Sets, Relations & Functions
Mathematical Reasoning / Tautology
Grade 11
<p>The statement \((p \to q) \to [(\sim p \to q) \to q]\) is</p>
Step-by-Step Solution
Key Concept: This is a tautology because whenever the antecedent (p → q) is true, the consequent [~p → q) → q] must also be true; when (p → q) is false, the entire implication becomes trivially true. Using logical equivalences: (p → q) ≡ (~p ∨ q), so the formula reduces to a logical truth.
<p><strong>Step 1:</strong> Let A = (p → q) and B = [(~p → q) → q]. We need to evaluate A → B.</p><p><strong>Step 2:</strong> Check all cases where (p → q) is true: When p = T, q = T: (~p → q) = (F → T) = T, so B = (T → T) = T. When p = F, q = T: (~p → q) = (T → T) = T, so B = (T → T) = T. When p = F, q = F: (~p → q) = (T → F) = F, so B = (F → F) = T.</p><p><strong>Step 3:</strong> When (p → q) is false: p = T, q = F. Then A = F, making A → B automatically true regardless of B.</p><p><strong>Step 4:</strong> In all cases, (p → q) → [(~p → q) → q] evaluates to TRUE.</p><p>∴ Answer: C (Tautology)</p>
Correct Answer: C