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

[Krzysztof Skrzętnicki]

It certainly makes perfect sense, because total order antisymmetry law
states that

IF a <= b AND b <= a THEN a = b

However it should rather be written

IF a <= b AND b <= a THEN a ~= b,

since = could be any equivalence class. However, we can also specify the Ord
on type

type Foo = Foo Int (Int->Int)

in this way:

instance Ord Foo where
   compare (Foo a _) (Foo b _) = compare a b

which yields equivalence relation that is not assuming equivalence of the
functions.
So this restriction does not seem to work on Adrian Hey's side.



Christopher Skrzętnicki


On Mon, Mar 10, 2008 at 8:06 PM, Dan Weston <westondan at imageworks.com>
wrote:

> On the other hand, though the behavior of == is not defined by the
> Report, it does require in 6.3.1 that if compare is defined, then ==
> must be defined. That strongly implies a semantic causal link (in the
> Free Theorem kind of way), that the semantics of Ord completely specify
> the semantics of Eq, and the only free and continuous way to specify
> this is to make == and EQ always agree.
>
> I would (almost) take this conclusion as normative as well.
>
> Dan
>
>
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