Consider the following premises. forall x_11: Q(x_11) -> Q_11(x_11). exists x_11: Q_11(x_11) -> Q(x_11). forall x_11: Q_1(x_11) -> Q(x_11). forall x_11: Q_11(x_11) -> Q_1(x_11). Can we infer forall x_11: ~(Q(x_11) <-> Q_11(x_11)) from them? If possible, name the inference rules being used at each step.
 From forall x_11: Q(x_11) -> Q_11(x_11), exists x_11: Q_11(x_11) -> Q(x_11) we can infer exists x_11: Q(x_11) <-> Q_11(x_11) via existential biconditional introduction. From exists x_11: Q(x_11) <-> Q_11(x_11) we can infer exists x_11: Q(x_11) -> Q_11(x_11) via existential biconditional elimination. From forall x_11: Q_11(x_11) -> Q_1(x_11), forall x_11: Q_1(x_11) -> Q(x_11) we can infer forall x_11: Q_11(x_11) -> Q(x_11) via universal transitivity. Finally, from exists x_11: Q(x_11) -> Q_11(x_11), forall x_11: Q_11(x_11) -> Q(x_11) we can infer exists x_11: Q(x_11) <-> Q_11(x_11) via existential biconditional introduction, which contradicts forall x_11: ~(Q(x_11) <-> Q_11(x_11)). Therefore, the answer is no.