Without using truth table, show that
~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p
SOLUTION
L.H.S.
≡ ~r → ~ (p ∧ q)
≡ ~(~ r) ∨ ~ (p ∧ q) ....[p → q ≡ ~ p ∨ q]
≡ r ∨ ~(p ∧ q) ....[Negation of negation]
≡ r ∨ (~p ∨ ~q) ....[De Morgan’s law]
≡ ~p ∨ (~q ∨ r) .....[Commutative and 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.
