Consider the following premises. exists x_7: Q_2(x_7) -> Q(x_7). forall x_7: R(x_7) or ~Q(x_7). forall x_7: R(x_7) -> Q_2(x_7). Can we infer forall x_7: ~(Q(x_7) <-> R(x_7)) from them?
 From forall x_7: R(x_7) or ~Q(x_7) we can infer forall x_7: Q(x_7) -> R(x_7). From forall x_7: R(x_7) -> Q_2(x_7), exists x_7: Q_2(x_7) -> Q(x_7) we can infer exists x_7: R(x_7) -> Q(x_7). Finally, from forall x_7: Q(x_7) -> R(x_7), exists x_7: R(x_7) -> Q(x_7) we can infer exists x_7: Q(x_7) <-> R(x_7), which contradicts forall x_7: ~(Q(x_7) <-> R(x_7)). Therefore, the answer is no.