Consider the following premises. forall x_14: Q_4(x_14) -> P_10(x_14). forall x_14: Q_14(x_14) or Q_8(x_14). forall x_14: P_10(x_14) or ~Q_10(x_14). forall x_14: P(x_14) or ~Q_14(x_14). forall x_14: Q_8(x_14) -> P(x_14). forall x_14: Q_10(x_14) or Q_4(x_14). Can we infer exists x_14: ~(P(x_14) and P_10(x_14)) from them?
 From forall x_14: P(x_14) or ~Q_14(x_14) we can infer forall x_14: Q_14(x_14) -> P(x_14). From forall x_14: P_10(x_14) or ~Q_10(x_14) we can infer forall x_14: Q_10(x_14) -> P_10(x_14). From forall x_14: Q_14(x_14) -> P(x_14), forall x_14: Q_8(x_14) -> P(x_14), forall x_14: Q_14(x_14) or Q_8(x_14) we can infer forall x_14: P(x_14). From forall x_14: Q_10(x_14) -> P_10(x_14), forall x_14: Q_4(x_14) -> P_10(x_14), forall x_14: Q_10(x_14) or Q_4(x_14) we can infer forall x_14: P_10(x_14). Finally, from forall x_14: P(x_14), forall x_14: P_10(x_14) we can infer forall x_14: P(x_14) and P_10(x_14), which contradicts exists x_14: ~(P(x_14) and P_10(x_14)). Therefore, the answer is no.