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