With dependent types you can. I'm not aware of whether that proof is possible in Haskell, however. Besides, that's not what I was saying. What I was saying it

With dependent types you can. I'm not aware of whether that proof is possible in Haskell, however. Besides, that's not what I was saying. What I was saying it that + in Haskell means more than an overloaded operator. It means the type has access to additional methods. This gives you more information.