
Jonathan Cast
On 27 Dec 2007, at 10:44 AM, Achim Schneider wrote:
Wolfgang Jeltsch
wrote: Am Donnerstag, 27. Dezember 2007 16:34 schrieb Cristian Baboi:
I'll have to trust you, because I cannot test it.
let x=(1:x); y=(1:y) in x==y .
I also cannot test this:
let x=(1:x); y=1:1:y in x==y
In these examples, x and y denote the same value but the result of x == y is _|_ (undefined) in both cases. So (==) is not really equality in Haskell but a kind of weak equality: If x doesn’t equal y, x == y is False, but if x equals y, x == y might be True or undefined.
[1..] == [1..] certainly isn't undefined, it always evaluates to True,
If something happens, it does eventually happen.
More importantly, we can prove that [1..] == [1..] = _|_, since
[1..] == [1..] = LUB (n >= 1) [1..n] ++ _|_ == [1..n] ++ _|_ = LUB (n >= 1) _|_ = _|_
As far as I understand http://www.haskell.org/haskellwiki/Bottom , only computations which cannot be successful are bottom, not those that can be successful, but aren't. Kind of idealizing reality, that is. Confusion of computations vs. reductions and whether time exists or not is included for free here. Actually, modulo mere words, I accept both your and my argument as true, but prefer mine. You _do_ accept that you won't ever see Prelude.undefined in ghci when evaluating let x=(1:x); y=(1:y) in x==y , and there won't ever be a False in the chain of &&'s, don't you? The question arises, which value is left from the possible values of Bool when you take away False and _|_? And now don't you dare to say that _|_ /= undefined.