Consider the following premises. exists x: T_2(x). forall x: S(x) -> ~T_2(x). forall x: T_4(x) -> S(x). forall x: S(x) -> T_4(x). Can we infer forall x: T_4(x) from them?
 From forall x: S(x) -> ~T_2(x), exists x: T_2(x) we can infer exists x: ~S(x). From forall x: S(x) -> T_4(x), forall x: T_4(x) -> S(x) we can infer forall x: S(x) <-> T_4(x). From forall x: S(x) <-> T_4(x) we can infer forall x: T_4(x) -> S(x). Finally, from forall x: T_4(x) -> S(x), exists x: ~S(x) we can infer exists x: ~T_4(x), which contradicts forall x: T_4(x). Therefore, the answer is no.