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