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Question
Without using the truth table show that P ↔ q ≡ (p ∧ q) ∨ (~ p ∧ ~ q)
Solution
L.H.S = p ↔ q
≡ (p → q) ∧ (q → p) ........(Biconditional Law)
≡ (∼ p ∨ q) ∧ (∼ q ∨ p) ........(Conditional Law)
≡ [∼ p ∧ (∼ q ∨ p)] ∨ [q ∧ (∼ q ∨ p)] ....(Distributive Law)
≡ [(∼ p ∧ ∼ q)] ∨ (∼ p ∧ p)] ∨ [(q ∧ ∼ q) ∨ (q ∧ p)] .........(Distributive Law)
≡ [(∼ p ∧ ∼ q) ∨ F] ∨ [F ∨ (q ∧ p)] ........(Complement Law)
≡ (∼ p ∧ ∼ q) ∨ (q ∧ p) .......(Identity Law)
≡ (∼ p ∧ ∼ q) ∨ (p ∧ q) ........(Commutative Law)
≡ (p ∧ q) ∨ (∼ p ∧ ∼ q) ........(Commutative Law)
≡ R.H.S.
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