Consider the following premises. forall y: T(y) -> Q(y). forall y: Q(y) -> R_1(y). exists y: R_1(y) -> T(y). Can we infer exists y: T(y) -> Q(y) from them?
 From forall y: Q(y) -> R_1(y), exists y: R_1(y) -> T(y) we can infer exists y: Q(y) -> T(y). From forall y: T(y) -> Q(y), exists y: Q(y) -> T(y) we can infer exists y: T(y) <-> Q(y). Finally, from exists y: T(y) <-> Q(y) we can infer exists y: T(y) -> Q(y). Therefore, the answer is yes.