Consider the following premises. forall x: Q(x) <-> Q_4(x). Q(f_4) -> t. Can we infer ~(Q_4(f_4) -> t) from them?
 From forall x: Q(x) <-> Q_4(x) we can infer Q(f_4) <-> Q_4(f_4). From Q(f_4) <-> Q_4(f_4) we can infer Q_4(f_4) -> Q(f_4). Finally, from Q_4(f_4) -> Q(f_4), Q(f_4) -> t we can infer Q_4(f_4) -> t, which contradicts ~(Q_4(f_4) -> t). Therefore, the answer is no.