Consider the following premises. forall x: P(x) <-> T(x). forall x: Q_1(x) -> T(x). forall x: P(x) -> Q_1(x). Can we infer exists x: ~(P(x) -> T(x)) from them?
 From forall x: P(x) -> Q_1(x), forall x: Q_1(x) -> T(x) we can infer forall x: P(x) -> T(x). From forall x: P(x) <-> T(x) we can infer forall x: T(x) -> P(x). From forall x: T(x) -> P(x), forall x: P(x) -> T(x) we can infer forall x: T(x) <-> P(x). Finally, from forall x: T(x) <-> P(x) we can infer forall x: P(x) -> T(x), which contradicts exists x: ~(P(x) -> T(x)). Therefore, the answer is no.