[Haskell-cafe] Re: Wikipedia on first-class object

[Achim Schneider]

Jonathan Cast <jonathanccast at fastmail.fm> wrote:

> On 27 Dec 2007, at 10:44 AM, Achim Schneider wrote:
> 
> > Wolfgang Jeltsch <g9ks157k at acme.softbase.org> 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.