Consider the following premises. forall x: R(x). forall x: R_1(x). q -> ~R(b_1). Can we infer q from them?
 From forall x: R(x), forall x: R_1(x) we can infer forall x: R(x) and R_1(x). From forall x: R(x) and R_1(x) we can infer R(b_1) and R_1(b_1). From R(b_1) and R_1(b_1) we can infer R(b_1). Finally, from q -> ~R(b_1), R(b_1) we can infer ~q, which contradicts q. Therefore, the answer is no.