Re: [Haskell-cafe] Re: (flawed?) benchmark : sort

[Dan Weston]

If x <= y && y <= x does not imply that x == y, then Ord has no business 
being a subclass of Eq. By your logic, there is absolutely no 
constructive subclassing going on here, only an existence proof of (==) 
given (<=). What is the rational basis of such an existence claim, 
unless == has the obvious meaning?

Or should I take it that you are suggesting we should move Ord up to be 
a peer of Eq?

Dan

Neil Mitchell wrote:
> Hi
>
> >  But antisymmetry means that (x <= y) && (y <= x) ==> x = y, where '=' means
> >  identity. Now what does (should) 'identity' mean?
>
> I think you are using the word identity when the right would would be
> equality. Hence, the answer is, without a doubt, (==). If you define
> equality, then you are defining equality.
>
> >  In short, I would consider code where for some x, y and a function f we have
> >  (x <= y) && (y <= x) [or, equivalently, compare x y == EQ] but f x /= f y
> >  broken indeed.
>
> I would consider it slightly bad code too. But not broken code. I
> think Ord functions can assume that Ord is a total ordering, nothing
> more. Nothing to do with the existence (or otherwise) of entirely
> unrelated code.
>
> Consider the following implementation of Data.Set, which *does* meet
> all the invariants in Data.Set:
>
> data Set a = Set [a]
> insert x (Set xs) = Set $ x : filter (/= x) xs
> elems (Set xs) = xs
>
> i.e. there is real code in the base libraries which breaks this notion
> of respecting classes etc. Is the interface to Data.Set broken? I
> would say it isn't.
>
> Thanks
>
> Neil


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