Consider the following premises. forall x_7: R(x_7) -> Q_1(x_7). forall x_7: R(x_7) and Q_7(x_7). Can we infer ~Q_1(a) from them?
 From forall x_7: R(x_7) and Q_7(x_7) we can infer forall x_7: R(x_7). Finally, from forall x_7: R(x_7) -> Q_1(x_7), forall x_7: R(x_7) we can infer forall x_7: Q_1(x_7), which contradicts ~Q_1(a). Therefore, the answer is no.