Consider the following premises. forall x_16: P_16(x_16) <-> Q_16(x_16). forall x_16: P_3(x_16) -> Q_3(x_16). forall x_16: P_16(x_16). exists x_16: P_3(x_16). Can we infer forall x_16: ~(Q_16(x_16) and Q_3(x_16)) from them?
 From forall x_16: P_16(x_16) <-> Q_16(x_16) we can infer forall x_16: P_16(x_16) -> Q_16(x_16). From forall x_16: P_16(x_16) -> Q_16(x_16), forall x_16: P_16(x_16) we can infer forall x_16: Q_16(x_16). From forall x_16: P_3(x_16) -> Q_3(x_16), exists x_16: P_3(x_16) we can infer exists x_16: Q_3(x_16). Finally, from forall x_16: Q_16(x_16), exists x_16: Q_3(x_16) we can infer exists x_16: Q_16(x_16) and Q_3(x_16), which contradicts forall x_16: ~(Q_16(x_16) and Q_3(x_16)). Therefore, the answer is no.