Using the algebra of statement, prove that (p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q)

EXERCISE 1.9Q 2.3   PAGE 22

Exercise 1.9 | Q 2.3 | Page 22

Using the algebra of statement, prove that

(p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q)

SOLUTION

(p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q)

L.H.S.

= (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) ∨ (~ p ∧ q)      ....[Commutative law]

= R.H.S.