Consider the following premises. forall x_14: P(x_14) -> U(x_14). forall x_14: ~P_14(x_14) or R_14(x_14). forall x_14: P_14(x_14) or Q_14(x_14). forall x_14: P(x_14). forall x_14: Q_14(x_14) -> U_2(x_14). forall x_14: U_2(x_14) or ~R_14(x_14). Can we infer exists x_14: ~(U(x_14) and U_2(x_14)) from them? If possible, name the inference rules being used at each step.
 From forall x_14: P_14(x_14) or Q_14(x_14), forall x_14: ~P_14(x_14) or R_14(x_14) we can infer forall x_14: Q_14(x_14) or R_14(x_14) via universal resolution. From forall x_14: U_2(x_14) or ~R_14(x_14) we can infer forall x_14: R_14(x_14) -> U_2(x_14) via universal rewriting. From forall x_14: Q_14(x_14) -> U_2(x_14), forall x_14: R_14(x_14) -> U_2(x_14), forall x_14: Q_14(x_14) or R_14(x_14) we can infer forall x_14: U_2(x_14) via universal disjunction elimination. From forall x_14: P(x_14) -> U(x_14), forall x_14: P(x_14) we can infer forall x_14: U(x_14) via universal modus ponens. Finally, from forall x_14: U(x_14), forall x_14: U_2(x_14) we can infer forall x_14: U(x_14) and U_2(x_14) via universal conjunction, which contradicts exists x_14: ~(U(x_14) and U_2(x_14)). Therefore, the answer is no.