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