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Without using truth table show that
(p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ ( ~ p ∧ q)
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Solution
L.H.S
= (p ∨ q) ∧ (~ p ∨ ~ q)
≡ [(p ∨ q) ∧ ~ p] ∨ [(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) ......[Complement law]
= R.H.S
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