Sum

Without using the truth table show that P ↔ q ≡ (p ∧ q) ∨ (~ p ∧ ~ q)

Advertisement Remove all ads

#### Solution

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)

≡ (p ∧ q) ∨ (∼ p ∧ ∼ q) ........(Commutative Law)

≡ R.H.S.

Concept: Algebra of Statements

Is there an error in this question or solution?

#### APPEARS IN

Advertisement Remove all ads