Consider the following premises. forall x: Q_1(x) -> R_4(x). exists x: Q_3(x). forall x: Q_1(x) <-> Q_3(x). Can we infer forall x: ~R_4(x) from them?
 From forall x: Q_1(x) <-> Q_3(x) we can infer forall x: Q_3(x) -> Q_1(x). From forall x: Q_3(x) -> Q_1(x), exists x: Q_3(x) we can infer exists x: Q_1(x). Finally, from forall x: Q_1(x) -> R_4(x), exists x: Q_1(x) we can infer exists x: R_4(x), which contradicts forall x: ~R_4(x). Therefore, the answer is no.