Sets, Relations & Functions
Mathematical Reasoning / Logic
Grade 11
Question:
<p>Which of the following is correct?</p><p>(i) \((A \wedge B) \wedge (\sim A \vee B) \equiv A \wedge (B \wedge (\sim A \vee B))\)</p><p>(ii) \((A \vee B) \wedge (\sim A \wedge B) \equiv (A \wedge \sim A) \wedge B \equiv F \wedge B \equiv F\)</p><p>(iii) \((A \vee B) \wedge (\sim A \vee B) \equiv B\)</p><p>(iv) \((A \vee B) \wedge (\sim A \vee B) \equiv B \vee (A \wedge \sim A) \equiv B \vee F \equiv F\)</p>
<p>Only (i)</p>
<p>Only (ii)</p>
<p>Only (iii)</p>
<p>Only (iv)</p>
Step-by-Step Solution
Key Concept: Use the associativity, commutativity, and distributive laws of Boolean algebra to simplify logical expressions. Recognize that (X ∨ Y) ∧ (¬X ∨ Y) ≡ Y by factoring out Y, and that A ∧ ¬A ≡ F (contradiction) and A ∨ F ≡ A.
<p><strong>Step 1: Verify statement (i) - Associativity</strong></p><p>(A ∧ B) ∧ (¬A ∨ B) ≡ A ∧ (B ∧ (¬A ∨ B)) is clearly TRUE by associativity of ∧. ✓</p><p><strong>Step 2: Verify statement (ii) - Contradiction Law</strong></p><p>(A ∨ B) ∧ (¬A ∧ B): Here ¬A ∧ B cannot coexist with A from the first term's A part. Expanding: (A ∧ ¬A ∧ B) ∨ (B ∧ ¬A ∧ B) = F ∨ (¬A ∧ B) ≠ F in general. Statement is FALSE. ✗</p><p><strong>Step 3: Verify statement (iii) - Consensus/Absorption</strong></p><p>(A ∨ B) ∧ (¬A ∨ B): Apply distributive law: (A ∧ ¬A) ∨ (A ∧ B) ∨ (B ∧ ¬A) ∨ (B ∧ B) = F ∨ (A ∧ B) ∨ (B ∧ ¬A) ∨ B = B ∨ [(A ∧ B) ∨ (¬A ∧ B)] = B ∨ [B ∧ (A ∨ ¬A)] = B ∨ (B ∧ T) = B ∨ B = B. Statement is TRUE. ✓</p><p><strong>Step 4: Verify statement (iv) - Incorrect grouping</strong></p><p>(A ∨ B) ∧ (¬A ∨ B) ≡ B ∨ (A ∧ ¬A) is FALSE grouping. From Step 3, we know this equals B, not F. Statement is FALSE. ✗</p><p><strong>Correct statements: (i) and (iii)</strong></p><p>∴ Answer: A</p>
Correct Answer: A