Re: [Haskell-cafe] Questions about the Functor class and it's use in "Data types à la carte"

[Jonathan Cast]

On 16 Dec 2007, at 3:21 AM, Dominic Steinitz wrote:

>>> Do you have a counter-example of (.) not being function  
>>> composition in
>>> the categorical sense?
>>
>> Let bot be the function defined by
>>
>> bot :: alpha -> beta
>> bot = bot
>>
>> By definition,
>>
>> (.) = \ f -> \ g -> \ x -> f (g x)
>>
>> Then
>>
>>    bot . id
>> = ((\ f -> \ g -> \ x -> f (g x)) bot) id
>> = (\ g -> \ x -> bot (g x)) id
>> = \ x -> bot (g x)
>
> I didn't follow the reduction here. Shouldn't id replace g everywhere?

Yes, sorry.

> This would give
>
> = \x -> bot x
>
> and by eta reduction

This is the point --- by the existence of seq, eta reduction is  
unsound in Haskell.

>
> = bot
>
>>
>> which /= bot since (seq bot () = bot) but (seq (\ x -> M) () = ())
>> regardless of what expression we substitute for M.
>>
>
> Why is seq introduced?

Waiting for computers to get fast enough to run Haskell got old.

Oh, you mean here?  Equality (=) for pickier Haskellers always means  
Leibnitz' equality:

Given x, y :: alpha

x = y if and only if for all functions f :: alpha -> (), f x = f y

f ranges over all functions definable in Haskell, (for some version  
of the standard), and since Haskell 98 defined seq, the domain of f  
includes (`seq` ()).  So since bot and (\ x -> bot x) give different  
results when handed to (`seq` ()), they must be different.

The `equational reasoning' taught in functional programming courses  
is unsound, for this reason.  It manages to work as long as  
everything terminates, but if you want to get picky you can find  
flaws in it (and you need to get picky to justify extensions to  
things like infinite lists).

jcc