Consider the following premises. forall x_8: Q_2(x_8) -> Q_8(x_8). forall y: Q_2(y). forall x_8: Q_8(x_8) -> Q_2(x_8). Can we infer ~Q_8(a_9) from them?
 From forall y: Q_2(y) we can infer Q_2(a_9). From forall x_8: Q_2(x_8) -> Q_8(x_8), forall x_8: Q_8(x_8) -> Q_2(x_8) we can infer forall x_8: Q_2(x_8) <-> Q_8(x_8). From forall x_8: Q_2(x_8) <-> Q_8(x_8) we can infer Q_2(a_9) <-> Q_8(a_9). From Q_2(a_9) <-> Q_8(a_9) we can infer Q_2(a_9) -> Q_8(a_9). Finally, from Q_2(a_9) -> Q_8(a_9), Q_2(a_9) we can infer Q_8(a_9), which contradicts ~Q_8(a_9). Therefore, the answer is no.