\[P:2.4\] Assume that for all limit ordinals $\gamma{\!\leq\!}\alpha$ satisfying Case1 the approximating sequences $\big(\lambda(\gamma,n):n{\!\in\!}{\mathbb{N}}\big)$ satisfies the above condition . It follows for all $\gamma\le \alpha$, with $$\gamma=\omega^{\xi_l}m_l+\omega^{\xi_{l-1}}m_{l-1}+ \ldots +\omega^{\xi_1}m_1,$$ being the Cantor Normal Form, that $$\begin{aligned}
\label{E:2.4.1} &{\mathcal S}_\gamma= {\mathcal S}_{\gamma_2}[{\mathcal S}_{\gamma_1}], \text{ where for some $j=1,\ldots l$}\\
& \gamma_1 =\omega^{\xi_l}m_l+\omega^{\xi_{l-1}}m_{l-1}+ \ldots \omega^{\xi_j}m^{(1)}_j \text{ and }
\gamma_2= \omega^{\xi_j}m_j^{(2)}+\omega^{\xi_{j-1}}m_{j-1}+ \ldots \omega^{\xi_1}m_1,\notag\\
&\qquad \text{with $m^{(1)}_j,m^{(2)}_j\in{\mathbb{N}}\cup\{0\}$,   $m_j=m^{(1)}_j+m_j^{(2)}$.}\notag\end{aligned}$$

We will show by transfinite induction for all $\gamma\le \alpha$. Assume that holds for all $\tilde \gamma<\gamma$. If $\gamma=\omega^\xi$, then is trivially satisfied. Indeed, in that case $\gamma=\gamma+ 0=0+\gamma$, are the only two choices to write $\gamma$ as the sum of two ordinals, and we observe that ${\mathcal S}_0[{\mathcal S}_\gamma]={\mathcal S}_\gamma[{\mathcal S}_0]={\mathcal S}_\gamma$.

It is left to verify in the case that $l\ge 2$ or $m_l\ge 2$. Let $\gamma=\gamma_1+\gamma_2$ be a decomposition of $\gamma$ as in the statement of . We can without loss of generality assume that $\gamma_2>0$.

If $\gamma_2=\beta+1$ for some $\beta$ (which implies that $\gamma$ itself is a successor ordinal) it follows from the induction hypothesis and that ${\mathcal S}_{\gamma_1+\beta+1}= {\mathcal S}_1\big[{\mathcal S}_\beta[{\mathcal S}_{\gamma_1}]\big]$, so we need to show that $${\mathcal S}_1\big[{\mathcal S}_\beta[{\mathcal S}_{\gamma_1}]\big] ={\mathcal S}_{\beta+1}[{\mathcal S}_{\gamma_1}].$$ If $A\in  {\mathcal S}_1\big[{\mathcal S}_\beta[{\mathcal S}_{\gamma_1}]\big]$, 