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Question
Without using truth table, show that
p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
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Solution
L.H.S.
≡ p ↔ q
≡ (p → q) ∧ (q → p)
≡ (~p ∨ q) ∧ (~q ∨ p)
≡ [~ 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]
≡ R.H.S.
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