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