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

More interesting: foldl (flip const) whatever (repeat 1 ++ [1,2,3]) Daniel Fischer wrote on 18.03.2009 15:17:

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

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