Consider the following premises. exists x_15: Q_15(x_15) -> P_15(x_15). forall x_15: P_8(x_15) -> Q_15(x_15). forall x_15: P_15(x_15) or Q_8(x_15). forall x_15: P_15(x_15) -> Q_15(x_15). forall x_15: P_8(x_15) <-> Q_8(x_15). Can we infer forall x_15: ~Q_15(x_15) from them? If possible, name the inference rules being used at each step.
 From forall x_15: P_15(x_15) -> Q_15(x_15), exists x_15: Q_15(x_15) -> P_15(x_15) we can infer exists x_15: P_15(x_15) <-> Q_15(x_15) via existential biconditional introduction. From exists x_15: P_15(x_15) <-> Q_15(x_15) we can infer exists x_15: P_15(x_15) -> Q_15(x_15) via existential biconditional elimination. From forall x_15: P_8(x_15) <-> Q_8(x_15) we can infer forall x_15: Q_8(x_15) -> P_8(x_15) via universal biconditional elimination. From forall x_15: Q_8(x_15) -> P_8(x_15), forall x_15: P_8(x_15) -> Q_15(x_15) we can infer forall x_15: Q_8(x_15) -> Q_15(x_15) via universal transitivity. Finally, from exists x_15: P_15(x_15) -> Q_15(x_15), forall x_15: Q_8(x_15) -> Q_15(x_15), forall x_15: P_15(x_15) or Q_8(x_15) we can infer exists x_15: Q_15(x_15) via existential disjunction elimination, which contradicts forall x_15: ~Q_15(x_15). Therefore, the answer is no.