Consider the following premises. R_2(a_7) -> Q_4(a_7). forall y_4: Q_4(y_4) <-> R_2(y_4). Can we infer R_2(a_7) <-> Q_4(a_7) from them?
 From forall y_4: Q_4(y_4) <-> R_2(y_4) we can infer Q_4(a_7) <-> R_2(a_7). From Q_4(a_7) <-> R_2(a_7) we can infer Q_4(a_7) -> R_2(a_7). Finally, from R_2(a_7) -> Q_4(a_7), Q_4(a_7) -> R_2(a_7) we can infer R_2(a_7) <-> Q_4(a_7). Therefore, the answer is yes.