Neil Mitchell wrote:
> instance Eq Foo where
> (==) (Foo a _) (Foo b _) = (==) a b
[...]
> Please give the sane law that this ordering violates. I can't see any!
The (non-existant) law would be
(Eq1) x == y => f x == f y, for all f of appropriate type
which is analogous to this (existant) law about observational equality:
(Eq2) x = y => f x = f y, for all f of appropriate type
Kalman
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