[proofplan] We prove by transfinite induction on the ordinal $\alpha$ that $L_\alpha$ is transitive. The base stage is empty. At a successor stage, definability over $L_\alpha$ ensures every element of $L_{\alpha+1}$ is a subset of $L_\alpha$, while the induction hypothesis ensures elements of $L_\alpha$ are themselves definable subsets of $L_\alpha$, so they lie in $L_{\alpha+1}$. At a limit stage, membership in $L_\lambda$ occurs already at some earlier stage, where the induction hypothesis applies. Finally, transitivity of the class $L$ follows by choosing a stage containing the outer element. [/proofplan] [step:Establish the base stage] The set $L_0 = \varnothing$ is transitive. Indeed, to verify transitivity, suppose $x \in y \in L_0$. Since no such $y$ exists, the implication defining transitivity is satisfied. [guided] We recall that a set $A$ is transitive if for all sets $x$ and $y$, \begin{align*} x \in y \in A \implies x \in A. \end{align*} For $A = L_0 = \varnothing$, there is no set $y$ with $y \in L_0$. Hence the antecedent $x \in y \in L_0$ is never satisfied. Therefore the transitivity condition holds for $L_0$. [/guided] [/step] [step:Pass transitivity from $L_\alpha$ to $L_{\alpha+1}$] Let $\alpha$ be an ordinal and assume as the induction hypothesis that $L_\alpha$ is transitive. We prove that $L_{\alpha+1}$ is transitive. Let $x$ and $y$ be sets with $x \in y \in L_{\alpha+1}$. Since \begin{align*} L_{\alpha+1} = \operatorname{Def}(L_\alpha), \end{align*} the set $y$ is a definable subset of $L_\alpha$. In particular, \begin{align*} y \subset L_\alpha. \end{align*} Because $x \in y$, it follows that $x \in L_\alpha$. It remains to place $x$ in $L_{\alpha+1}$. Since $L_\alpha$ is transitive and $x \in L_\alpha$, every element of $x$ lies in $L_\alpha$, so \begin{align*} x \subset L_\alpha. \end{align*} Moreover, $x$ is definable over $(L_\alpha,\in)$ using the parameter $x \in L_\alpha$: it is the subset of $L_\alpha$ defined by the formula $z \in x$. Hence \begin{align*} x \in \operatorname{Def}(L_\alpha) = L_{\alpha+1}. \end{align*} Therefore $L_{\alpha+1}$ is transitive. [guided] Assume $L_\alpha$ is transitive. We must show that $L_{\alpha+1}$ is transitive, so take arbitrary sets $x$ and $y$ satisfying \begin{align*} x \in y \in L_{\alpha+1}. \end{align*} By definition of the constructible hierarchy, \begin{align*} L_{\alpha+1} = \operatorname{Def}(L_\alpha). \end{align*} Every member of $\operatorname{Def}(L_\alpha)$ is, by definition, a subset of $L_\alpha$ definable over the structure $(L_\alpha,\in)$ with parameters from $L_\alpha$. Since $y \in \operatorname{Def}(L_\alpha)$, we have \begin{align*} y \subset L_\alpha. \end{align*} Now $x \in y$, so the inclusion $y \subset L_\alpha$ gives \begin{align*} x \in L_\alpha. \end{align*} The last point is slightly important: from $x \in L_\alpha$ we still need $x \in L_{\alpha+1}$. This is where the induction hypothesis is used a second time. Since $L_\alpha$ is transitive and $x \in L_\alpha$, every element of $x$ belongs to $L_\alpha$. Thus \begin{align*} x \subset L_\alpha. \end{align*} Because parameters from $L_\alpha$ are allowed in the definition of $\operatorname{Def}(L_\alpha)$, the subset $x$ of $L_\alpha$ is definable over $(L_\alpha,\in)$ using the parameter $x$ itself, namely by the formula $z \in x$. Therefore \begin{align*} x \in \operatorname{Def}(L_\alpha) = L_{\alpha+1}. \end{align*} Since arbitrary $x \in y \in L_{\alpha+1}$ implies $x \in L_{\alpha+1}$, the set $L_{\alpha+1}$ is transitive. [/guided] [/step] [step:Handle limit stages by returning to an earlier stage] Let $\lambda$ be a limit ordinal and assume that $L_\beta$ is transitive for every ordinal $\beta < \lambda$. We prove that $L_\lambda$ is transitive. Let $x$ and $y$ be sets with $x \in y \in L_\lambda$. Since \begin{align*} L_\lambda = \bigcup_{\beta < \lambda} L_\beta, \end{align*} there exists an ordinal $\beta < \lambda$ such that $y \in L_\beta$. By the induction hypothesis, $L_\beta$ is transitive. Hence $x \in y \in L_\beta$ implies \begin{align*} x \in L_\beta. \end{align*} Since $\beta < \lambda$, we have $L_\beta \subset L_\lambda$ by the definition of $L_\lambda$ as the union of the earlier stages. Therefore $x \in L_\lambda$, proving that $L_\lambda$ is transitive. [guided] Let $\lambda$ be a limit ordinal, and suppose that every earlier stage $L_\beta$ with $\beta < \lambda$ is transitive. We want to prove that $L_\lambda$ is transitive. Take arbitrary sets $x$ and $y$ such that \begin{align*} x \in y \in L_\lambda. \end{align*} The defining property of a limit stage is \begin{align*} L_\lambda = \bigcup_{\beta < \lambda} L_\beta. \end{align*} Thus membership in $L_\lambda$ means membership in some earlier stage. Since $y \in L_\lambda$, there exists an ordinal $\beta < \lambda$ such that \begin{align*} y \in L_\beta. \end{align*} By the induction hypothesis, $L_\beta$ is transitive. Applying transitivity of $L_\beta$ to $x \in y \in L_\beta$, we obtain \begin{align*} x \in L_\beta. \end{align*} Finally, because $L_\lambda$ is the union of all $L_\gamma$ with $\gamma < \lambda$, the inclusion $L_\beta \subset L_\lambda$ holds for this particular $\beta < \lambda$. Therefore \begin{align*} x \in L_\lambda. \end{align*} Since arbitrary $x \in y \in L_\lambda$ implies $x \in L_\lambda$, the limit stage $L_\lambda$ is transitive. [/guided] [/step] [step:Conclude transitivity for every stage and for $L$] By transfinite induction, $L_\alpha$ is transitive for every ordinal $\alpha$. Now let $x$ and $y$ be sets with $x \in y \in L$. Since \begin{align*} L = \bigcup_{\alpha \in \operatorname{Ord}} L_\alpha, \end{align*} there exists an ordinal $\alpha$ such that $y \in L_\alpha$. The set $L_\alpha$ is transitive, so $x \in y \in L_\alpha$ implies $x \in L_\alpha$. Hence $x \in L$. Therefore $L$ is a transitive class. [/step]