Consider the following premises. exists y_1: S(y_1) <-> Q_3(y_1). forall y_1: ~Q_3(y_1). forall y_1: Q_3(y_1) -> S(y_1). Can we infer forall y_1: S(y_1) from them?
 From exists y_1: S(y_1) <-> Q_3(y_1) we can infer exists y_1: S(y_1) -> Q_3(y_1). From forall y_1: Q_3(y_1) -> S(y_1), exists y_1: S(y_1) -> Q_3(y_1) we can infer exists y_1: Q_3(y_1) <-> S(y_1). Finally, from exists y_1: Q_3(y_1) <-> S(y_1), forall y_1: ~Q_3(y_1) we can infer exists y_1: ~S(y_1), which contradicts forall y_1: S(y_1). Therefore, the answer is no.