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Question
Using the algebra of statement, prove that
(p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∨ ~ q) ∧ (~ p ∨ q)
Solution
L.H.S. = (p ∨ q) ∧ (~ p ∨ ~ q)
≡ [p ∧ (~ p ∨ ~ q)] ∨ [q ∧ (~ p ∨ ~ q)] ...[Distributive law]
≡ [(p ∧ ~ p) ∨ (p ∧ ~ q)] ∨ [(q ∧ ~ p) ∨ (q ∧ ~ q)] ...[Distributive law]
≡ [c ∨ (p ∧ ~ q)] ∨ [(q ∧ ~ p) ∨ c] ...[Complement law]
≡ (p ∧ ~ q) ∨ (q ∧ ~ p) ...[Identity law]
≡ (p ∧ ~ q) ∨ (~ p ∧ q) ...[Commutative law]
= R.H.S.
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