re https://thimoteus.github.io/posts/2018-09-21-existential-types.html

@thimoteus asks

```
newtype Showable = Showable (forall r. (forall a. Show a => a -> r) -> r)
(Question: can we write this using the type Exists from above?)
```

I think we cannot refactor this using `type Exists f = ∀ r. (∀ a. f a -> r) -> r`

, right?

This doesn’t work `newtype Showable = Showable (Exists (forall a . Show a => a))`