Consider the following premises. forall x_12: Q_2(x_12) <-> T(x_12). forall x_12: P_6(x_12) or ~Q_6(x_12). forall x_12: Q_2(x_12) -> Q_6(x_12). Can we infer exists x_12: ~(T(x_12) -> P_6(x_12)) from them?
 From forall x_12: Q_2(x_12) <-> T(x_12) we can infer forall x_12: T(x_12) -> Q_2(x_12). From forall x_12: P_6(x_12) or ~Q_6(x_12) we can infer forall x_12: Q_6(x_12) -> P_6(x_12). From forall x_12: Q_2(x_12) -> Q_6(x_12), forall x_12: Q_6(x_12) -> P_6(x_12) we can infer forall x_12: Q_2(x_12) -> P_6(x_12). Finally, from forall x_12: T(x_12) -> Q_2(x_12), forall x_12: Q_2(x_12) -> P_6(x_12) we can infer forall x_12: T(x_12) -> P_6(x_12), which contradicts exists x_12: ~(T(x_12) -> P_6(x_12)). Therefore, the answer is no.