Sets, Relations & Functions
Mathematical Reasoning / Logic
Grade 11
Question:
<p>The logic statement \([\sim (\sim p \vee q) \wedge (p \wedge r)] \wedge (\sim p \wedge r)\) is equivalent to:</p>
<p>\((p \wedge r) \wedge \sim q\)</p>
<p>\(\sim p \vee r\)</p>
<p>\(p \wedge (\sim q \wedge r)\)</p>
<p>\((\sim p \wedge \sim q) \wedge r\)</p>
Step-by-Step Solution
Key Concept: Apply De Morgan's laws systematically to simplify ¬(¬p ∨ q) first, then use associativity and idempotence of conjunction to combine identical terms. The key is recognizing that (p ∧ r) ∧ (¬p ∧ r) = ∅ because p and ¬p cannot both be true.
<p><strong>Step 1:</strong> Simplify ¬(¬p ∨ q) using De Morgan's Law:</p><p>¬(¬p ∨ q) = ¬¬p ∧ ¬q = p ∧ ¬q</p><p><strong>Step 2:</strong> Substitute back into the original expression:</p><p>[p ∧ ¬q ∧ (p ∧ r)] ∧ (¬p ∧ r)</p><p><strong>Step 3:</strong> Regroup using associativity of conjunction:</p><p>= [(p ∧ r) ∧ (¬p ∧ r)] ∧ (p ∧ ¬q)</p><p><strong>Step 4:</strong> Analyze (p ∧ r) ∧ (¬p ∧ r):</p><p>= (p ∧ ¬p) ∧ r = ⊥ ∧ r = ⊥ (contradiction)</p><p><strong>Step 5:</strong> Any statement conjoined with a contradiction yields a contradiction:</p><p>⊥ ∧ (p ∧ ¬q) = ⊥</p><p>∴ The statement is equivalent to <strong>False (or Contradiction)</strong>. Answer: C</p>
Correct Answer: C