<p>\({\sim}(p \lor q) \lor ({\sim}p \land q)\) is equivalent to:</p>
Step-by-Step Solution
Key Concept: Use De Morgan's Law to expand ∼(p ∨ q) = (∼p ∧ ∼q), then apply Boolean algebra (absorption and complement laws) to simplify the entire expression.
<p><strong>Step 1:</strong> Apply De Morgan's Law to ∼(p ∨ q)</p><p>∼(p ∨ q) = (∼p ∧ ∼q)</p><p><strong>Step 2:</strong> Rewrite the full expression</p><p>(∼p ∧ ∼q) ∨ (∼p ∧ q)</p><p><strong>Step 3:</strong> Factor out ∼p using distributive law</p><p>= ∼p ∧ (∼q ∨ q)</p><p><strong>Step 4:</strong> Apply complement law: (∼q ∨ q) = True (T)</p><p>= ∼p ∧ T</p><p><strong>Step 5:</strong> Apply identity law: ∼p ∧ T = ∼p</p><p>∴ Answer: <strong>∼p</strong> (Option D)</p>
Correct Answer: D