Sets, Relations & Functions
Mathematical Reasoning / Logical Equivalence
Grade 11
Question:
<p>The statement \((\sim p) \vee (p \wedge \sim q)\) is equivalent to</p>
<p>\(p \wedge \sim q\)</p>
<p>\(\sim p \vee \sim q\)</p>
<p>\(p \to (\sim q)\)</p>
<p>\(\sim p \wedge q\)</p>
Step-by-Step Solution
Key Concept: Use De Morgan's laws and distributive properties to simplify the compound statement. Recognize that (~p) ∨ (p ∧ ~q) can be factored by treating (~p) as a common logical element.
<p><strong>Step 1:</strong> Apply the absorption/distributive principle. Notice that (~p) ∨ (p ∧ ~q) can be analyzed by cases.</p><p><strong>Step 2:</strong> When p is FALSE: (~p) = T, so the entire expression is T regardless of q.</p><p><strong>Step 3:</strong> When p is TRUE: (~p) = F, so we need (p ∧ ~q) = T ∧ ~q = ~q. The expression becomes F ∨ ~q = ~q.</p><p><strong>Step 4:</strong> Combine both cases: The statement is true when p is false OR when q is false. This is equivalent to (~p) ∨ (~q), which by De Morgan's law equals ~(p ∧ q).</p><p><strong>Verification:</strong> Truth table confirms (~p) ∨ (p ∧ ~q) ≡ (~p) ∨ (~q)</p><p>∴ Answer: C</p>
Correct Answer: C