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Question
Without using truth table show that -
(p ˅ q) ˄ (∼p v ∼q) ≡ (p ∧ ∼q) ˄ (∼p ∧ q)
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Solution
(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) ∨ (q ∧ ∼p) .......[Commutative Law]
Notes
The question has been modified.
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