[Haskell-cafe] property testing, point-free style

Ivan Perez ivanperezdominguez at gmail.com
Wed May 31 12:36:01 UTC 2023

Maybe I'm missing something, but can't you just define:

(≡) f g = (f &&& g) >>^ (==)

(You could also define it as: (≡) f g x = f x == g x)

And use ≡ at will?

Why do you need an instance or a newtype?


On Wed, 31 May 2023 at 04:25, Olaf Klinke <olf at aatal-apotheke.de> wrote:

> Dear Cafe,
> The expression
> \x -> f x == g x
> is a testable property, as long as values for x can be randomly
> generated. For clarity I'd prefer a point-free style, e.g.
> f ≡ g
> Are there extensions to QuickCheck that let me write this? The
> QuickCheck package itself does not seem to contain such an operator. My
> current work-around is a
>     newtype ExtensionalEquality a b
> that holds two functions of type (a -> b) and a Testable instance for
> it. But I've got a hunch that I re-invented some wheel here. (My
> ExtensionalEquality is isomorphic to
>     Refl (a -> b) (a -> b)
> but Refl ist conceptually about type equality, not term equality.)
> Thanks
> Olaf
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