sol1 - 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 propositions: (a) ( p q ) ( p ⊕¬ q ) , (b) ( ¬ p ↔ ¬ q ) ( q r ) . ( a ) p q ( p q ) ( p ⊕¬ q ) 0 0 1 0 1 0 1 0 0 1 1 1 ( b ) p q r ( ¬ p ↔ ¬ q ) ( q r ) 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 1 0 1 1 1 1 0 0 1 1 1 1 2. Show that the following compound statement is a tautology ¬ ( p q ) → ¬ q. Method 1. Construct a truth table. Method 2. Use logical equivalences: ¬ ( p q ) → ¬ q ⇐⇒ ¬¬ ( p q ) ∨¬ q expression for implications ⇐⇒ ¬ p q ∨¬ q expression for implications + double negation law ⇐⇒ T law of excluded middle + domination law. 3. Show that ( p q ) ( p 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 q ) ( p r ) ⇐⇒ ¬ p q ∨¬ p r expression for implications ⇐⇒ ¬ p ( q r ) idempotent law + associative law ⇐⇒ p ( q r ) expression for implication 4. Show that ( p q ) r and p ( q r ) are not logically equivalent. Method 1. Construct the truth tables for both statements and compare. Method 2. 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 = q = r = 0 then ( p q ) r = 0 while p ( q r ) = 1 . Method 3. Use logical equivalences to simplify the biconditional (( p q ) r ) ( p ( q r )) . It can be shown to be equivalent to p r , that is not a tautology. 1
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5. Simplify the compound statement (((( p q ) r ) (( p q ) ∧¬ r )) ∨¬ q ) s. Use logical equivalences: (((( p q ) r ) (( p q ) ∧¬ r )) ∨¬ q ) s ⇐⇒ ((( p q ) ( r ∨¬ r )) ∨¬ q ) s distributive law ⇐⇒ (( p q ) ∨¬ q ) s law of excluded middle + domination law ⇐⇒ (( p ∨¬ q ) ( q ∨¬ q )) s distributive law ⇐⇒ ( p ∨¬ q ) s . law of excluded middle + domination law
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sol1 - MACM 101 — Discrete Mathematics I Outline...

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