Consider the following premises. forall z_2: P(z_2) -> R(z_2). R(d) -> Q_2(d). forall z_2: Q_2(z_2) -> P(z_2). Can we infer ~(Q_2(d) <-> R(d)) from them?
 From forall z_2: Q_2(z_2) -> P(z_2), forall z_2: P(z_2) -> R(z_2) we can infer forall z_2: Q_2(z_2) -> R(z_2). From forall z_2: Q_2(z_2) -> R(z_2) we can infer Q_2(d) -> R(d). Finally, from Q_2(d) -> R(d), R(d) -> Q_2(d) we can infer Q_2(d) <-> R(d), which contradicts ~(Q_2(d) <-> R(d)). Therefore, the answer is no.