<p>The following statement \((p \to q) \to [(\sim p \to q) \to q]\) is</p>
Step-by-Step Solution
Key Concept: Recognize this as a tautology by analyzing the truth table or using logical equivalences. The statement is true regardless of the truth values of p and q because whenever the antecedent (p → q) is true, the consequent [¬p → q) → q] is also true.
<p><strong>Step 1:</strong> Construct truth table for all combinations of p and q.</p><p><strong>Step 2:</strong> Evaluate (p → q) first.</p><p><strong>Step 3:</strong> Evaluate (¬p → q) → q using the definition: if (¬p → q) is true and q is true, result is true; if (¬p → q) is true and q is false, result is false; if (¬p → q) is false, result is true.</p><p><strong>Step 4:</strong> Evaluate the main implication (p → q) → [(¬p → q) → q] for all cases:</p><p>• When p=T, q=T: (T→T) → [(F→T)→T] = T → [T→T] = T → T = T</p><p>• When p=T, q=F: (T→F) → [(F→F)→F] = F → [T→F] = F → F = T</p><p>• When p=F, q=T: (F→T) → [(T→T)→T] = T → [T→T] = T → T = T</p><p>• When p=F, q=F: (F→F) → [(T→F)→F] = T → [F→F] = T → T = T</p><p><strong>Step 5:</strong> All rows yield T, therefore the statement is a <strong>tautology</strong>.</p><p>∴ Answer: D (Tautology)</p>
Correct Answer: D