Consider the following premises. forall x_4: ~S(x_4). forall x_4: U(x_4) -> S(x_4). exists x_4: Q_4(x_4) -> U(x_4). forall x_4: S(x_4) -> Q_4(x_4). Can we infer forall x_4: U(x_4) from them?
 From forall x_4: S(x_4) -> Q_4(x_4), exists x_4: Q_4(x_4) -> U(x_4) we can infer exists x_4: S(x_4) -> U(x_4). From exists x_4: S(x_4) -> U(x_4), forall x_4: U(x_4) -> S(x_4) we can infer exists x_4: S(x_4) <-> U(x_4). Finally, from exists x_4: S(x_4) <-> U(x_4), forall x_4: ~S(x_4) we can infer exists x_4: ~U(x_4), which contradicts forall x_4: U(x_4). Therefore, the answer is no.