Sets, Relations & Functions
Mathematical Logic
Grade 11
Question:
<p>The proposition \({\sim}(p \lor {\sim}q) \lor {\sim}(p \lor q)\) is equivalent to:</p>
<p>(1) \(p \lor q\)</p>
<p>(2) \({\sim}p \land q\)</p>
<p>(3) \({\sim}p \land ({\sim}p \lor {\sim}q)\), which is equivalent to \({\sim}p\) tautology</p>
<p>(4) \({\sim}p\)</p>
Step-by-Step Solution
Key Concept: Apply De Morgan's Laws systematically to each negated disjunction: ~(A ∨ B) ≡ (~A ∧ ~B), then simplify using the distributive property and absorption laws to identify the final form.
<p><strong>Step 1:</strong> Apply De Morgan's Law to ~(p ∨ ~q):</p><p>~(p ∨ ~q) = ~p ∧ ~~q = ~p ∧ q</p><p><strong>Step 2:</strong> Apply De Morgan's Law to ~(p ∨ q):</p><p>~(p ∨ q) = ~p ∧ ~q</p><p><strong>Step 3:</strong> Substitute back into the original expression:</p><p>(~p ∧ q) ∨ (~p ∧ ~q)</p><p><strong>Step 4:</strong> Factor out ~p using distributive property:</p><p>~p ∧ (q ∨ ~q)</p><p><strong>Step 5:</strong> Simplify using (q ∨ ~q) = T (Tautology):</p><p>~p ∧ T = ~p</p><p>∴ Answer: D (~p or ¬p)</p>
Correct Answer: D