Consider the following premises. forall x_2: S(x_2) <-> S_2(x_2). Can we infer ~(S(e) -> S_2(e)) from them?
 From forall x_2: S(x_2) <-> S_2(x_2) we can infer S(e) <-> S_2(e). Finally, from S(e) <-> S_2(e) we can infer S(e) -> S_2(e), which contradicts ~(S(e) -> S_2(e)). Therefore, the answer is no.