Consider the following premises. forall z: Q(z) <-> R(z). R(a) -> s. s -> Q(a). Can we infer R(a) -> Q(a) from them?
 From forall z: Q(z) <-> R(z) we can infer Q(a) <-> R(a). Finally, from R(a) -> s, s -> Q(a) we can infer R(a) -> Q(a). Therefore, the answer is yes.