Consider the following premises. forall x: U_4(x) <-> Q_3(x). S(j) -> r. forall x: U(x) -> U_4(x). forall x: Q_3(x) -> S(x). Can we infer U(j) -> r from them?
 From forall x: U_4(x) <-> Q_3(x) we can infer forall x: U_4(x) -> Q_3(x). From forall x: U(x) -> U_4(x), forall x: U_4(x) -> Q_3(x) we can infer forall x: U(x) -> Q_3(x). From forall x: U(x) -> Q_3(x), forall x: Q_3(x) -> S(x) we can infer forall x: U(x) -> S(x). From forall x: U(x) -> S(x) we can infer U(j) -> S(j). Finally, from U(j) -> S(j), S(j) -> r we can infer U(j) -> r. Therefore, the answer is yes.