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

[MigMit]

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
> 
> foldl (flip const) whatever (undefined ++ [1,2,3])
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