1.

Without using truth table prove that :(p ∧ q) ∨ (~ p ∧ q) ∨ (p ∧ ~q) ≡ p ∨ q

Answer»

LHS = (p ∧ q) v (~p ∧ q) ∨ (p ∧ ~q)

≡ [(p ∨ ~p) ∧ q] ∨ (p ∧ ~q) … (Distributive Law) 

≡ (T ∧ q) ∨ (p ∧ ~q) … (Complement Law) 

≡ q ∨ (p ∧ ~q) … (Identity Law) 

≡ (q ∨ p) ∧ (q ∨ ~q) … (Distributive Law) 

≡ (q ∨ p) ∧ T .. (Complement Law) 

≡ q ∨ p … (Identity Law)

≡ p ∨ q … (Commutative Law) 

≡ RHS.



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