Consider the following premises. exists z_2: S(z_2) -> R_1(z_2). forall z_2: R_1(z_2) <-> S(z_2). forall z_2: ~S(z_2). Can we infer exists z_2: ~R_1(z_2) from them?
 From forall z_2: R_1(z_2) <-> S(z_2) we can infer forall z_2: R_1(z_2) -> S(z_2). From exists z_2: S(z_2) -> R_1(z_2), forall z_2: R_1(z_2) -> S(z_2) we can infer exists z_2: S(z_2) <-> R_1(z_2). Finally, from exists z_2: S(z_2) <-> R_1(z_2), forall z_2: ~S(z_2) we can infer exists z_2: ~R_1(z_2). Therefore, the answer is yes.