[Haskell-cafe] Re: Re: A free monad theorem?
Udo Stenzel
u.stenzel at web.de
Sun Sep 3 11:46:37 EDT 2006
Lennart Augustsson wrote:
> Well, bind is extracting an 'a'. I clearly see a '\ a -> ...'; it
> getting an 'a' so it can give that to g. Granted, the extraction is
> very convoluted, but it's there.
Oh, that can be remedied...
> m >>= g = m . flip g
In fact, why even mention m?
> (>>=) = (. flip) . (.)
Anyway, there's no "a extracted from m", since a function cannot be
deconstructed. That lets the "free theorem" degenerate into "m >>= k
does something with m and/or k, most of the time", which is kinda
meaningless and explains exactly nothing.
Udo.
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