Consider the following premises. S_12(a_12) <-> S(a_12). forall x_12: S(x_12) -> S_12(x_12). Can we infer S_12(a_12) -> S(a_12) from them?
 From forall x_12: S(x_12) -> S_12(x_12) we can infer S(a_12) -> S_12(a_12). Finally, from S_12(a_12) <-> S(a_12) we can infer S_12(a_12) -> S(a_12). Therefore, the answer is yes.