Consider the following premises. forall z_3: Q_1(z_3) <-> U(z_3). forall z_3: U(z_3) -> S_1(z_3). exists z_3: S_1(z_3) -> Q_1(z_3). Can we infer exists z_3: S_1(z_3) -> Q_1(z_3) from them?
That is one of the premises. Therefore, the answer is yes.