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