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