Consider the following premises. forall y_3: R(y_3) <-> Q(y_3). forall z: Q(z). Can we infer R(a_6) from them?
 From forall y_3: R(y_3) <-> Q(y_3) we can infer forall y_3: Q(y_3) -> R(y_3). From forall z: Q(z) we can infer Q(a_6). From forall y_3: Q(y_3) -> R(y_3) we can infer Q(a_6) -> R(a_6). Finally, from Q(a_6) -> R(a_6), Q(a_6) we can infer R(a_6). Therefore, the answer is yes.