There's always this, for ALL types a :(  So even where you would think that the documentation can claim that a given Eq instance follows the law of commutativity, it really cannot.

Prelude Control.Concurrent System.IO.Unsafe> do m <- newEmptyMVar ;putMVar m 1 ; print $ (unsafePerformIO (do v <- takeMVar m; putMVar m 2; return v)) == (unsafePerformIO (do v <- takeMVar m; putMVar m 1 ; return v)) ; print $ (unsafePerformIO (do v <- takeMVar m; putMVar m 1; return v)) == (unsafePerformIO (do v <- takeMVar m; putMVar m 2 ; return v))
False
True

---------- Původní zpráva ----------
Od: Jonas Almström Duregård <jonas.duregard@chalmers.se>
Datum: 24. 7. 2012
Předmět: Re: [Haskell-cafe] Is (==) commutative?

Hi,

I suppose you need to define what commutativity means in the presence
of undefined values. For instance if error "0" is different from error
"1" then you do not have commutativity. Also nontermination needs to
be considered, for instance "(fix $ \x->x) == undefined" typically
fails to terminate but "undefined == (fix $ \x->x)" typically yields
an error.

Regards,
Jonas

On 24 July 2012 10:10, Christian Sternagel <c.sternagel@gmail.com> wrote:
> Dear all,
>
> with respect to formal verification of Haskell code I was wondering whether
> (==) of the Eq class is intended to be commutative (for many classes such
> requirements are informally stated in their description, since Eq does not
> have such a statement, I'm asking here). Or are there any known cases where
> commutativity of (==) is violated (due to strictness issues)?
>
> cheers
>
> chris
>
> _______________________________________________
> Haskell-Cafe mailing list
> Haskell-Cafe@haskell.org
> http://www.haskell.org/mailman/listinfo/haskell-cafe

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