[proofplan] We prove first that a proper subset relation between ordinals is the same as membership: if $\gamma \subsetneq \delta$, then $\gamma \in \delta$. Applying this twice reduces trichotomy to ordinary set inclusion: either $\alpha \subset \beta$, $\beta \subset \alpha$, or each has an element outside the other. The last case is impossible by choosing least elements in the two differences and using transitivity of ordinals. Finally, Foundation rules out overlap among the three alternatives. [/proofplan] [step:Show that a proper subset of an ordinal is an element of it] Let $\gamma$ and $\delta$ be ordinals, and assume $\gamma \subsetneq \delta$. Since $\delta$ is well-ordered by $\in$ and $\delta \setminus \gamma$ is a nonempty subset of $\delta$, there exists an $\in$-least element $\eta \in \delta \setminus \gamma$. We claim that $\eta = \gamma$. First let $\xi \in \eta$. Since $\eta \in \delta$ and $\delta$ is transitive, $\xi \in \delta$. By minimality of $\eta$ in $\delta \setminus \gamma$, we cannot have $\xi \in \delta \setminus \gamma$, because $\xi \in \eta$. Hence $\xi \in \gamma$. Thus $\eta \subset \gamma$. Conversely let $\xi \in \gamma$. Since $\gamma \subset \delta$, we have $\xi \in \delta$. If $\xi = \eta$, then $\eta \in \gamma$, contradicting $\eta \in \delta \setminus \gamma$. If $\eta \in \xi$, then transitivity of $\gamma$ from $\xi \in \gamma$ would imply $\eta \in \gamma$, again contradicting $\eta \in \delta \setminus \gamma$. Since $\delta$ is well-ordered by $\in$, the elements $\xi,\eta \in \delta$ are comparable, so the only remaining possibility is $\xi \in \eta$. Hence $\gamma \subset \eta$. Therefore $\eta = \gamma$, and since $\eta \in \delta$, we obtain $\gamma \in \delta$. [guided] The point of this step is to prove the basic structural fact that ordinal inclusion has no hidden intermediate form: for ordinals, being a proper subset is exactly being an element. Let $\gamma$ and $\delta$ be ordinals with $\gamma \subsetneq \delta$. The set difference $\delta \setminus \gamma$ is nonempty. Because $\delta$ is an ordinal, the relation $\in$ well-orders $\delta$, so every nonempty subset of $\delta$ has an $\in$-least element. Choose the $\in$-least element $\eta \in \delta \setminus \gamma$. We prove $\eta = \gamma$ by double inclusion. First let $\xi \in \eta$. Since $\eta \in \delta$ and $\delta$ is transitive, membership descends inside $\delta$, so $\xi \in \delta$. If $\xi$ were not in $\gamma$, then $\xi \in \delta \setminus \gamma$. But $\xi \in \eta$, contradicting the choice of $\eta$ as the $\in$-least element of $\delta \setminus \gamma$. Therefore $\xi \in \gamma$, and hence $\eta \subset \gamma$. Now let $\xi \in \gamma$. Since $\gamma \subset \delta$, also $\xi \in \delta$. The elements $\xi$ and $\eta$ both lie in the well-ordered set $\delta$, so exactly one of $\xi \in \eta$, $\xi=\eta$, or $\eta \in \xi$ holds inside that well-order. The equality $\xi=\eta$ is impossible because $\xi \in \gamma$ but $\eta \notin \gamma$. The alternative $\eta \in \xi$ is also impossible: from $\xi \in \gamma$ and transitivity of the ordinal $\gamma$, it would follow that $\eta \in \gamma$, again contradicting $\eta \notin \gamma$. Therefore $\xi \in \eta$. Since this holds for every $\xi \in \gamma$, we have $\gamma \subset \eta$. Combining $\eta \subset \gamma$ and $\gamma \subset \eta$ gives $\eta=\gamma$. Since $\eta \in \delta$, this proves $\gamma \in \delta$. [/guided] [/step] [step:Exclude the possibility that neither ordinal contains the other as a subset] Assume, for contradiction, that $\alpha \nsubseteq \beta$ and $\beta \nsubseteq \alpha$. Since $\alpha$ is well-ordered by $\in$, choose the $\in$-least element $\xi \in \alpha \setminus \beta$. Since $\beta$ is well-ordered by $\in$, choose the $\in$-least element $\eta \in \beta \setminus \alpha$. For every $\zeta \in \xi$, transitivity of $\alpha$ gives $\zeta \in \alpha$, and minimality of $\xi$ gives $\zeta \in \beta$. Hence $\xi \subset \beta$. Since $\xi \in \alpha$ and every element of an ordinal is an ordinal, $\xi$ is an ordinal. Since $\xi \notin \beta$, the previous step applied to the ordinals $\xi$ and $\beta$ shows that $\xi$ cannot be a proper subset of $\beta$. Therefore $\xi = \beta$. Similarly, for every $\zeta \in \eta$, transitivity of $\beta$ gives $\zeta \in \beta$, and minimality of $\eta$ gives $\zeta \in \alpha$. Hence $\eta \subset \alpha$. Since $\eta \in \beta$ and every element of an ordinal is an ordinal, $\eta$ is an ordinal. Since $\eta \notin \alpha$, the previous step applied to the ordinals $\eta$ and $\alpha$ gives $\eta = \alpha$. Thus $\xi=\beta$ and $\eta=\alpha$. But $\xi \in \alpha$ and $\eta \in \beta$, so this gives $\beta \in \alpha$ and $\alpha \in \beta$, contradicting Foundation. Hence at least one of $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$ holds. [guided] We now prove that two ordinals cannot fail to contain one another as subsets. Suppose the contrary: $\alpha \nsubseteq \beta$ and $\beta \nsubseteq \alpha$. Then both differences $\alpha \setminus \beta$ and $\beta \setminus \alpha$ are nonempty. Because $\alpha$ is well-ordered by $\in$, choose the $\in$-least element $\xi \in \alpha \setminus \beta$. Because $\beta$ is well-ordered by $\in$, choose the $\in$-least element $\eta \in \beta \setminus \alpha$. We first analyze $\xi$. Let $\zeta \in \xi$. Since $\xi \in \alpha$ and $\alpha$ is transitive, $\zeta \in \alpha$. By the minimality of $\xi$ in $\alpha \setminus \beta$, no element of $\xi$ can lie in $\alpha \setminus \beta$. Therefore $\zeta \in \beta$. This proves $\xi \subset \beta$. Now $\beta$ is an ordinal by hypothesis, and $\xi$ is an ordinal because $\xi \in \alpha$ and every element of an ordinal is an ordinal. Therefore the previous step applies to the pair $\xi,\beta$. If $\xi \subsetneq \beta$, that step would imply $\xi \in \beta$, contradicting $\xi \in \alpha \setminus \beta$. Therefore $\xi$ is not a proper subset of $\beta$. Since we already proved $\xi \subset \beta$, we must have $\xi=\beta$. The same argument with $\alpha$ and $\beta$ interchanged gives $\eta=\alpha$. Indeed, every element of $\eta$ lies in $\alpha$ by transitivity of $\beta$ and minimality of $\eta$, so $\eta \subset \alpha$. Also $\eta$ is an ordinal because $\eta \in \beta$ and every element of an ordinal is an ordinal. If the inclusion $\eta \subset \alpha$ were proper, the previous step would give $\eta \in \alpha$, contradicting $\eta \in \beta \setminus \alpha$. Thus $\xi=\beta$ and $\eta=\alpha$. Since $\xi \in \alpha$ and $\eta \in \beta$, we obtain $\beta \in \alpha$ and $\alpha \in \beta$. This is a membership cycle, forbidden by the Axiom of Foundation. The contradiction proves that one of $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$ must hold. [/guided] [/step] [step:Convert subset comparability into membership comparability] By the previous step, either $\alpha \subseteq \beta$ or $\beta \subseteq \alpha$. If $\alpha \subseteq \beta$, then either $\alpha=\beta$ or $\alpha \subsetneq \beta$. In the second case, the first step gives $\alpha \in \beta$. If $\beta \subseteq \alpha$, then either $\beta=\alpha$ or $\beta \subsetneq \alpha$. In the second case, the first step gives $\beta \in \alpha$. Therefore at least one of \begin{align*} \alpha \in \beta, \qquad \alpha=\beta, \qquad \beta \in \alpha \end{align*} holds. [/step] [step:Use Foundation to prove the alternatives are mutually exclusive] It remains to show that no two alternatives can hold simultaneously. If $\alpha=\beta$ and $\alpha \in \beta$, then $\alpha \in \alpha$, contradicting Foundation. If $\alpha=\beta$ and $\beta \in \alpha$, then again $\alpha \in \alpha$, contradicting Foundation. Finally, if $\alpha \in \beta$ and $\beta \in \alpha$, then the two-element set $\{\alpha,\beta\}$ has no $\in$-minimal element, contradicting Foundation. Thus the three alternatives are pairwise mutually exclusive. Since the previous step proved that at least one holds, exactly one of them holds. [/step]