Transfinite Recursion Theorem/Corollary

Theorem
Let $x$ be an ordinal.

Let $G$ be a mapping

There exists a unique mapping $f$ that satisfies the following properties:


 * The domain of $f$ is $x$
 * $\forall y \in x: f(y) = G(f \restriction y)$

Proof
Construct $K$ and $F$ just as in the First Principle of Transfinite Recursion. Set $f = ( F \restriction x )$.

Then, the domain of $f$ is $x$, since $x \subseteq \operatorname{Dom} F$.

Thus, such a function $f$ exists. Now, suppose there are two functions $f$ and $g$ that satisfy these conditions. We will use the first principle of transfinite induction to show that $f = g$. Take $y \in x$. Suppose $\forall z \in y: f(z) = g(z)$. Then $(f \restriction y) = (g \restriction y)$

So $f(y)=g(y)$ for all $y \in x$ by transfinite induction. Since $x$ is the domain of $f$ and $g$, it follows that $f=g$.