Consider the following premises. forall z: ~P_1(z). forall z: R_2(z) <-> P_1(z). forall z: T_1(z) -> R_2(z). forall z: S_3(z) -> T_1(z). Can we infer exists z: S_3(z) from them?
 From forall z: S_3(z) -> T_1(z), forall z: T_1(z) -> R_2(z) we can infer forall z: S_3(z) -> R_2(z). From forall z: R_2(z) <-> P_1(z) we can infer forall z: R_2(z) -> P_1(z). From forall z: R_2(z) -> P_1(z), forall z: ~P_1(z) we can infer forall z: ~R_2(z). Finally, from forall z: S_3(z) -> R_2(z), forall z: ~R_2(z) we can infer forall z: ~S_3(z), which contradicts exists z: S_3(z). Therefore, the answer is no.