Sets, Relations & Functions
Mathematical Logic - Tautology and Fallacy
Grade 11
Question:
<p>Consider:<br>
<strong>Statement I:</strong><br>
\((p \wedge \sim q) \wedge (\sim p \wedge q)\) is a fallacy.<br>
<strong>Statement II:</strong><br>
\((p \to q) \leftrightarrow (\sim q \to \sim p)\) is a tautology.</p>
<p>(1) Statement I is true; statement II is true; statement II is a correct explanation for statement I.<br>
(2) Statement I is true; statement II is true; statement II is not a correct explanation for statement I.<br>
(3) Statement I is true; statement II is false.<br>
(4) Statement I is false; statement II is true.</p>
<p>Statement I is true; statement II is true; statement II is a correct explanation for statement I.</p>
<p>Statement I is true; statement II is true; statement II is not a correct explanation for statement I.</p>
<p>Statement I is true; statement II is false.</p>
<p>Statement I is false; statement II is true.</p>
Step-by-Step Solution
Key Concept: A fallacy is a formula that is always false (contradiction), while a tautology is always true. Statement I is a conjunction of mutually exclusive conditions (p true AND q false) AND (p false AND q true), which can never both hold simultaneously. Statement II relies on the logical equivalence that p→q is logically equivalent to its contrapositive ¬q→¬p, making it a tautology.
<p><strong>Step 1: Analyze Statement I</strong></p><p>Expression: (p ∧ ¬q) ∧ (¬p ∧ q)</p><p>This requires BOTH (p AND not-q) AND (not-p AND q) to be true simultaneously.</p><p>This means: p is true AND p is false at the same time—a logical contradiction.</p><p>Therefore, this expression is <strong>always false</strong> → it is a <strong>fallacy (true)</strong></p><p><strong>Step 2: Analyze Statement II</strong></p><p>Expression: (p → q) ↔ (¬q → ¬p)</p><p>The contrapositive law states: p → q ≡ ¬q → ¬p (they are logically equivalent)</p><p>Therefore, (p → q) ↔ (¬q → ¬p) is always true → it is a <strong>tautology (true)</strong></p><p><strong>Step 3: Check if Statement II explains Statement I</strong></p><p>Statement I being a fallacy stems from the structure containing p ∧ ¬p (contradiction).</p><p>Statement II (about contrapositive equivalence) does NOT explain why (p ∧ ¬q) ∧ (¬p ∧ q) is a fallacy.</p><p>They test different logical principles → <strong>NOT a correct explanation</strong></p><p>∴ Answer: <strong>(2)</strong> Statement I is true; statement II is true; statement II is not a correct explanation for statement I.</p>
Correct Answer: B