Consider the following premises. u. forall z_4: S_1(z_4) <-> P(z_4). ~S_1(g) -> q. ~P(g). Can we infer ~q from them?
 From forall z_4: S_1(z_4) <-> P(z_4) we can infer S_1(g) <-> P(g). From S_1(g) <-> P(g), ~P(g) we can infer ~S_1(g). Finally, from ~S_1(g) -> q, ~S_1(g) we can infer q, which contradicts ~q. Therefore, the answer is no.