Consider the following premises. forall y: Q_3(y) -> R(y). forall y: R(y) -> Q_3(y). ~Q_3(f_3). Can we infer R(f_3) from them?
 From forall y: R(y) -> Q_3(y), forall y: Q_3(y) -> R(y) we can infer forall y: R(y) <-> Q_3(y). From forall y: R(y) <-> Q_3(y) we can infer R(f_3) <-> Q_3(f_3). Finally, from R(f_3) <-> Q_3(f_3), ~Q_3(f_3) we can infer ~R(f_3), which contradicts R(f_3). Therefore, the answer is no.