[Haskell] (.) . (.)

[Brian Hulley]

Lennart Augustsson wrote:
> Why not ask the computer?
>
> -- Lennart
>
> bamse% ghci
>    ___         ___ _
>   / _ \ /\  /\/ __(_)
>  / /_\// /_/ / /  | |      GHC Interactive, version 6.4.1, for
> Haskell 98. / /_\\/ __  / /___| |      http://www.haskell.org/ghc/
> \____/\/ /_/\____/|_|      Type :? for help.
>
> Loading package base-1.0 ... linking ... done.
> Prelude> :t (.) . (.)
> (.) . (.) :: (b -> c) -> (a -> a1 -> b) -> a -> a1 -> c
> Prelude> Leaving GHCi.
> bamse% djinn
> Welcome to Djinn version 2005-12-12.
> Type :h to get help.
> Djinn> f ? (b -> c) -> (a -> a1 -> b) -> a -> a1 -> c
> f :: (b -> c) -> (a -> a1 -> b) -> a -> a1 -> c
> f x1 x2 x3 x4 = x1 (x2 x3 x4)
> Djinn> :q
> Bye.

Thanks! Although I note that Djinn will give a representative function 
corresponding to the type which is not necessarily (.) . (.), and the type 
given by GHCi doesn't tell me how it got there...

>
>
> 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)
>> === \x -> (\y z -> (c x) (y z))
>> === \x -> (\y z -> (\p q -> c x q p) (y z))
>> === \x -> (\y z -> (\p q -> x (q p) (y z))
>> === \x -> (\y z -> (\q -> x (q (y z))))
>>
>> === \x y z q -> x (q (y z))
>>
>> Have I made an error somewhere above?

I see my error was that I was reversing the args in eta expansion, so the 
correct derivation is:

(.) . (.)
=== c c c
=== \x -> c c c x
=== \x -> c (c x)
=== \x -> (\y z -> c (c x) y z)
=== \x -> (\y z -> (c x) (y z))
=== \x -> (\y z -> (\p q -> c x p q)(y z))  -- where I went wrong before
=== \x -> (\y z -> (\p q -> x (p q))(y z))
=== \x -> (\y z -> (\q -> x ((y z) q))

=== \x y z q -> x (y z q)

My apologies to Taral for doubting the original simplification...

Regards, Brian.

-
Logic empowers us and Love gives us purpose.
But societal laws, and religious dogma,
empower the dead,
to destroy us.

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