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