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

Subject: Re: [Haskell-cafe] Re: (flawed?) benchmark : sort
From: Dan Weston
Date: Mon, 10 Mar 2008 14:01:53 -0700
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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