Question

The expression ~(~p → q) is logically equivalent to
1. p ∧ ~q
2. p ∧ q
3. ~p ∧ q
4. ~p ∧ ~q

Answer 1

Correct Answer - Option 4 : ~p ∧ ~q

Since, the expression

P → q = ~p ∨ q

So, ~p → q = p ∨ q

and therefore ~(~p → q) ≡ ~(p ∨ q)

≡ (~p) ∧ (~q) [by De Morgan's law]