Some "Theorem" on the generalized superfunction

[Leo.W]

You ... equation.

Can you expand on this?

...

OMG I'm sorry I forgot to answer this... so many things happened recently...(my hometown was struck by heavy flood last month, losing electricity and water, it was hard time )

Quick check:
Given \( \zeta(f(z))=g(\zeta(z)) \)
Take derivative \( \zeta'(f(z))f'(z)=g'(\zeta(z))\zeta'(z) \)
Notice that if we want the equation ONLY contains zeta'(z) without involving any other functions, we should let g'(zeta(z))=constant, so that g(z) is a linear function.
And as Law 3 stated, all g can be mapped to g(z)=z+1, which is definitely the Abel equation.
or more precisely, \( \zeta(f(z))=k\zeta(z)+s \), let \( P(z)=\frac{\log\left(\frac{kz+s-z}{Ck-C+s}\right)}{\log(k)},P({C})=0 \) which is the superfunction of g(z), we have \( P(\zeta(z))=\alpha(z) \) is the Abel function
And the derivative of abel function is the reciprocal of Julia function. So the whole RS definition is a derivation from Abel's equation.

Regards
Leo