On Sun, 2021-11-14 at 20:24 +0100, Sven Panne wrote:
Am So., 14. Nov. 2021 um 20:05 Uhr schrieb Olaf Klinke < olf@aatal-apotheke.de>:
That module defines two intervals to be equal if their bounds are equal, but to be unequal only if they are disjoint.
Well, this is simply a broken implementation. Quick question: Is the Ord instance lawful or not? Hard to tell... (at least without pencil & paper)
No, it's not. All four relations (>), (>=), (<), (<=) are transitive, but (>=) and (<=) should be reflexive, which they aren't: let i = I 0 1 :: Interval Int in i <= i = 1 <= 0 = False Comparability fails as well, which is why the author did not implement `compare`: j = I 0 2 :: Interval Int k = I 1 3 :: Interval Int (j <= k) || (k <= j) = (2 <= 1) || (3 <= 0) = False || False = False Antisymmetry is particularly interesting: It holds, but only for singleton intervals.
In my eyes this speaks for the proposal, because the library author would have been unable to write the broken instance. [...]
But this is an extremely weak argument for the proposal: One can easily write broken instances even with a single (==) method, e.g. violating symmetry etc. As another example: Given some e.g. Monad instance, can one quickly see if it is lawful? I very much doubt so, unless one tries to prove it. Laws of type classes are very much "outside of the Haskell language", anyway, so I see nothing special for Eq.
Indeed in general one can't tell whether a monad instance is lawful (Rice's theorem applies), but a least the class methods aren't trivially redundant. That removes one possible source of non-lawful- ness. Olaf