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# 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 6. Prove that the Rule for Proof by Cases is a valid argument.
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sol1 - MACM 101 — Discrete Mathematics I Outline...

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