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