Sets, Relations & Functions
Mathematical Reasoning / Logical Equivalence
Grade 11
Question:
<p>Given \(p \rightarrow (\sim q \vee r) \equiv \sim p \wedge (\sim q \vee r)\). The truth table for \(\sim(p \wedge q) \vee r\) is given. Which option correctly represents the equivalent statement?</p>
<p>\(\equiv (\sim p \vee \sim q) \vee r \equiv \sim(p \wedge q) \vee r\)</p>
<p>\(\equiv p \wedge q \wedge r\)</p>
<p>\(\equiv \sim p \vee q \vee r\)</p>
<p>\(\equiv p \wedge (\sim q \vee r)\)</p>
Step-by-Step Solution
Key Concept: The given equivalence p → (∼q ∨ r) ≡ ∼p ∧ (∼q ∨ r) is false by standard logic; this signals the question tests recognition that the implication p → S is equivalent to ∼p ∨ S, not ∼p ∧ S. Students must identify what the correct logical form should be and match it to the truth table of ∼(p ∧ q) ∨ r.
<p><strong>Step 1:</strong> Recognize the error in the given statement. The correct equivalence is: p → (∼q ∨ r) ≡ ∼p ∨ (∼q ∨ r)</p><p><strong>Step 2:</strong> Simplify ∼p ∨ (∼q ∨ r) = ∼p ∨ ∼q ∨ r, which by De Morgan's law equals ∼(p ∧ q ∧ ∼r)</p><p><strong>Step 3:</strong> Construct truth table for ∼(p ∧ q) ∨ r: This is true when p ∧ q is false OR r is true. It is false only when (p ∧ q) is true AND r is false (i.e., p=T, q=T, r=F)</p><p><strong>Step 4:</strong> Compare with the truth table of ∼p ∨ ∼q ∨ r, which has identical rows. Both expressions are logically equivalent to the correct form of the implication.</p><p>∴ Answer: A</p>
Correct Answer: A