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