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