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प्रश्न
Without using truth table prove that:
(p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ p ∨ q
उत्तर
L.H.S. = (p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q)
≡ [(p ∨ ∼ p) ∧ q] ∨ (p ∧ ∼ q) ..........(Distributive Law)
≡ (T ∧ q) ∨ (p ∧ ∼ q) .........(Complement Law)
≡ q ∨ (p ∧ ∼ q) ...........(Identity Law)
≡ (q ∨ p) ∧ (q ∨ ∼ q) .........(Distributive Law)
≡ (q ∨ p) ∧ T ..........(Complement Law)
≡ q ∨ p ..........(Identity Law)
≡ p ∨ q ......(Commutative Law)
= R.H.S.
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