Consider the following premises. forall x: Q_1(x). forall x: Q(x) -> ~Q_1(x). forall x: P_1(x). Can we infer exists x: Q(x) from them?
 From forall x: Q_1(x), forall x: P_1(x) we can infer forall x: Q_1(x) and P_1(x). From forall x: Q_1(x) and P_1(x) we can infer forall x: Q_1(x). Finally, from forall x: Q(x) -> ~Q_1(x), forall x: Q_1(x) we can infer forall x: ~Q(x), which contradicts exists x: Q(x). Therefore, the answer is no.