Consider the following premises. forall x: S(x) -> T_3(x). exists x: U(x) -> S(x). forall x: T_3(x) -> U(x). Can we infer forall x: ~(T_3(x) <-> U(x)) from them?
 From exists x: U(x) -> S(x), forall x: S(x) -> T_3(x) we can infer exists x: U(x) -> T_3(x). Finally, from forall x: T_3(x) -> U(x), exists x: U(x) -> T_3(x) we can infer exists x: T_3(x) <-> U(x), which contradicts forall x: ~(T_3(x) <-> U(x)). Therefore, the answer is no.