[Haskell] (.) . (.)

[Christophe Poucet]

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). 
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)

Cheers
Christophe

Brian Hulley wrote:

> Brian Hulley wrote:
>
>> Taral wrote:
>>
>>> On 5/28/06, Dominic Steinitz <dominic.steinitz at blueyonder.co.uk>
>>> wrote:
>>>
>>>> Is this defined in some library? Thanks, Dominic.
>>>
>>>
>>> Don't think so. I use:
>>>
>>> \a b -> f (g a b)
>>
>>
>> I don't see how (.) . (.) translates into something so simple.
>> Using c for (.) to make things easier to write, I get:
>>
>> (.) . (.)
>> === c c c
>> === \x -> (c c c x)
>> === \x -> (c (c x))
>> === \x -> (\y z -> c (c x) y z)
>
>
> Here was one error ^^^^
> so fixing it and continuing:
>
> \x -> (c (c x))
> === \x -> (\y z -> c (c x) z y)
> === \x -> (\y z -> (c x) (z y))
> === \x -> (\y z -> (\p q -> c x q p) (z y))
> === \x -> (\y z -> (\p q -> x (q p)) (z y))
> === \x -> (\y z -> (\q -> x (q (z y))))
>
> === \x y z q -> x (q (z y))
>
> But it still doesn't match f (g a b)...
>
> Regards, Brian.
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-- 
Christophe Poucet
Ph.D. Student
Phone:+32 16 28 87 20
E-mail: Christophe (dot) Poucet (at) imec (dot) be
Website: http://notvincenz.com/  
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