10 Mar
2008
10 Mar
'08
5:08 p.m.
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