Consider the following premises. exists y: ~T(y). forall y: S_1(y) -> T(y). forall y: R(y) <-> Q_1(y). forall y: Q_1(y) -> S_1(y). Can we infer exists y: ~R(y) from them?
 From forall y: Q_1(y) -> S_1(y), forall y: S_1(y) -> T(y) we can infer forall y: Q_1(y) -> T(y). From forall y: Q_1(y) -> T(y), exists y: ~T(y) we can infer exists y: ~Q_1(y). Finally, from forall y: R(y) <-> Q_1(y), exists y: ~Q_1(y) we can infer exists y: ~R(y). Therefore, the answer is yes.