Consider the following premises. s <-> ~S_16(a_16). ~R_16(a_16). forall x_16: R_16(x_16) <-> S_16(x_16). Can we infer ~s from them?
 From forall x_16: R_16(x_16) <-> S_16(x_16) we can infer R_16(a_16) <-> S_16(a_16). From R_16(a_16) <-> S_16(a_16), ~R_16(a_16) we can infer ~S_16(a_16). From s <-> ~S_16(a_16) we can infer ~S_16(a_16) -> s. Finally, from ~S_16(a_16) -> s, ~S_16(a_16) we can infer s, which contradicts ~s. Therefore, the answer is no.