Consider the following premises. forall x_14: Q_14(x_14) or Q_8(x_14). forall x_14: Q_10(x_14) or Q_4(x_14). forall x_14: Q_4(x_14) -> T_4(x_14). forall x_14: T_4(x_14) or ~Q_10(x_14). forall x_14: Q(x_14) or ~Q_14(x_14). forall x_14: Q_8(x_14) -> Q(x_14). Can we infer exists x_14: ~(Q(x_14) and T_4(x_14)) from them? If possible, name the inference rules being used at each step.
 From forall x_14: Q(x_14) or ~Q_14(x_14) we can infer forall x_14: Q_14(x_14) -> Q(x_14) via universal rewriting. From forall x_14: T_4(x_14) or ~Q_10(x_14) we can infer forall x_14: Q_10(x_14) -> T_4(x_14) via universal rewriting. From forall x_14: Q_14(x_14) -> Q(x_14), forall x_14: Q_8(x_14) -> Q(x_14), forall x_14: Q_14(x_14) or Q_8(x_14) we can infer forall x_14: Q(x_14) via universal disjunction elimination. From forall x_14: Q_10(x_14) -> T_4(x_14), forall x_14: Q_4(x_14) -> T_4(x_14), forall x_14: Q_10(x_14) or Q_4(x_14) we can infer forall x_14: T_4(x_14) via universal disjunction elimination. Finally, from forall x_14: Q(x_14), forall x_14: T_4(x_14) we can infer forall x_14: Q(x_14) and T_4(x_14) via universal conjunction, which contradicts exists x_14: ~(Q(x_14) and T_4(x_14)). Therefore, the answer is no.