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