+Returning to our investigation of order, now that we have `proven'
+antisymmetry, our universe has a partial order. The only thing required to
+upgrade to a total order is proof that every number is comparable, but first a
+lemma.
+
+\begin{theorem} \label{thm:weak-self-relation}
+For $x = \surreal{X_L}{X_R}$, it holds that $X_L \leq x \leq X_R$.
+\end{theorem}
+
+\begin{proof}
+Seeking a contradiction, assume $\exists x_L \in X_L$ such that $x_L \nleq x$.
+
+By Axiom \autoref{ax:leq-relation}, this creates two possible cases. Either
+$\exists (x_L)_L \in (X_L)_L$ such that $(x_L)_L \geq x$ or $\exists x_R \in
+X_R$ such that $x_R \leq x_L$. By Axiom \autoref{ax:number-definition}, the
+latter case is impossible.
+
+We address the remaining case by inductively noting that $(x_L)_L \leq x_L$,
+thus $x \leq (x_L)_L$. However, since this means that $x \leq (x_L)_L \leq
+x_L$, by \autoref{thm:transitive}, $x \leq x_L$. By definition, this means $X_L
+\nleq x_L$, a contradiction since $x_L \in X_L$.
+
+The proof of $x \leq X_R$ follows similarly.
+\end{proof}
+
+\begin{theorem} \label{thm:all-comparable}
+For every $x$ and $y$ in the universe, if $y \nleq x$, then $x \leq y$.
+\end{theorem}
+
+\begin{proof}
+Seeking a contradiction, let $x$ and $y$ be numbers such that $y \nleq x$ and
+$x \nleq y$.
+
+By Axiom \autoref{ax:leq-relation}, this creates two possible cases. Either
+$\exists x_L \in X_L$ such that $x_L \geq y$ or $\exists y_R \in Y_R$ such that
+$x \geq y_R$. In the first case, by \autoref{thm:weak-self-relation}, $x_L \leq
+x$, so by \autoref{thm:transitive}, $y \leq x$, a contradiction. A similar
+contradiction is reached in the other case.
+\end{proof}
+
+Well, that's it. We have a total order, but we're no longer in the same
+universe as Knuth. We can even go one step further and note that, by only
+considering a finite number of generations, we are guaranteed to only generate
+a finite quantity of numbers, and a finite totally ordered set is automatically
+well ordered.
+
+With a total order in hand, we can strengthen $x \nleq y$ into $x > y$,
+motivating the following theorem.
+
+\begin{theorem} \label{thm:strong-self-relation}
+For any number $x = \surreal{X_L}{X_R}$, it holds that $X_L < x < X_R$.
+\end{theorem}
+
+\begin{proof}
+Simply recognize why `linear order' is a synonym for `total order'.
+\end{proof}
+
+Moving away from considering order, note that creating Definition
+\autoref{defi:similar} imposed order on our universe at the cost of some
+uniqueness. Let's examine that loss of uniqueness a bit further.
+
+The twenty numbers from generations 0-2 break down into seven similarity-based
+equivalence classes, as shown below.
+
+$$\surreal{}{-1} \similar \surreal{}{-1,0} \similar \surreal{}{-1,1} \similar \surreal{}{-1,0,1}$$
+$$\surreal{-1}{0} \similar \surreal{-1}{0,1} \similar \surreal{-1}{}$$