=== Strict Ordering === Let $\tuple {x, y} \mathrel {R_0} \tuple {x, y}$. Then: :$\map \max {x, y} < \map \max {x, y} \lor \tuple {x, y} \mathrel {\operatorname {Le} } \tuple {x, y}$ {{explain|Where is the construct $\operatorname{Le}$ defined, and why can $\le$ not be used instead?}} Both lead to contradictions, so: :$\neg \tuple {x, y} \mathrel {R_0} \tuple {x, y}$ and $R_0$ is antireflexive. {{qed|lemma}} Let: :$(1): \quad \tuple {\alpha, \beta} \mathrel {R_0} \tuple {\gamma, \delta}$ and: :$(2): \quad \tuple {\gamma, \delta} \mathrel {R_0} \tuple {\epsilon, \zeta}$ There are two cases: :$\map \max {\alpha, \beta} < \map \max {\gamma, \delta}$ or: :$\map \max {\alpha, \beta} = \map \max {\gamma, \delta}$ {{begin-eqn}} {{eqn | l = \map \max {\alpha, \beta} < \map \max {\gamma, \delta}       | o = \leadsto       | r = \map \max {\alpha, \beta} < \map \max {\epsilon, \zeta}       | c = from $(2): \quad \map \max {\gamma, \delta} \le \map \max {\epsilon, \zeta}$ }} {{eqn | o = \leadsto       | r = \tuple {\alpha, \beta} \mathrel {R_0} \tuple {\epsilon, \zeta}       | c = {{Defof|Canonical Order}} }} {{end-eqn}} {{explain|Where does it follow that $\map \max {\gamma, \delta} \le \map \max {\epsilon, \zeta}$? All we have to work on is the relation $R_0$.}} {{begin-eqn}} {{eqn | l = \map \max {\alpha, \beta} = \map \max {\gamma, \delta}       | o = \leadsto       | r = \tuple {\alpha, \beta} \mathrel {\operatorname {Le} } \tuple {\gamma, \delta} }} {{eqn | o = \leadsto       | r = \map \max {\alpha, \beta} < \max \map {\epsilon, \zeta} \lor \tuple {\alpha, \beta} \mathrel {\operatorname {Le} } \tuple {\epsilon, \zeta}       | c = Transitivity of $\operatorname{Le}$ }} {{eqn | o = \leadsto       | r = \tuple {\alpha, \beta} \mathrel {R_0} \tuple {\epsilon, \zeta}       | c = {{Defof|Canonical Order}} }} {{end-eqn}} In either case: :$\tuple {\alpha, \beta} \mathrel {R_0} \tuple {\epsilon, \zeta}$ and $R_0$ is transitive. {{qed|lemma}} === Strict Total Ordering === Suppose: :$\neg \tuple {\alpha, \beta} \mathrel {R_0} \tuple {\gamma, \delta}$ and: :$\neg \tuple {\gamma, \delta} \mathrel {R_0} \tuple {\alpha, \beta}$ Then: :$\map \max {\alpha, \beta} \le \map \max {\gamma, \delta}$ and: :$\map \max {\gamma, \delta} \le \map \max {\alpha, \beta}$ So: :$\map \max {\alpha, \beta} = \map \max {\gamma, \delta}$ Therefore: :$\neg \tuple {\alpha, \beta} \mathrel {\operatorname {Le} } \tuple {\gamma, \delta}$ and: :$\neg \tuple {\gamma, \delta} \mathrel {\operatorname {Le} } \tuple {\alpha, \beta}$ From Lexicographic Order forms Well-Ordering on Ordered Pairs of Ordinals: :$\tuple {\alpha, \beta} = \tuple {\gamma, \delta}$ {{qed|lemma}} === Well-Ordering === Take any nonempty subset $A$ of $\paren {\On \times \On}$. We shall allow $A$ to be any class. This isn't strictly necessary, but it will not alter the proof. The $\max$ mapping sends each element in $A$ to an element of $\On$. Therefor the image of $\max$ has a minimal element, $N$. Take $B$ to be the class of all ordered pairs $\tuple {x, y}$, such that $\map \max {x, y} = N$. Let the $\operatorname {Le}$-minimal element of $B$ be denoted $C$. Then: :$\map \max C = N$ and $C$ is seen to be $\operatorname {Le}$-minimal. Therefore $C$ is the $R_0$-minimal element of $A$. {{qed}}