Consider the following premises. forall z: U_2(z) or ~S(z). forall z: S(z) <-> U_2(z). Can we infer exists z: ~(U_2(z) <-> S(z)) from them?
 From forall z: U_2(z) or ~S(z) we can infer forall z: S(z) -> U_2(z). From forall z: S(z) <-> U_2(z) we can infer forall z: U_2(z) -> S(z). Finally, from forall z: U_2(z) -> S(z), forall z: S(z) -> U_2(z) we can infer forall z: U_2(z) <-> S(z), which contradicts exists z: ~(U_2(z) <-> S(z)). Therefore, the answer is no.