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