Haskell Functor implied law

Question

Typeclassopedia says:

\"A similar argument also shows that any Functor instance satisfying the first law (fmap id = id) will automatically satisfy the second law as well

Answer 1

It seems that it was realized quite recently that law 2 follows from law 1. Thus, when the documentation was originally written, it was probably thought to be an independent requirement.

(Personally, I'm not convinced by the argument, but since I haven't had the time to work out the details myself, I'm giving it the benefit of doubt here.)