# Exercise 1.9 (1)

1. Without using truth table, show that
i) p ↔ q ≡ (p ∧ q) ∨ (∼ p ∧ ∼ q)
Ans:
L.H.S.
=p ↔ q
≡(p → q) ∧ (q → p)……………….[Biconditional Law]
≡(~p ∨ q) ∧ (~q ∨ p)……………….….[Conditional Law]
≡[~p ∧ (~q ∨ p)] ∨ [q ∧ (~q∨p)].[Distributive Law]
≡[(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)]  [Distributive Law]
≡[(~p ∧ ~q) ∨ F] ∨ [F ∨ (q ∧ p)]…..[Complement Law]
≡(~p ∧ ~q) ∨ (q ∧ p)     [Identity Law]
≡(p ∧ q) ∨ (~p ∧ ~q)     [Commutative Law]
≡R.H.S.

ii) p ∧ [(∼ p ∨ q) ∨ ∼ q] ≡ p
Ans:
L.H.S.
= p ∧ [(~p ∨ q) ∨ ~q]
≡ p ∧ [(~p ∨ (q ∨ ~q)]   [Associative Law]
≡ p ∧ (~p ∨ T)     [Complement Law]
≡ p ∧ T      [Identity Law]
≡ p    [Identity Law]
= R.H.S.

iii) ∼ [(p ∧ q) → ∼ q] ≡ p ∧ q
Ans:
L.H.S
= ∼ [(p ∧ q) → ∼ q]
≡(p ∧ q) ∧ ~(~q)…………………(Negation of Implication)
≡(p ∧ q) ∧ q…………………(Negation of Negation)
≡p ∧ (q ∧ q)…………………(Associative Law)
≡p ∧ q …………………….(Identity Law)
= R.H.S

iv) ∼ r → ∼ (p ∧ q) ≡ [∼ (q → r)] → ∼ p
Ans:
L.H.S
= ∼ r → ∼ (p ∧ q)
≡ ~(~r) ∨ ~(p ∧ q)…………….[Conditional Law]
≡ r ∨ ~(p ∧ q)…………………[Negation of negation]
≡ r ∨ (~p ∨ ~q)…………………[De Morgan’s Law]
≡r ∨ ~p ∨ ~q……………..[Associative Law]
≡~p ∨ (~q ∨ r)……………..[Associative Law]
≡~p ∨ (q → r)………………..[p→q≡~p∨q]
≡(q → r) ∨ ~p………….[Commutative Law]
≡~[~(q → r)] ∨ ~p…………[Negation of negation]
≡[~(q → r)]→ ~p……..[p → q≡~p ∨ q]
= R.H.S

v) (p ∨ q) → r ≡ (p → r) ∧ (q → r)
Ans:
L.H.S
=(p ∨ q) → r
≡~(p ∨ q) ∨ r………..[p→q≡~p∨q]
≡(~p ∧ ~q) ∨ r……..[De Morgan’s Law]
≡(~p ∨ r) ∧ (~q ∨ r)….[Distributive Law]
≡(p → r) ∧ (q → r)……[p → q≡ ~p ∨ q]
= R.H.S