 Solution

We prove simultaneously that each $L_{\alpha }$ is a transitive set and that $L_{\alpha }\subseteq L_{\alpha +1}$. The key observation is that if $A$ is transitive and $x\in A$, then $x\subseteq A$ andThis is an internal definition with parameter $x$, so $x\in \mathrm{Def}(A,1)$. Hence $A\subseteq \mathrm{Def}(A,1)$. Every $z\in \mathrm{Def}(A,1)$ is a subset of $A$; if $y\in z$, then $y\in A\subseteq \mathrm{Def}(A,1)$. Thus the definable power set is transitive too.

Start with the empty set. At successor stages the observation proves transitivity and inclusion. At a limit stage the union of the earlier transitive sets is transitive, and the same observation then proves its inclusion in its own definable power set. This completes the transfinite induction and gives