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