Transitivity, growth, ordinals and rank in L

If $A$ is transitive, every $a\in A$ is the subset of $A$ defined by $x\in a$, so $A\subseteq \mathrm{Def}(A)$. Each $b\in \mathrm{Def}(A)$ is a subset of $A$; hence $z\in b$ implies $z\in A\subseteq \mathrm{Def}(A)$. This proves transitivity of Def(A), including $A=\mathrm{\varnothing }$ by its special clause. Set transfinite induction along each ordinal interval now proves transitivity of the levels and nesting: successors use this observation; limits are increasing unions.

Induct simultaneously against the cumulative hierarchy. At zero the inclusion is equality. If $L_{\beta }\subseteq V_{\beta }$, every member of $L_{\beta +1}$ is a subset of $L_{\beta }$, hence is in $V_{\beta +1}$. At limits take unions. Thus $L_{\alpha }\subseteq V_{\alpha }$; in particular any ordinal in $L_{\alpha }$ is less than $\alpha$.

Induction proves that every ordinal below $\alpha$ belongs to $L_{\alpha }$. Zero is immediate. Given $L_{\beta }\cap \mathrm{Ord}=\beta$, for $\beta >0$ the bounded ordinalhood formula over the transitive nonempty $L_{\beta }$ defines precisely the subset $\beta$, so $\beta \in L_{\beta +1}$. For $\beta =0$, $0\in L_1=\{0\}$. Old ordinals remain by nesting, and limit stages take unions. This proves the ordinal intersection identity and $\alpha \in L_{\alpha +1}$.

The formula $x=x$ defines the whole set $L_{\alpha }$ over itself when nonempty, and Def(empty) contains empty. Hence $L_{\alpha }\in L_{\alpha +1}$. For $x\in L$, its first membership stage is ${\rho }_L(x)+1$. If $x\in L_{\alpha }$, leastness gives ${\rho }_L(x)+1\le \alpha$ and so ${\rho }_L(x)<\alpha$. Conversely that inequality and nesting place $x$ in $L_{\alpha }$. Transitivity of the class union follows from transitivity of each level; step 3.1 puts every ordinal in that union.