1.

Without using truth table prove that:(p ∧ q) ∨ (∼ p ∧ q) ∨ (p ∧ ∼ q) ≡ p ∨ q

Answer»

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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