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