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