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