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