Consider the following premises. exists x_2: Q_2(x_2) -> R_4(x_2). forall x_2: R_4(x_2) -> Q_2(x_2). forall x_2: ~Q_2(x_2). Can we infer forall x_2: R_4(x_2) from them?
 From forall x_2: R_4(x_2) -> Q_2(x_2), exists x_2: Q_2(x_2) -> R_4(x_2) we can infer exists x_2: R_4(x_2) <-> Q_2(x_2). Finally, from exists x_2: R_4(x_2) <-> Q_2(x_2), forall x_2: ~Q_2(x_2) we can infer exists x_2: ~R_4(x_2), which contradicts forall x_2: R_4(x_2). Therefore, the answer is no.