1 1 1 1 1 1 2. Which of the following statements are tautologies? ( p → q ) ↔ ( ¬ p ∨ q ) ( p → q ) ↔ ( ¬ ( p ∧¬ q )) ( p ∨ q ) → ( p ∧ q ) Solution Method 1: Use truth tables. Method 2: Reason about it: 1) Yes, each of ( p → q ) and ( ¬ p ∨ q ) is false if and only if p is true and q is false. 2) Yes, it follows from 1), De Morgan’s law and double negation law: ( p → q ) ↔ ( ¬ p ∨ q ) ⇔ ( p → q ) ↔ ( ¬¬ ( ¬ p ∨ q )) ⇔ ( p → q ) ↔ ( ¬ ( ¬¬ p ∧¬ q )) ⇔ ( p → q ) ↔ ( ¬ ( p ∧¬ q )) 3) No, if we set p = 0 ,q = 1 then the formula turns into 0. 3. Show that ( p → r ) ∧ ( q → r ) and ( p ∨ q ) → r are logically equivalent. Method 1: Use truth table. 1 Method 2: Use the laws of logic. ( p → r ) ∧ ( q → r ) ⇔ (definition of → ) ( ¬ p ∨ r ) ∧ ( ¬ q ∨ r ) ⇔ (distributivity) ( ¬ p ∧¬ q ) ∨ r ⇔ (De Morgan) ( ¬ ( p ∨ q )) ∨ r ⇔ (definition of → ) ( p ∨ q ) → r 4. Show that ( a ∧ b ∧ c ∧ d ∧ e ) → f and (...
View
Full
Document
 Spring '09
 jcliu
 Math, Logic

Click to edit the document details