Consider the following premises. forall x: R(x) or ~S(x). exists x: R(x) <-> R_4(x). Can we infer exists x: S(x) -> R_4(x) from them?
 From exists x: R(x) <-> R_4(x) we can infer exists x: R(x) -> R_4(x). From forall x: R(x) or ~S(x) we can infer forall x: S(x) -> R(x). Finally, from forall x: S(x) -> R(x), exists x: R(x) -> R_4(x) we can infer exists x: S(x) -> R_4(x). Therefore, the answer is yes.