Consider the following premises. forall x: R_2(x) <-> S(x). exists x: R_2(x) -> Q_2(x). forall x: ~Q_2(x). Can we infer forall x: S(x) from them?
 From exists x: R_2(x) -> Q_2(x), forall x: ~Q_2(x) we can infer exists x: ~R_2(x). Finally, from forall x: R_2(x) <-> S(x), exists x: ~R_2(x) we can infer exists x: ~S(x), which contradicts forall x: S(x). Therefore, the answer is no.