Consider the following premises. forall y: Q(y) or ~Q_3(y). forall y: Q(y) -> Q_3(y). Can we infer exists y: ~(Q_3(y) <-> Q(y)) from them?
 From forall y: Q(y) or ~Q_3(y) we can infer forall y: Q_3(y) -> Q(y). Finally, from forall y: Q_3(y) -> Q(y), forall y: Q(y) -> Q_3(y) we can infer forall y: Q_3(y) <-> Q(y), which contradicts exists y: ~(Q_3(y) <-> Q(y)). Therefore, the answer is no.