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