The statement  ~(p↔∼q) is

~(p↔∼q)≡∼((p→∼q)∧(~q→p)≡∼((~p∨~q)∧(q∨p))≡∼(~q∨~p)∨(~(q∨p))≡(~(q)∧~(~p))∨(~q∧~p)≡(q∧p)∨(~q∧~p)≡[(q∧p)∨(~q)]∧[(q∧p)∨(~p)]≡[(q∨(~q))∧(p∨(~q))]∧[(q∨(~p)) ∧(p∨(~p))]≡[t∧(~q∨p)]∧[((~p)∨q))∧t]≡(~q∨p)∧(~p∨q)≡(q→p)∧(p→q)≡p↔qAlternative Solution. Use the following table. From the last two colums, we get t~(p↔(~q))≡p↔q