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