I am studying "A Mathematical Introduction to Logic" by Herbert Enderton, and I have a question about this theorem which is proved in 4 different steps in the book, and the proof is rather long. I was thinking of a much shorter proof, but I'm not sure if my proof is rigorous. It might have some errors.
Recursion Theorem Assume that the subset C of U is freely generated from B by f and g, where
$f: \mathbb U \times \mathbb U \to \mathbb U$
$g: \mathbb U \to \mathbb U$
Further assume that $\mathbb V$ is a set and F, G, and h functions such that
$h: \mathbb B \to \mathbb V$
$F: \mathbb V \times \mathbb V \to \mathbb V$
$G: \mathbb V \to \mathbb V$
Then, there is a unique function $\overline{h}:\mathbb C \to \mathbb V$ such that
(i) $\overline{h}(x) = h(x) \ \forall x \in B$
(ii) for x and y in C,
$\overline{h}(g(x)) = G(\overline{h}(x))$
$\overline{h}(f(x,y)) = F(\overline{h}(x), \overline{h}(y))$
By freely generated, we mean that the restrictions of f and g, $f_{C} , g_{C}$ are one to one, and the ranges of $f_C$, $g_C$ and the set B are pairwise disjoint.
My Proof
Define $ S = \{x \in U |\ \overline{h}(x)\ \text{is uniquely determined by (i) and (ii)}\}$.
We claim that S = C. if $x \in B$, then by (i) we have $\overline{h}(x) = h(x)$, and since the ranges of $f_C$, $g_C$, and B are mutually exclusive, $h(x)$ is the only possilbe assignment to $\overline{h}(x)$.Suppose $x, y \in \mathbb U $ such that $\overline{h}(x)$ and $\overline{h}(y)$ have been uniquely determined by (i) and (ii), then by (ii), we assinge $\overline{h}(f(x,y)) = F(\overline{h}(x), \overline{h}(y))$. Since F is one-to-one, and the ranges $f_C$, $g_C$, and B are disjoint, the assingment is unique. The same arguement works also for $\overline{h}(g(x))$ when $\overline{h}(x)$ has been uniquely determined. Therefore, we have inductively proved that $S=C$.