Consider the following premises. ~R_1(d). forall x_1: S_2(x_1) <-> R_1(x_1). S_2(d) <-> q. Can we infer ~q from them?
 From forall x_1: S_2(x_1) <-> R_1(x_1) we can infer S_2(d) <-> R_1(d). From S_2(d) <-> R_1(d), ~R_1(d) we can infer ~S_2(d). Finally, from S_2(d) <-> q, ~S_2(d) we can infer ~q. Therefore, the answer is yes.