Consider the following premises. forall z_4: T(z_4) -> Q_2(z_4). forall z_4: Q_2(z_4) -> W(z_4). forall z_4: T(z_4) -> W(z_4). forall z_4: W(z_4) -> T(z_4). Can we infer exists z_4: ~(Q_2(z_4) -> T(z_4)) from them? If possible, name the inference rules being used at each step.
 From forall z_4: T(z_4) -> W(z_4), forall z_4: W(z_4) -> T(z_4) we can infer forall z_4: T(z_4) <-> W(z_4) via universal biconditional introduction. From forall z_4: T(z_4) <-> W(z_4) we can infer forall z_4: W(z_4) -> T(z_4) via universal biconditional elimination. From forall z_4: Q_2(z_4) -> W(z_4), forall z_4: W(z_4) -> T(z_4) we can infer forall z_4: Q_2(z_4) -> T(z_4) via universal transitivity. From forall z_4: T(z_4) -> Q_2(z_4), forall z_4: Q_2(z_4) -> T(z_4) we can infer forall z_4: T(z_4) <-> Q_2(z_4) via universal biconditional introduction. Finally, from forall z_4: T(z_4) <-> Q_2(z_4) we can infer forall z_4: Q_2(z_4) -> T(z_4) via universal biconditional elimination, which contradicts exists z_4: ~(Q_2(z_4) -> T(z_4)). Therefore, the answer is no.