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Using truth table prove that \( \sim(p \leftrightarrow q) \equiv(p \wedge \sim q) \vee(q \wedge \sim p) \). |
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Answer» p ↔ q ≡ (p → q) ʌ (q → p) ≡ (∼p v q) ʌ (∼q v p) \(\therefore\) ∼ (p ↔ q) ≡ ∼((∼p v q) ʌ (∼q v p)) ≡ ∼ (∼ p v q) v ∼(∼ q v p) (\(\because\) ∼(A ʌ B) ≡ ∼ A v ∼B) ≡ (∼(∼p) ʌ ∼ q) v (∼(∼q) ʌ ∼p) ≡ (p ʌ ∼q) v (q ʌ ∼p) Hence, p ↔ q is logically equivalent to (p ʌ ∼q) v (q ʌ ∼p) |
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