Closed christiaanb closed 7 years ago
test22 :: forall x y . (KnownNat x, KnownNat y) => Proxy x -> Proxy y -> Integer test22 _ _ = natVal (Proxy :: Proxy (y*x*y))
> test22 (Proxy @3) (Proxy @4) 16
Wow, nasty. Sounds like asserting that after finding out the unknowns and substituting them back into the original equation the proposition should still hold would be a good idea.