Exercise 1.9 (2)

2. Using the algebra of statement, prove that
i) [p ∧ (q ∨ r)] ∨ [∼ r ∧ ∼ q ∧ p] ≡ p
Ans: 
L.H.S .
=[p ∧ (q ∨ r)] ∨ [~r ∧ ~q ∧ p]
≡[p ∧ (q ∨ r)] ∨ [(~r ∧ ~q) ∧ p]……[Associative Law]
≡[p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p]…….[Commutative Law]
≡[p ∧ (q ∨ r) ∨ [~(q ∨ r) ∧ p]…..[De Morgan’s Law]
≡[p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)]……[Commutative Law]
≡p ∧ [(q ∨ r) ∨ ~(q ∨ r)]…….[Distributive Law]
≡p ∧ t……………[Complement Law]
≡p………………..[Identity Law]
=R.H.S

ii) (p ∧ q) ∨ (p ∧ ∼ q) ∨ (∼ p ∧ ∼ q) ≡ (p ∨ ∼ q)
Ans: 
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

iii) (p ∨ q) ∧ (∼ p ∨ ∼ q) ≡ (p ∨ ∼ q) ∧ (∼ p ∨ q)
Ans: 
L.H.S
= (p ∨ q) ∧ (∼ p ∨ ∼ q)
≡[P ∧ (∼P ∨ ∼q)] ∨ [q ∧ (∽p ∨ ∽q)]….(Distributive Law)
≡[(p ∧ ∼p) ∨ (p ∧ ∼q)] ∨ [(q ∧ ∼p) ∨ (q ∧ ∼q)]..(Distributive Law)
≡[F ∨ (p ∧ ∼q)] ∨ [(q ∧ ∼p) ∨ F]..(Complement Law)
≡(p ∧ ∼q) ∨ (q ∧ ∼p)….(Identity Law)
≡(p ∧ ∼q) ∨ (∼p ∧ q)…(Commutative Law)
≡RHS.