Consider the following premises. forall z: S_1(z) <-> U(z). forall z: U(z) -> S_1(z). forall z: S_1(z) -> U(z). Can we infer exists z: ~(U(z) <-> S_1(z)) from them?
 From forall z: S_1(z) <-> U(z) we can infer forall z: S_1(z) -> U(z). From forall z: U(z) -> S_1(z), forall z: S_1(z) -> U(z) we can infer forall z: U(z) <-> S_1(z). From forall z: U(z) <-> S_1(z) we can infer forall z: U(z) -> S_1(z). Finally, from forall z: U(z) -> S_1(z), forall z: S_1(z) -> U(z) we can infer forall z: U(z) <-> S_1(z), which contradicts exists z: ~(U(z) <-> S_1(z)). Therefore, the answer is no.