1.

Using truth table prove that \( \sim(p \leftrightarrow q) \equiv(p \wedge \sim q) \vee(q \wedge \sim p) \).

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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