Consider the following premises. forall x: S(x) <-> S_2(x). forall x: S(x) -> S_2(x). exists x: ~S_2(x). Can we infer forall x: S(x) from them?
 From forall x: S(x) <-> S_2(x) we can infer forall x: S_2(x) -> S(x). From forall x: S_2(x) -> S(x), forall x: S(x) -> S_2(x) we can infer forall x: S_2(x) <-> S(x). Finally, from forall x: S_2(x) <-> S(x), exists x: ~S_2(x) we can infer exists x: ~S(x), which contradicts forall x: S(x). Therefore, the answer is no.