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 ---------------------------------------------------------------------- Finally - A spam blocker that actually works. http://www.bluebottle.com/tag/4 _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe