Consider the following premises. exists x_7: Q_7(x_7) -> R_7(x_7). forall x_7: Q_2(x_7) -> R_7(x_7). forall x_7: R_7(x_7) -> Q_2(x_7). forall x_7: Q_2(x_7) -> Q_7(x_7). Can we infer forall x_7: ~(Q_2(x_7) <-> R_7(x_7)) from them?
 From forall x_7: Q_2(x_7) -> Q_7(x_7), exists x_7: Q_7(x_7) -> R_7(x_7) we can infer exists x_7: Q_2(x_7) -> R_7(x_7). Finally, from forall x_7: R_7(x_7) -> Q_2(x_7), forall x_7: Q_2(x_7) -> R_7(x_7) we can infer forall x_7: R_7(x_7) <-> Q_2(x_7), which contradicts forall x_7: ~(Q_2(x_7) <-> R_7(x_7)). Therefore, the answer is no.