[Haskell-cafe] Re: [Haskell] (.) . (.)

[Brian Hulley]

[Moved to cafe]
Christophe Poucet wrote:
> Hello,
>
> I think you look at the (.) . (.) as all being c's in the wrong way.
> You have to consider what will evaluate first (syntactical priority
> rules).

(.) . (.)
=== (.) (.) (.) -- a + b === (+) a b
=== c c c

> Anyways the simplest case that this is equivalent can be seen
> here: 22:33 < vincenz> @pl \f g a b -> f (g a b)
> 22:33 < lambdabot> (.) . (.)
>
> Writing it explicilty:
> (.) f g a = f (g a)
>
>
> so we get
> (.) . (.) = \f -> (.) ((.) f)
> (.) . (.) f = (.) ((.) f)
> (.) . (.) f g = (.) ((.) f) g
> (.) . (.) f g = ((.) f) . g
> (.) . (.) f g a = ((.) f) (g a)
> (.) . (.) f g a = (.) f (g a)
> (.) . (.) f g a = f . (g a)
> (.) . (.) f g a b = f . (g a) $ b
> (.) . (.) f g a b = f ((g a) b)
> (.) . (.) f g a b = f (g a b)

Thanks for the proof. I see it's got the advantage over mine that it doesn't 
need to use eta expansion and is also much clearer.

Regards, Brian.