Using the algebra of statement, prove that

[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p

#### SOLUTION

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.