+\begin{defi} \label{defi:younger}
+Given two sets of surreal numbers $X$ and $Y$ with equal cardinality, we say
+that $X$ is \emph{younger} than $Y$ iff, $\forall x_i \in X$ and $\forall y_j
+\in Y$ it holds that $\sum g(x_i) < \sum g(y_j)$. Similarly, we say that $Y$ is
+\emph{older} than $X$.
+\end{defi}
+
+With these definitions in place, let's start pursuing an order.
+
+\begin{theorem} \label{thm:transitive}
+Axiom \autoref{ax:leq-relation} is transitive. That is, if $x \leq y$ and $y
+\leq z$, then $x \leq z$.
+\end{theorem}
+
+\begin{proof}
+Seeking a contradiction, let $x = \surreal{X_L}{X_R}$, $y =
+\surreal{Y_L}{Y_R}$, and $z = \surreal{Z_L}{Z_R}$ be the youngest set of
+numbers satisfying $x \leq y$, $y \leq z$ and $x \nleq z$. Note by inspection
+that generation-0 and generation-1 obey transitivity.
+
+By Axiom \autoref{ax:leq-relation}, this creates two possible cases. There
+could exist an $x_L \in X_L$ such that $x_L \geq z$ or there could exist a $z_R
+\in Z_R$ such that $z_R \leq x$.
+
+In either case, we end up with a set of three numbers which do not obey
+transitivity, namely \set{y, z, x_L} or \set{z_R, x, y}. But, since $x_L
+\in X_L$ and $z_R \in Z_R$, we know that $g(x_L) < g(x)$ and $g(z_R) < g(z)$.
+In other words, if we consider the sum $g(x) + g(y) + g(z)$, it must be greater
+than both the sum $g(y) + g(z) + g(x_L)$ and the sum $g(z_R) + g(x) + g(y)$.
+
+Since this means a younger set of numbers exist for which transitivity fails,
+we have reached the desired contradiction.
+\end{proof}
+
+\begin{theorem} \label{thm:reflexive}
+Axiom \autoref{ax:leq-relation} is reflexive. That is, $x \leq x$.
+\end{theorem}
+
+\begin{proof}
+Let $x$ be the youngest number satisfying $x \nleq x$. Note that $0 \leq 0$ by
+inspection.
+
+By Axiom \autoref{ax:leq-relation}, this creates two possible cases. Either
+$\exists x_L \in X_L$ such that $x \leq x_L$ or $\exists x_R \in X_R$ such that
+$x_R \leq x$.
+
+In the first case, by Axiom \autoref{ax:leq-relation}, we find that $X_L \ngeq
+x_L$, implying that $x_L \ngeq x_L$, a contradiction since $x_L \in X_L$ from
+the start. We arrive at a similar contradiction for the second case.
+\end{proof}
+
+Per \autoref{thm:transitive} and \autoref{thm:reflexive}, we find that Axiom
+\autoref{ax:leq-relation} is at least a preorder on our universe. Normally we
+would move on to proving that Axiom \autoref{ax:leq-relation} is antisymmetric,
+however, there is a problem. Working by hand with Axioms
+\autoref{ax:symbol-generation} and \autoref{ax:number-definition}, generation-2
+contains the following numbers.