Consider the following premises. forall z: R(z) <-> S(z). ~S(i_3) -> q. ~R(i_3). Can we infer q from them?
 From forall z: R(z) <-> S(z) we can infer R(i_3) <-> S(i_3). From R(i_3) <-> S(i_3), ~R(i_3) we can infer ~S(i_3). Finally, from ~S(i_3) -> q, ~S(i_3) we can infer q. Therefore, the answer is yes.