Consider the following premises. forall x: R(x) <-> Q_2(x). forall x: ~S(x). forall x: R_1(x) <-> Q_2(x). forall x: S(x) <-> R(x). Can we infer exists x: R_1(x) from them?
 From forall x: S(x) <-> R(x), forall x: ~S(x) we can infer forall x: ~R(x). From forall x: R(x) <-> Q_2(x), forall x: ~R(x) we can infer forall x: ~Q_2(x). Finally, from forall x: R_1(x) <-> Q_2(x), forall x: ~Q_2(x) we can infer forall x: ~R_1(x), which contradicts exists x: R_1(x). Therefore, the answer is no.