[Haskell-cafe] Re: (flawed?) benchmark : sort

Neil Mitchell ndmitchell at gmail.com
Mon Mar 10 17:08:29 EDT 2008


> 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.



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