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Is my proof draft below already a proof? How can the proof be completed?

Definition:
A unary complex function is a function from a subset of $\mathbb{C}$ into $\mathbb{C}$.
A binary complex function is a function from a subset of $\mathbb{C}^2$ into $\mathbb{C}$.
A finitary complex function is a function from a subset of $\mathbb{C}^n$ into $\mathbb{C}$ wherein $n\in\mathbb{N}_{\ge 1}$.

Conjecture:
Let $k,\kappa\in\mathbb{N}_{\ge 1}$,
$F_1,...,F_k$ and $\Phi_1,...,\Phi_\kappa$ unary transcendental or finitary algebraic complex functions,
$F$ the set of all functions that are generated by applying finite numbers of $F_1,...,F_k$,
$\Phi$ the set of all functions that are generated by applying finite numbers of $\Phi_1,...,\Phi_\kappa$,
and let
$n\in\{1,2\}$,
$A$ a unary or binary algebraic complex function,
$f_1,f_2$ unary transcendental complex functions
so that
$f\colon z\mapsto A(f_1(z),...,f_n(z))$ is a bijective complex function.
Let $f^{-1}$ be the inverse of $f$.
If $f\in F$ and $f^{-1}\in \Phi$, there exist $m$ functions $f_1,...,f_m\in F$ so that $f=f_m\circ\ ...\circ\ f_1$.

Proof draft:
Let $^{-1}$ denote the respective inverse.
If $n=1$, the assumption of the conjecture is obviously.
Assume $n=2$.
a) Assume $f_1,f_2$ are algebraically dependent.
Because $f_1,f_2$ are algebraically dependent, there is an algebraic function $A_1$ so that $f_2(z)=A_1(f_1(z))$ for all $z\in \text{dom}(f)$. Therefore there is an algebraic function $A_2$ so that $f(z)=A_2(f_1(z))$ for all $z\in \text{dom}(f)$, and the assumption of the conjecture follows.
b) Assume $f_1,f_2$ are algebraically independent.
Let's draw how the functions are applied for generating $f$:

$$f\colon\ \ \ z\ \ {^{\nearrow\ f_1(z)}_{\searrow\ f_2(z)}}^{\searrow}_{\nearrow}\ \ A(f_1(z),f_2(z))$$

Because $f$ is bijective, $A$ is bijective, and therefore $A^{-1}$ exists. Because $A$ is binary, $A^{-1}$ is a function into $\mathbb{C}^2$.
Let's draw how the inverse of $f$ is generated through that:

$$f^{-1}\colon\ \ \ A(f_1(z),f_2(z))\ \ {^{\nearrow\ f_1(z)}_{\searrow\ f_2(z)}}^{\searrow}_{\nearrow}\ \ z$$

We see from the figures: If $f$ is represented as above, $A^{-1}$, a function into $\mathbb{C}^2$, is needed for representing $f^{-1}$.
But $\Phi$ doesn't contain a function into $\mathbb{C}^2$. Therefore $n\neq 2$.
Because $A$ is unary or binary according to the preconditions, $A$ is unary, and the assumption of the conjecture follows.
That proofs the conjecture.
q.e.d.

Clearly, the conjecture and its proof can easily be generalized to

  • finitary $A$,
  • functions contained in $F$ with representations containing finitary $A$,
  • $f_1,f_2$ finitary complex functions,
  • arbitrary fields $\mathbb{K}$ instead of the field $\mathbb{C}$,
  • arbitrary sets $S$ with arbitrary finitary operations into $S$ instead of the field $\mathbb{C}$.
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