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Question : 111 of 200
Marks:
+1,
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Solution:
p ↔ q ≡ (p → q) ∧ (q → p) ≡ (~p ∨ q) ∧ (~q ∨ p)
p ↔ q ≡ (~p ∨ q) ∧ (~q ∨ p)
Equivalent statements
∴ ~(p ↔ q) ≡ ~[(~p ∨ q) ∧ (~q ∨ p)]
≡ ~(~p ∨ q) ∨ ~(~q ∨ p)
≡ (p ∧ ~q) ∨ (q ∧ ~p)
≡ (~q ∧ p) ∨ (~p ∧ q)
≡ (~p ∧ q) ∨ (~q ∧ p)
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