Using the algebra of statement, prove that
(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)
SOLUTION
L.H.S.
= (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q)
≡ (p ∧ q) ∨ [(p ∧ ~ q) ∨ (~ p ∧ ~ q)] ....[Associative Law]
≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~ q ∧ ~ p)] ....[Commutative Law]
≡ (p ∧ q) ∨ [~q ∧ (p ∨ ~ p)] ....[Distributive Law]
≡ (p ∧ q) ∨ (~q ∧ t) .....[Complement Law]
≡ (p ∧ q) ∨ (~q) .....[Identity Law]
≡ (p ∨ ~ q) ∧ (q ∨ ~q) .....[Distributive Law]
≡ (p ∨ ~ q) ∧ t ....[Complement Law]
≡ p ∨ ~ q .....[Identity Law]
= R.H.S.
