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