Consider the following premises. forall x: Q_3(x) <-> Q(x). u_4 -> Q_3(h_3). Can we infer ~(u_4 -> Q(h_3)) from them?
 From forall x: Q_3(x) <-> Q(x) we can infer Q_3(h_3) <-> Q(h_3). From Q_3(h_3) <-> Q(h_3) we can infer Q_3(h_3) -> Q(h_3). Finally, from u_4 -> Q_3(h_3), Q_3(h_3) -> Q(h_3) we can infer u_4 -> Q(h_3), which contradicts ~(u_4 -> Q(h_3)). Therefore, the answer is no.