For the purposes of this proof, a relation $<_U$ on a subset $U$ of $S$ will be considered '''compatible''' with $\prec$ {{iff}}: :$\forall a, b \in U: a \prec b \implies a < b$ Let $M$ be the set of partial mappings $f$ from $S \times S$ to $\left\{ {0, 1}\right\}$ such that for all $x, y, z \in S$: :$(a): \quad \left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom} \left({f}\right) \implies \left({f \left({x, y}\right) = 1 \land f \left({y, z}\right) = 1 \implies f \left({x, z}\right) = 1}\right)$ :$(b): \quad \left({x, y}\right), \left({y, x}\right) \in \operatorname{Dom} \left({f}\right) \implies \left({f \left({x, y}\right) = 1 \iff f \left({y, x}\right) = 0}\right)$ :$(c): \quad \left({x,y}\right) \in \operatorname{Dom} \left({f}\right) \implies \left({x \prec y \implies f \left({x, y}\right) = 1}\right)$ {{explain|Is it worth extracting this definition and setting it up in its own definition page?}} We will first show that for some $\phi \in M$, the domain of $f$ is $S \times S$. We will then show that $\phi$ is the characteristic function of a strict total ordering on $S$ that extends $\prec$. We will apply the Cowen-Engeler Lemma to show that $\phi$ exists, so we must show that $M$ satisfies the premises of that theorem. Specifically, we must show that: :$(1): \quad \left\{{f \left({x, y}\right): f \in M}\right\}$ is finite for each $\left({x, y}\right) \in S \times S$ :$(2): \quad$ For each finite subset $F$ of $S \times S$, there exists an element $f \in M$ such that the domain of $f$ is $F$ :$(3): \quad  M$ has finite character. === $M$ satisfies premise $(1)$ === For each $g \in M$ and each $x \in \operatorname{Dom} \left({g}\right)$, $g \left({x}\right) \in \left\{{0, 1}\right\}$, so requirement $(1)$ is trivially satisfied. {{qed|lemma}} === $M$ satisfies premise $(2)$ === Let $F$ be a finite subset of $S \times S$. Let $S_F = \operatorname{Dom} \left({F}\right) \cup \operatorname{Img} \left({F}\right)$. Since $F$ is finite, so is $S_F$. {{explain|Trivial, I know, but express it formally: it rests on (a) $\operatorname{Dom} \left({F}\right)$ being finite (needs to be formally established), $\operatorname{Img} \left({F}\right)$ being finite (also needs to be formally established) and their union being finite.}} Thus by the Order-Extension Principle for Finite Sets there is a strict total ordering $<_F$ on $S_F$ which is '''compatible''' with $\prec$. Let $f_F$ be the restriction of the characteristic function of $<_F$ to $F$. Then since $F \subseteq S_F \times S_F$, $\operatorname{Dom} \left({f_F}\right) = F$ by the definition of restriction. Then $f_F \in M$: Let: : $\left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom} \left({f_F}\right) = F$ : $f_F \left({x, y}\right) = 1 \land f_F \left({y, z}\right) = 1$ Then $x <_F y \land y <_F z$. Thus since $<_F$ is transitive: :$x <_F z$ Thus: :$f_F \left({x, z}\right) = 1$ Similarly, let $\left({x, y}\right), \left({y, x}\right) \in \operatorname{Dom} \left({f_F}\right) = F$. Then since $<_F$ is asymmetric and connected: : $x <_F y \iff \neg (y <_F x)$ Thus: :$f_F \left({x, y}\right) = 1 \iff f_F \left({y, x}\right) = 0$. If $\left({x, y}\right) \in \operatorname{Dom}(f_F)$, then since $<_F$ is '''compatible''' with $\prec$: :$x \prec y \implies f \left({x, y}\right) = 1$ So $f_F \in M$. As such an $f_F$ exists for each such $F$, $M$ satisfies premise $(2)$. {{qed|lemma}} === $M$ satisfies premise $(3)$ === We must show that $M$ has finite character. That is, a mapping $f$ from some subset of $S \times S$ to $\left\{{0, 1}\right\}$ is an element of $M$ {{iff}}: :its restriction to each finite subset of $\operatorname{Dom} \left({f}\right)$ is in $M$. Let $f$ be a partial mapping from $S \times S$ to $\left\{{0, 1}\right\}$. Suppose that $f \in M$. Let $F$ be a finite subset of $\operatorname{Dom} \left({f}\right)$. Let $x, y, z \in S$. Let $\left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom}(f \restriction F)$. Then: :$\left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom} \left({f}\right)$ Thus since $f \in M$: :$f \left({x, y}\right) = 1 \land f \left({y, z}\right) = 1 \implies f \left({x, z}\right) = 1$ But by the definition of restriction: :$f \restriction F \left({x, y}\right) = f \left({x, y}\right)$ :$f \restriction F \left({y, z}\right) = f \left({y, z}\right)$ :$f \restriction F \left({x, z}\right) = f \left({x, z}\right)$ Thus: :$f \restriction F \left({x, y}\right) = 1 \land f \restriction F \left({y, z}\right) = 1 \implies f \restriction F \left({x, z}\right) = 1$ So $f \restriction F$ satisfies $(a)$. A similar argument shows that $f \restriction F$ satisfies $(b)$ and $(c)$, so $f \restriction F \in M$. Suppose instead that for every finite subset $F$ of $\operatorname{Dom} \left({f}\right)$, $f \restriction F \in M$. Let $x, y \in S$. Suppose that $\left({x, y}\right), \left({y, x}\right) \in \operatorname{Dom} \left({f}\right)$. Then $K = \left\{ {\left({x, y}\right), \left({y, x}\right)}\right\}$ is a finite subset of $\operatorname{Dom} \left({f}\right)$. Thus $f \restriction K \in M$. So: :$f \restriction K \left({x, y}\right) = 1 \iff f \restriction K \left({y, x}\right) = 0$ Since $f \restriction K \left({x, y}\right) = f \left({x, y}\right)$ and $f \restriction K \left({y, x}\right) = f \left({y, x}\right)$: :$f \left({x, y}\right) = 1 \iff f \left({y, x}\right) = 0$ Thus $f$ satisfies $(b)$ for all pairs. A similar argument shows that $f$ satisfies $(a)$ for all pairs. Thus $f \in M$. {{qed|lemma}} Since $M$ satisfies $(1)$, $(2)$, and $(3)$, by the Cowen-Engeler Lemma there is a $\phi \in M$ whose domain is $S \times S$. === $\phi$ is Characteristic Function which is Compatible with $\prec$ === Since $\operatorname{Dom} \left({\phi}\right) = S \times S$ and $\phi \in M$, for all $x, y, z \in S$: :$(a): \quad \phi \left({x, y}\right) = 1 \land \phi \left({y, z}\right) = 1 \implies \phi \left({x, z}\right) = 1$ :$(b): \quad \phi \left({x, y}\right) = 1 \iff \phi \left({y, x}\right) = 0$ :$(c): \quad x \prec y \implies \phi \left({x, y}\right) = 1$ By an argument similar to that given for premise $(2)$, above, $\phi$ is the characteristic function of a strict total ordering on $S$ '''compatible''' with $\prec$. {{qed}}