solution1-11

# solution1-11 - MACM 101 Discrete Mathematics I Outline...

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MACM 101 — Discrete Mathematics I Outline Solutions to Exercises on Propositional Logic 1. Construct a truth table for the following compound statement: ( p q ) ( q ⊕ ¬ r ) . p q r ( p q ) ( q ⊕ ¬ r ) 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 1 2. Determine whether the following compound statement is a tautology ( ¬ p ( p q )) → ¬ q. No, it is not. Method 1. Construct a truth table. Method 2. Use logical equivalences: ( ¬ p ( p q )) → ¬ q ⇐⇒ ( ¬ p ( ¬ p q )) → ¬ q expression for implications ⇐⇒ ¬ p → ¬ q absorption law ⇐⇒ q p contrapositive. Now, if q = 1 and p = 0 , the implication is false. 3. Show that ( p r ) ( q r ) and ( p q ) r are logically equivalent. Method 1. Construct the truth tables for both statements and compare. Method 2. Use logical equivalences. ( p r ) ( q r ) ⇐⇒ ( ¬ p r ) ( ¬ q r ) expression for implications ⇐⇒ ( ¬ p ∨ ¬ q ) r idempotent law + associative law ⇐⇒ ¬ ( p q ) r De Morgan’s law ⇐⇒ ( p q ) r expression for implication 4. Show that ( p q ) r and ( p r ) ( q r ) are not logically equivalent. Do not use truth tables. Method 1. It is sufficient to find one assignment of values to p, q, r such that the two statements get different truth values. For instance if p = 0 , q = 1 , r = 0 then p q ) r = 1 while ( p r ) ( q r ) = 0 . Method 2. (Not recommended.) Use logical equivalences to simplify the biconditional (( p q ) r ) (( p r ) ( q r )) . It can be shown to be equivalent to ( p q ) ( ¬ p ∧ ¬ q ) r , that is, as is easily seen, not a tautology. 1

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5. Simplify the compound statement ¬ ( p ( q r ) (( p q ) r )) . Use logical equivalences: ¬ ( p ( q r ) (( p q ) r )) ⇐⇒ ¬ p ∨ ¬ ( q r ) (( p q ) r )) De Morgan’s law ⇐⇒ ¬ p ∨ ¬ ( q r ) ( ¬ ( p q ) r )) expression for implication ⇐⇒ ¬ p ( ¬ q ∧ ¬ r ) ( p
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