Consider the following premises. forall x_6: P_6(x_6) -> ~Q_6(x_6). exists x_6: Q(x_6). forall x_6: P_6(x_6) or P_3(x_6). forall x_6: Q_6(x_6). Can we infer forall x_6: ~(P_3(x_6) and Q(x_6)) from them?
 From forall x_6: P_6(x_6) -> ~Q_6(x_6), forall x_6: Q_6(x_6) we can infer forall x_6: ~P_6(x_6). From forall x_6: P_6(x_6) or P_3(x_6), forall x_6: ~P_6(x_6) we can infer forall x_6: P_3(x_6). Finally, from forall x_6: P_3(x_6), exists x_6: Q(x_6) we can infer exists x_6: P_3(x_6) and Q(x_6), which contradicts forall x_6: ~(P_3(x_6) and Q(x_6)). Therefore, the answer is no.