I have a question about a proof from the book 'A First Journey through Logic' by M. Hils and F. Loeser:
Lemma 1.6.3: Let $\alpha$ and $\beta$ be ordinals, with $\alpha \neq 0$. Then there exists a unique pair of ordinals $(\rho,\mu)$ such that $\rho < \alpha$ and $\beta=\alpha \mu + \rho$.
The proof for the existence goes as follows: When $\beta=0$ there is nothing to prove. Assume $\beta \neq 0$. The mapping $f_0:\beta \rightarrow \alpha \times \beta:x \mapsto (0,x)$ is strictly increasing, hence $\beta \leq \alpha \beta$ by Lemma 1.5.8. If $\beta = \alpha \beta$, one sets $\mu=\beta$ and $\rho=0$. Otherwise, we have $\beta \in \alpha \beta$. Let $f$ be the unique isomorphism of ordered sets between $\alpha \beta$ and $\alpha \times \beta$ (Theorem 1.5.9). One sets $(\rho,\mu)=f(\beta)$ and one checks that $\beta=\alpha \mu + \rho$.
I have no idea why $\beta=\alpha \mu + \rho$ if $(\rho,\mu)$ is defined as $f(\beta)$. Can someone help me?
The results mentioned in the proof are:
Lemma 1.5.8: Let $f: \alpha \rightarrow \alpha'$ be a strictly increasing map between two ordinals. Then $f(\beta)\geq \beta$ for every $\beta \in \alpha$.
Theorem 1.5.9: Every well-ordered set $X$ is isomorphic, as an ordered set, to some ordinal. Furthermore, the ordinal and the isomorphism are both unique.
Also, $\alpha \beta$ is defined as the unique ordinal isomorphic to the reverse lexicographic product $\alpha \times \beta$.