Consider the following premises. forall x_20: R_20(x_20) -> S_20(x_20). s -> u. s <-> S_20(a_20). u -> R_20(a_20). Can we infer ~(R_20(a_20) <-> S_20(a_20)) from them?
 From forall x_20: R_20(x_20) -> S_20(x_20) we can infer R_20(a_20) -> S_20(a_20). From s -> u, u -> R_20(a_20) we can infer s -> R_20(a_20). From s <-> S_20(a_20) we can infer S_20(a_20) -> s. From S_20(a_20) -> s, s -> R_20(a_20) we can infer S_20(a_20) -> R_20(a_20). Finally, from R_20(a_20) -> S_20(a_20), S_20(a_20) -> R_20(a_20) we can infer R_20(a_20) <-> S_20(a_20), which contradicts ~(R_20(a_20) <-> S_20(a_20)). Therefore, the answer is no.