Consider the following premises. forall x_1: ~Q_1(x_1). forall x_1: R_3(x_1) -> Q_1(x_1). forall x_1: R_3(x_1) <-> Q(x_1). Can we infer exists x_1: Q(x_1) from them?
 From forall x_1: R_3(x_1) -> Q_1(x_1), forall x_1: ~Q_1(x_1) we can infer forall x_1: ~R_3(x_1). Finally, from forall x_1: R_3(x_1) <-> Q(x_1), forall x_1: ~R_3(x_1) we can infer forall x_1: ~Q(x_1), which contradicts exists x_1: Q(x_1). Therefore, the answer is no.