Consider the following premises. exists y: R_4(y) or ~Q_4(y). forall y: Q_4(y) or ~R_4(y). Can we infer forall y: ~(Q_4(y) -> R_4(y)) from them?
 From exists y: R_4(y) or ~Q_4(y) we can infer exists y: Q_4(y) -> R_4(y). Finally, from forall y: Q_4(y) or ~R_4(y) we can infer forall y: R_4(y) -> Q_4(y), which contradicts forall y: ~(Q_4(y) -> R_4(y)). Therefore, the answer is no.