Consider the following premises. forall x_12: P_12(x_12) <-> Q_12(x_12). forall x_12: P_7(x_12) or ~P_12(x_12). forall x_12: Q_12(x_12) -> P_12(x_12). Can we infer exists x_12: ~(Q_12(x_12) -> P_7(x_12)) from them? If possible, name the inference rules being used at each step.
 From forall x_12: P_12(x_12) <-> Q_12(x_12) we can infer forall x_12: P_12(x_12) -> Q_12(x_12) via universal biconditional elimination. From forall x_12: P_7(x_12) or ~P_12(x_12) we can infer forall x_12: P_12(x_12) -> P_7(x_12) via universal rewriting. From forall x_12: Q_12(x_12) -> P_12(x_12), forall x_12: P_12(x_12) -> Q_12(x_12) we can infer forall x_12: Q_12(x_12) <-> P_12(x_12) via universal biconditional introduction. From forall x_12: Q_12(x_12) <-> P_12(x_12) we can infer forall x_12: Q_12(x_12) -> P_12(x_12) via universal biconditional elimination. Finally, from forall x_12: Q_12(x_12) -> P_12(x_12), forall x_12: P_12(x_12) -> P_7(x_12) we can infer forall x_12: Q_12(x_12) -> P_7(x_12) via universal transitivity, which contradicts exists x_12: ~(Q_12(x_12) -> P_7(x_12)). Therefore, the answer is no.