Jonathan Cast wrote:
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.
Am I correct in assuming that if my program doesn't contain seq then I can reason using eta reduction?
Why is seq introduced?
Waiting for computers to get fast enough to run Haskell got old.
I'm guessing you were not being entirely serious here but I think that's a good answer.
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).
Reasoning as though you were in a category with a bottom should be ok as long as seq isn't present? I'm recalling a paper by Freyd on CPO categories which I can't lay my hands on at the moment or find via a search engine. I suspect Haskell (without seq) is pretty close to a CPO category.