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