[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.
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