Consider the following premises. t -> Q(c). ~S(c). forall z_4: Q(z_4) <-> S(z_4). Can we infer t from them?
 From forall z_4: Q(z_4) <-> S(z_4) we can infer Q(c) <-> S(c). From Q(c) <-> S(c), ~S(c) we can infer ~Q(c). Finally, from t -> Q(c), ~Q(c) we can infer ~t, which contradicts t. Therefore, the answer is no.