Comparability of well-orders

By Separation applied to $V\times W$, the collection $f=\{(v,w)\in V\times W:V_{<v}\cong W_{<w}\}$ is a set.

$f$ is a function: if $V_{<v}\cong W_{<w}$ and $V_{<v}\cong W_{<w^{\mathrm{\prime }}}$ with $w\mathrm{\ne }{w}^{\mathrm{\prime }}$, say $w{<}_W{w}^{\mathrm{\prime }}$, then $W_{<w}=(W_{<w^{\mathrm{\prime }}}{)}_{<w}$ is a proper initial segment of the well-order $W_{<w^{\mathrm{\prime }}}$ and $W_{<w^{\mathrm{\prime }}}\cong V_{<v}\cong W_{<w}$, contradicting [L4]; hence $w=w^{\mathrm{\prime }}$.

$f$ is injective: the same argument with the roles of $V$ and $W$ exchanged shows that $V_{<v}\cong W_{<w}\cong V_{<v^{\mathrm{\prime }}}$ forces $v=v^{\mathrm{\prime }}$.

Let $v{<}_V{v}^{\mathrm{\prime }}$ with $v^{\mathrm{\prime }}\in \mathrm{d}\mathrm{o}\mathrm{m}(f)$, and let $g:V_{<v^{\mathrm{\prime }}}\to W_{<f(v^{\mathrm{\prime }})}$ be an order isomorphism; then $g$ carries $(V_{<v^{\mathrm{\prime }}}{)}_{<v}=V_{<v}$ onto $(W_{<f(v^{\mathrm{\prime }})}{)}_{<g(v)}=W_{<g(v)}$, so $V_{<v}\cong W_{<g(v)}$, whence $v\in \mathrm{d}\mathrm{o}\mathrm{m}(f)$ with $f(v)=g(v)$, and $g(v)\in W_{<f(v^{\mathrm{\prime }})}$ gives $f(v){<}_{W}f(v^{\mathrm{\prime }})$.

Consequently $\mathrm{d}\mathrm{o}\mathrm{m}(f)$ is an initial segment of $V$ and $f$ is strictly increasing on it.

By the same argument with the roles of $V$ and $W$ exchanged, applied to the transpose of $f$, which is a function because $f$ is injective, $\mathrm{r}\mathrm{a}\mathrm{n}(f)$ is an initial segment of $W$.

$f$ is therefore a strictly increasing bijection from the initial segment $\mathrm{d}\mathrm{o}\mathrm{m}(f)$ of $V$ onto the initial segment $\mathrm{r}\mathrm{a}\mathrm{n}(f)$ of $W$, hence an order isomorphism between them.

$\mathrm{d}\mathrm{o}\mathrm{m}(f)$ and $\mathrm{r}\mathrm{a}\mathrm{n}(f)$ are not both proper: if $\mathrm{d}\mathrm{o}\mathrm{m}(f)=V_{<a}$ and $\mathrm{r}\mathrm{a}\mathrm{n}(f)=W_{<b}$ then step 5.1 gives $V_{<a}\cong W_{<b}$, so $(a,b)\in f$ and therefore $a\in \mathrm{d}\mathrm{o}\mathrm{m}(f)=V_{<a}$, that is $a{<}_{V}a$, which is impossible.

Since every initial segment is either the whole set or of the form $V_{<a}$ respectively $W_{<b}$, exactly three configurations remain, and they give (i) $V\cong W$, (ii) $V\cong W_{<b}$, and (iii) $V_{<a}\cong W$ respectively.

The three are mutually exclusive: (i) with (ii) gives $W\cong W_{<b}$; (i) with (iii) gives $V\cong V_{<a}$; and (ii) with (iii) give an isomorphism $\phi :V\to W_{<b}$ whose restriction carries $V_{<a}$ onto $W_{<\phi (a)}$, so $W\cong V_{<a}\cong W_{<\phi (a)}$; each conclusion contradicts [L4].

The witnesses are unique: $W_{<b}\cong V\cong W_{<b^{\mathrm{\prime }}}$ forces $b=b^{\mathrm{\prime }}$ by the argument of step 2.1, and $V_{<a}\cong W\cong V_{<a^{\mathrm{\prime }}}$ forces $a=a^{\mathrm{\prime }}$ by the argument of step 2.2.

Exactly one of (i), (ii), (iii) holds, with a unique witness in cases (ii) and (iii).