Consider the following premises. forall x: Q_1(x) -> S_4(x). forall x: S_4(x) -> T(x). exists x: S(x) <-> T(x). Can we infer exists x: Q_1(x) -> S(x) from them?
 From exists x: S(x) <-> T(x) we can infer exists x: T(x) -> S(x). From forall x: S_4(x) -> T(x), exists x: T(x) -> S(x) we can infer exists x: S_4(x) -> S(x). Finally, from forall x: Q_1(x) -> S_4(x), exists x: S_4(x) -> S(x) we can infer exists x: Q_1(x) -> S(x). Therefore, the answer is yes.