Consider the following premises. forall z: S(z) -> U(z). exists z: ~T(z). forall z: S(z) -> T(z). forall z: U(z) -> S(z). Can we infer exists z: ~U(z) from them?
 From forall z: S(z) -> T(z), exists z: ~T(z) we can infer exists z: ~S(z). From forall z: S(z) -> U(z), forall z: U(z) -> S(z) we can infer forall z: S(z) <-> U(z). Finally, from forall z: S(z) <-> U(z), exists z: ~S(z) we can infer exists z: ~U(z). Therefore, the answer is yes.