Consider the following premises. forall z: Q(z) <-> R_2(z). forall z: Q(z) or ~R_2(z). Can we infer exists z: ~(Q(z) -> R_2(z)) from them?
 From forall z: Q(z) <-> R_2(z) we can infer forall z: Q(z) -> R_2(z). From forall z: Q(z) or ~R_2(z) we can infer forall z: R_2(z) -> Q(z). From forall z: Q(z) -> R_2(z), forall z: R_2(z) -> Q(z) we can infer forall z: Q(z) <-> R_2(z). Finally, from forall z: Q(z) <-> R_2(z) we can infer forall z: Q(z) -> R_2(z), which contradicts exists z: ~(Q(z) -> R_2(z)). Therefore, the answer is no.