Sets, Relations & Functions
Mathematical Logic
Grade 11
Question:
<p>Let the statements be expressed as \(r = {\sim}p \leftrightarrow q\). Statement-1: \(r_1 = (p \land {\sim}q) \lor ({\sim}p \land q)\) and Statement-2: \(r_2 = {\sim}(p \leftrightarrow q) = (p \land q) \lor ({\sim}q \land {\sim}p)\). Which of the following is correct?</p>
<p>(1) Statement-1 is false and Statement-2 is true</p>
<p>(2) Both Statement-1 and Statement-2 are true</p>
<p>(3) Both Statement-1 and Statement-2 are false</p>
<p>(4) Statement-1 is true and Statement-2 is false</p>
Step-by-Step Solution
Key Concept: Recognize that ~p ↔ q is equivalent to p XOR q (exclusive or), which means exactly one of p or q is true. The XOR operation equals (p ∧ ~q) ∨ (~p ∧ q), while its negation ~(p ↔ q) equals (p ∧ q) ∨ (~p ∧ ~q) (both true or both false).
<p><strong>Step 1:</strong> Analyze r = ~p ↔ q. The biconditional A ↔ B is true when both have the same truth value.</p><p><strong>Step 2:</strong> So ~p ↔ q is true when: (~p is true AND q is true) OR (~p is false AND q is false), which gives (p is false AND q is true) OR (p is true AND q is false).</p><p><strong>Step 3:</strong> This simplifies to r = (~p ∧ q) ∨ (p ∧ ~q)</p><p><strong>Step 4:</strong> Compare with Statement-1: r₁ = (p ∧ ~q) ∨ (~p ∧ q). This is identical to our result for r.</p><p><strong>Step 5:</strong> For Statement-2: ~(p ↔ q) means the negation of the standard biconditional. The biconditional p ↔ q is true when both have same truth value: (p ∧ q) ∨ (~p ∧ ~q). Its negation gives the opposite: (~p ∧ q) ∨ (p ∧ ~q).</p><p><strong>Step 6:</strong> Statement-2 claims r₂ = ~(p ↔ q) = (p ∧ q) ∨ (~q ∧ ~p), which is INCORRECT. The correct negation should be (~p ∧ q) ∨ (p ∧ ~q).</p><p><strong>Conclusion:</strong> Statement-1 is correct (r₁ correctly represents ~p ↔ q), and Statement-2 is incorrect (the given expression is actually p ↔ q, not its negation).</p><p>∴ Answer: D</p>
Correct Answer: D