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Unformatted text preview: MACM 101 — Discrete Mathematics I Outline Solutions to Exercises on Propositional Logic 1. Construct a truth table for the following compound proposition: ( ¬ p ↔ ¬ q ) ↔ ( q ↔ r ) . p q r ( ¬ p ↔ ¬ q ) ↔ ( q ↔ r ) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2. Show that the following compound statement is a contradiction (( p → q ) ∨ ( p → r )) ⊕ ( p → ( q ∨ r )) . Method 1. Construct a truth table. Method 2. If this statement is not a contradiction, it is true for some truth values of p , q , and r . We try to find these values. Since the statement is an exclusive OR, it is true only if one part is true and the other is false. Suppose (( p → q ) ∨ ( p → r )) is false. It is a disjunction, therefore both part of it must be false. If p → q is false, p must be true, q must be true. Similarly, if p → r is false, p must be true, r must be true. It is easy to check that with this truth values the second part, p → ( q ∨ r ) is also false, therefore the whole statement is false. The case when p → ( q ∨ r ) is false is similar. 3. Show that ( p ↔ q ) and ( ¬ p ↔ ¬ q ) are logically equivalent. Method 1. Construct the truth tables for both statements and compare. Method 2. Use logical equivalences. ( p ↔ q ) ⇐⇒ ( p → q ) ∧ ( q → p ) expression for biconditional ⇐⇒ ( ¬ p ∨ q ) ∧ ( ¬ q ∨ p ) expression for implication, twice ⇐⇒ ( q ∨¬ p ) ∧ ( p ∨¬ q ) Commutative Law ⇐⇒ ( ¬ ( ¬ q ) ∨¬ p ) ∧ (( ¬ ( ¬ p ) ∨¬ q ) double negation ⇐⇒ (( ¬ q ) → ¬ p ) ∧ (( ¬ p ) → ¬ q ) expression for implication ⇐⇒ ( ¬ q ↔ ¬ p ) expression for biconditional 4. Show that ( p ∧ q ) → r and ( p → r ) ∧ ( q → r ) are not logically equivalent. Do not use truth tables....
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This note was uploaded on 12/12/2010 for the course MACM 101 taught by Professor Pearce during the Summer '08 term at Simon Fraser.
 Summer '08
 PEARCE

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