Consider the following premises. forall z_1: S(z_1) <-> Q(z_1). exists z_1: S(z_1) <-> P_2(z_1). Can we infer forall z_1: ~(P_2(z_1) -> Q(z_1)) from them?
 From exists z_1: S(z_1) <-> P_2(z_1) we can infer exists z_1: P_2(z_1) -> S(z_1). Finally, from forall z_1: S(z_1) <-> Q(z_1) we can infer forall z_1: S(z_1) -> Q(z_1), which contradicts forall z_1: ~(P_2(z_1) -> Q(z_1)). Therefore, the answer is no.