On Thu, Jun 21, 2012 at 7:00 AM, George Giorgidze
Regarding antisymmetry, if I also define
instance Ord Foo where (==) = (==) `on` a
then would that count as satisfying the law?
Probably, you mean here Eq instead of Ord.
If a <= b and b <= a then a = b (antisymmetry)
It all depends on what one means by "=". Opinions differ on this, see [1]. If we mean "==", which is a function to Bool, from the Eq class, then it does satisfy the antisymmetry law.
But in this case, the question arises what are the laws that Eq instances should satisfy.
Should it be that x == y implies x = y? But once again what does "=" mean here. It depends on who you ask (once again see [1]).
These nits don't need to be picked. The theory of equivalence relations is extremely well understood. This is a topic covered by the sophomore mathematics level.
But, my opinion is that if you can write a pure function f that can tell them apart (i.e., f x /= f y), I find it a very strange notion of equality.
There's your problem. It is straight-forward to generate an equivalence relation that can tell things apart with respect to a different equivalence relation. Equivalence relations do not need to agree! If you really want to keep distinct equivalence relations apart semantically, just use a newtype wrapper to explicitly "name" the relation.
I am not saying that this is a trivial undertaking, but would be a more principled approach. It would require the Haskell community and, especially, standard library and language specification developers to agree on a suitable notion of equality that Eq instances should satisfy.
That has already happened. A relation merely has to satisfy the equivalence relation axioms.