Consider the following premises. S_5(a_5) -> R_5(a_5). forall x_5: R_5(x_5) -> S_5(x_5). Can we infer ~(R_5(a_5) <-> S_5(a_5)) from them?
 From forall x_5: R_5(x_5) -> S_5(x_5) we can infer R_5(a_5) -> S_5(a_5). Finally, from R_5(a_5) -> S_5(a_5), S_5(a_5) -> R_5(a_5) we can infer R_5(a_5) <-> S_5(a_5), which contradicts ~(R_5(a_5) <-> S_5(a_5)). Therefore, the answer is no.