Consider the following premises. forall z_2: T(z_2) -> S(z_2). forall z_2: T(z_2) -> S(z_2). forall z_2: T(z_2) <-> S(z_2). Can we infer exists z_2: ~(T(z_2) -> S(z_2)) from them?
 From forall z_2: T(z_2) <-> S(z_2) we can infer forall z_2: S(z_2) -> T(z_2). From forall z_2: T(z_2) -> S(z_2), forall z_2: S(z_2) -> T(z_2) we can infer forall z_2: T(z_2) <-> S(z_2). From forall z_2: T(z_2) <-> S(z_2) we can infer forall z_2: S(z_2) -> T(z_2). From forall z_2: S(z_2) -> T(z_2), forall z_2: T(z_2) -> S(z_2) we can infer forall z_2: S(z_2) <-> T(z_2). Finally, from forall z_2: S(z_2) <-> T(z_2) we can infer forall z_2: T(z_2) -> S(z_2), which contradicts exists z_2: ~(T(z_2) -> S(z_2)). Therefore, the answer is no.