Consider the following premises. forall z_4: R(z_4) -> S(z_4). exists z_4: S(z_4) -> Q_2(z_4). forall z_4: Q_2(z_4) -> R(z_4). forall z_4: ~S(z_4). Can we infer exists z_4: ~Q_2(z_4) from them?
 From forall z_4: Q_2(z_4) -> R(z_4), forall z_4: R(z_4) -> S(z_4) we can infer forall z_4: Q_2(z_4) -> S(z_4). From exists z_4: S(z_4) -> Q_2(z_4), forall z_4: Q_2(z_4) -> S(z_4) we can infer exists z_4: S(z_4) <-> Q_2(z_4). Finally, from exists z_4: S(z_4) <-> Q_2(z_4), forall z_4: ~S(z_4) we can infer exists z_4: ~Q_2(z_4). Therefore, the answer is yes.