Consider the following premises. exists z_4: R_3(z_4) <-> S(z_4). forall z_4: S(z_4) <-> R_3(z_4). Can we infer forall z_4: ~(R_3(z_4) -> S(z_4)) from them?
 From exists z_4: R_3(z_4) <-> S(z_4) we can infer exists z_4: R_3(z_4) -> S(z_4). Finally, from forall z_4: S(z_4) <-> R_3(z_4) we can infer forall z_4: S(z_4) -> R_3(z_4), which contradicts forall z_4: ~(R_3(z_4) -> S(z_4)). Therefore, the answer is no.