Consider the following premises. forall x: S_3(x) -> T(x). forall x: T(x) <-> R(x). forall x: T(x) -> S_3(x). exists x: ~S_3(x). Can we infer forall x: R(x) from them?
 From forall x: S_3(x) -> T(x), forall x: T(x) -> S_3(x) we can infer forall x: S_3(x) <-> T(x). From forall x: S_3(x) <-> T(x), exists x: ~S_3(x) we can infer exists x: ~T(x). Finally, from forall x: T(x) <-> R(x), exists x: ~T(x) we can infer exists x: ~R(x), which contradicts forall x: R(x). Therefore, the answer is no.