Consider the following premises. forall x: ~Q_2(x). forall x: Q_2(x) -> P_2(x). forall x: P_2(x) <-> T(x). exists x: P_2(x) -> Q_2(x). Can we infer forall x: T(x) from them?
 From exists x: P_2(x) -> Q_2(x), forall x: Q_2(x) -> P_2(x) we can infer exists x: P_2(x) <-> Q_2(x). From exists x: P_2(x) <-> Q_2(x), forall x: ~Q_2(x) we can infer exists x: ~P_2(x). Finally, from forall x: P_2(x) <-> T(x), exists x: ~P_2(x) we can infer exists x: ~T(x), which contradicts forall x: T(x). Therefore, the answer is no.