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Question
Without using truth table prove that (p ∧ q) ∨ (∼ p ∧ q) v (p∧ ∼ q) ≡ p ∨ q
Solution
L.H.S. = (p ∧ q) ∨ (∼ p ∧ q) v (p∧ ∼ q)
= [(p∨ ∼ p) ∧ q] ∨ (p∧ ∼ q) ......(Distributive law)
≡ (T ∧ q) ∨ (p∧ ∼ q) .......(Complement law)
≡ q ∨ (p∧ ∼ q) ......(Identity law)
≡ (q ∨ p) ∧ T .......(Complement law)
= q ∨ p ......(Identity law)
= p ∨ q ......(Commutative law)
= R.H.S
Hence proved.
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