Question

Without using truth table prove that:

∼ [(p ∨ ∼ q) → (p ∧ ∼ q)] ≡ (p ∨ ∼ q) ∧ (∼ p ∨ q)

Answer 1

L.H.S. = ∼ [(p ∨ ∼ q) → (p ∧ ∼ q)]

≡ (p ∨ ∼ q) → (p ∧ ∼ q) .......(Negation of implication)

≡ (p ∨ ∼ q) ∧ [∼ p ∨ ∼ (∼ q)] ......(Negation of conjunction)

≡ (p ∨ ∼ q) ∧ (∼ p ∨ q) .......(Negation of negation)

= R.H.S.