Consider the following premises. R(b) -> S_1(b). forall x: S_1(x) -> R(x). ~R(b). Can we infer S_1(b) from them?
 From forall x: S_1(x) -> R(x) we can infer S_1(b) -> R(b). From R(b) -> S_1(b), S_1(b) -> R(b) we can infer R(b) <-> S_1(b). From R(b) <-> S_1(b) we can infer S_1(b) -> R(b). Finally, from S_1(b) -> R(b), ~R(b) we can infer ~S_1(b), which contradicts S_1(b). Therefore, the answer is no.