Re: [Haskell-cafe] Proof of duality theorem of fold?

Am Mittwoch, 18. März 2009 13:10 schrieb Daniel Fischer:

Now prove the

Lemma:

foldl g e (ys ++ zs) = foldl g (foldl g e ys) zs

for all g, e, ys and zs of interest. (I don't see immediately under which conditions this identity could break, maybe there aren't any)

Of course, hit send and you immediately think of foldl (flip const) whatever (undefined ++ [1,2,3])