[Haskell-cafe] Re: Re: A free monad theorem?

[Udo Stenzel]

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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