Consider the following premises. forall x: T_2(x) <-> U(x). forall x: Q(x) or ~P_2(x). forall x: Q(x) -> U(x). Can we infer exists x: ~(P_2(x) -> T_2(x)) from them?
 From forall x: Q(x) or ~P_2(x) we can infer forall x: P_2(x) -> Q(x). From forall x: P_2(x) -> Q(x), forall x: Q(x) -> U(x) we can infer forall x: P_2(x) -> U(x). From forall x: T_2(x) <-> U(x) we can infer forall x: U(x) -> T_2(x). Finally, from forall x: P_2(x) -> U(x), forall x: U(x) -> T_2(x) we can infer forall x: P_2(x) -> T_2(x), which contradicts exists x: ~(P_2(x) -> T_2(x)). Therefore, the answer is no.