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