Transfinite Recursion Theorem/Uniqueness

Theorem
Let $G$ be a mapping.

Let $f$ be a mapping with a domain $y$ where $y$ is an ordinal.

Let $f$ satisfy the condition that:
 * $\forall x \in y: \map f x = \map G {f \restriction x}$

where $f \restriction x$ denotes the restriction of $f$ to $x$.

Let $g$ be a mapping with a domain $z$ where $z$ is an ordinal.

Let $g$ satisfy the condition that:
 * $\forall x \in z: \map g x = \map G {g \restriction x}$

Let $y \subseteq z$.

Then:
 * $\forall x \in y: \map f x = \map g x$

Proof
Proof by transfinite induction:

Suppose that:
 * $\forall x \in \alpha: \map f x = \map g x$

for some arbitrary ordinal $\alpha < y$.

Then $\alpha < z$.

Hence:

So applying induction:
 * $\forall \alpha < y: \map f \alpha = \map g \alpha$