Hi > 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?
Is this directed at me? I think x <= y && y <= x implies x == y. My point above was that where you have used x = y, I think = should be ==. I also think (compare x y == EQ) <=> (x == y), where <=> is bi-implication or boolean equality. i.e. Eq is a fine parent to Ord, but given a Eq/Ord pair they must be in agreement. Thanks Neil _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
