Which of the following is false?
(p⇒q)⇔(~q⇒∼p) is a contradiction
p∨(~p) is a tautology
~(~p)⇔p is a tautology
p∧(~p) is a contradiction
p⇒q is logically equivalent to ~q⇒∼p, therefore,
(p⇒q)⇔(~q⇒∼p) is a tautology but not a contradiction
So (a) is false.