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?

Ivan

On Wed, 31 May 2023 at 04:25, Olaf Klinke <olf@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

_______________________________________________
Haskell-Cafe mailing list
To (un)subscribe, modify options or view archives go to:
http://mail.haskell.org/cgi-bin/mailman/listinfo/haskell-cafe
Only members subscribed via the mailman list are allowed to post.