<p>The proposition \((\sim p) \vee (p \wedge \sim q)\) is equivalent to</p>
Step-by-Step Solution
Key Concept: Use De Morgan's laws and logical equivalences to simplify the compound proposition by distributing and reducing terms systematically.
<p><strong>Step 1:</strong> Start with (~p) ∨ (p ∧ ~q)</p><p><strong>Step 2:</strong> Apply the absorption law. Notice that when we have (~p) ∨ (p ∧ ~q), we can factor: (~p) ∨ (p ∧ ~q) = [(~p) ∨ p] ∧ [(~p) ∨ ~q]</p><p><strong>Step 3:</strong> Evaluate [(~p) ∨ p]. This is a tautology (always true), since either p is true or ~p is true.</p><p><strong>Step 4:</strong> Since [(~p) ∨ p] = True, the entire expression reduces to: True ∧ [(~p) ∨ ~q] = (~p) ∨ ~q</p><p><strong>Step 5:</strong> By De Morgan's law, (~p) ∨ ~q is equivalent to ~(p ∧ q)</p><p>∴ Answer: <strong>~(p ∧ q) or (~p) ∨ (~q)</strong></p>
Correct Answer: C