Step 1: Approximations. For each ordinal $\theta$, call a function $f$ a $\theta$-approximation if:

• $\mathrm{dom}(f) = \theta$,
• $f(\alpha) = G(f \restriction \alpha)$ for all $\alpha < \theta$.

Step 2: Uniqueness of approximations. We claim that if $f$ is a $\theta$-approximation and $g$ is a $\varphi$-approximation with $\theta \leq \varphi$, then $f = g \restriction \theta$.

Proof by transfinite induction on $\alpha < \theta$: if $f(\beta) = g(\beta)$ for all $\beta < \alpha$, then $f \restriction \alpha = g \restriction \alpha$, so $f(\alpha) = G(f \restriction \alpha) = G(g \restriction \alpha) = g(\alpha)$.

Step 3: Existence of approximations. We show by transfinite induction that for each $\theta$, a $\theta$-approximation exists.

• $\theta = 0$: The empty function $\emptyset$ is a $0$-approximation.
• $\theta = \gamma + 1$: By hypothesis, a $\gamma$-approximation $f$ exists. Define $f' = f \cup \{(\gamma, G(f))\}$, which is a $(\gamma+1)$-approximation.
• $\theta$ limit: For each $\gamma < \theta$, let $f_\gamma$ be a $\gamma$-approximation. By Step 2, these are compatible: $f_\gamma \restriction \beta = f_\beta$ for $\beta \leq \gamma$. Define $f = \bigcup_{\gamma < \theta} f_\gamma$. By replacement, this is a set. Check: $\mathrm{dom}(f) = \bigcup_{\gamma<\theta}\gamma = \theta$, and for $\alpha < \theta$, choosing $\gamma > \alpha$: $f(\alpha) = f_\gamma(\alpha) = G(f_\gamma \restriction \alpha) = G(f \restriction \alpha)$.

Step 4: Define $F$. For each ordinal $\alpha$, let $f_{\alpha+1}$ be an $(\alpha+1)$-approximation and define $F(\alpha) = f_{\alpha+1}(\alpha)$. By Step 2, this is independent of the choice of approximation.

Step 5: Verify the recursion. For any $\alpha$: $F(\alpha) = f_{\alpha+1}(\alpha) = G(f_{\alpha+1} \restriction \alpha) = G(F \restriction \alpha)$, using the compatibility from Step 2.

Step 6: Uniqueness. If $F'$ also satisfies the recursion, then by transfinite induction, $F(\alpha) = F'(\alpha)$ for all $\alpha$. $\blacksquare$