Sets, Relations & Functions
Mathematical Reasoning / Tautology and Fallacy
Grade 11
Question:
<p>Consider<br><strong>Statement-1:</strong> \((p \wedge \sim q) \wedge (\sim p \wedge q)\) is a fallacy.<br><strong>Statement-2:</strong> \((p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)\) is a tautology.</p>
<p>Statement-1 is True; Statement-2 is true; Statement-2 is not a correct explanation for Statement-1</p>
<p>Statement-1 is True; Statement-2 is False.</p>
<p>Statement-1 is False; Statement-2 is True</p>
<p>Statement-1 is True; Statement-2 is True; Statement-2 is a correct explanation for Statement-1.</p>
Step-by-Step Solution
Key Concept: Statement-1 is a fallacy because (p ∧ ¬q) ∧ (¬p ∧ q) requires p and ¬p to be simultaneously true, which is impossible. Statement-2 is a tautology because a conditional and its contrapositive are logically equivalent.
<p><strong>Step 1: Analyze Statement-1</strong></p><p>(p ∧ ¬q) ∧ (¬p ∧ q) requires both (p ∧ ¬q) AND (¬p ∧ q) to be true simultaneously.</p><p>This means: p is true AND p is false (contradiction). This is impossible for any truth values of p and q.</p><p>Therefore, Statement-1 is always FALSE → it is a <strong>fallacy</strong>. ✓</p><p><strong>Step 2: Analyze Statement-2</strong></p><p>We verify if (p → q) ↔ (¬q → ¬p) is a tautology by checking all cases:</p><p><table border='1' cellpadding='5'><tr><th>p</th><th>q</th><th>p → q</th><th>¬q → ¬p</th><th>(p → q) ↔ (¬q → ¬p)</th></tr><tr><td>T</td><td>T</td><td>T</td><td>T</td><td><strong>T</strong></td></tr><tr><td>T</td><td>F</td><td>F</td><td>F</td><td><strong>T</strong></td></tr><tr><td>F</td><td>T</td><td>T</td><td>T</td><td><strong>T</strong></td></tr><tr><td>F</td><td>F</td><td>T</td><td>T</td><td><strong>T</strong></td></tr></table></p><p>The biconditional is true for all combinations → Statement-2 is a <strong>tautology</strong>. ✓</p><p><strong>Both statements are correct.</strong></p><p>∴ Answer: <strong>A</strong></p>
Correct Answer: A