[Haskell-beginners] Foldable of Foldable

Tony Morris tonymorris at gmail.com
Fri Jan 2 15:16:35 UTC 2015

A foldable of a foldable is itself foldable. That fact is expressed in 


instance (Foldable g, Foldable f, Functor g) => Foldable (:. g f)

On 03/01/15 01:14, Julian Birch wrote:
> Apologies if this isn't clear, I suspect if I understood the 
> terminology better I'd already have solved the problem. I've got a 
> foldable of a foldable of a.  (Specifically `[Set a]`) and I want to 
> be able to express the concept that I can treat `(Foldable f1, 
> Foldable f2) => f1 (f2 a)` as `f3 a` i.e. a foldable of a foldable of 
> a can be newtyped to a foldable of a.  At least, I think that's right.
> Sadly, my attempts at actually expressing this concept have met with 
> incomprehension from GHC, could someone help me out, please?
> Thanks,
> Julian.
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