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