Lemma 4

Let E, F be two well-ordered sets and let f, g be two increasing mappings of E into F such that $f(E)$ is a segment of F and g is strictly increasing; then $f(x)\leq g(x)$ for all $x\in E$.

Suppose, on the contrary, that the set of elements $y\in E$ such that $f(y)>g(y)$ is not empty; then this set will have a least element $a$. If $x<a$, we have then $f(x)\leq g(x)<g(a)<f(a)$ since $g$ is strictly increasing. Since $f(E)$ is a segment of F, there exists $z\in E$ such that $g(a)=f(z)$; $f$ is increasing, so that $f(z)<f(a)$ implies $z<a$. Hence

which is absurd.