Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable

Theorem
Let $B$ be a well-orderable class.

Let $A$ be a class such that there exists an injection $f: A \to C$, where $C$ is a subclass of $B$.

Then $A$ is a well-orderable class.

Proof
Let $\RR$ be a well-ordering that can be established on $B$.

This can always be done, as $B$ is a well-orderable class.

Let $F$ be an injection that maps each element $x$ of $A$ to an element $\map F x$ of $B$.

Let $\preccurlyeq$ be the class of all ordered pairs $\tuple {x, y}$ of elements of $A$ such that $\tuple {\map F x, \map F y} \in \RR$.

Thus:
 * $(1): \quad x \preccurlyeq y \iff \tuple {\map F x, \map F y} \in \RR$

It will be demonstrated that $A$ is well-ordered under $\preccurlyeq$.

Let $\Img F$ denote the image of $F$.

By Subclass of Well-Ordered Class is Well-Ordered, $\Img F$ is well-ordered under $\RR$.

We have that $\RR$ is a well-ordering

Hence :
 * $\RR$ is reflexive, transitive and antisymmetric
 * Every pair of elements of $B$ is comparable under $\RR$

From reflexivity:
 * $\forall a \in A: \forall a \in A: \tuple {\map F a, \map F a} \in \RR$

and so from $(1)$:
 * $\forall a \in A: a \preccurlyeq a$

Thus $\preccurlyeq$ is reflexive on $A$.

Let $a, b, c, \in A$ such that $a \preccurlyeq b$ and $b \preccurlyeq c$.

Then from $(1)$:
 * $\tuple {\map F a, \map F b} \in \RR$

and:
 * $\tuple {\map F b, \map F c} \in \RR$

As $\RR$ is transitive:
 * $\tuple {\map F a, \map F c} \in \RR$

Hence from $(1)$:
 * $a \preccurlyeq c$

Thus $\preccurlyeq$ is transitive on $A$.

Let $a, b \in A$ such that $a \preccurlyeq b$ and $b \preccurlyeq a$.

Then from $(1)$:
 * $\tuple {\map F a, \map F b} \in \RR$

and:
 * $\tuple {\map F b, \map F a} \in \RR$

As $\RR$ is antisymmetric:
 * $\map F a = \map F b$

Because $F$ is an injection:
 * $a = b$

Thus $\preccurlyeq$ is antisymmetric on $A$.

Let $a, b \in A$ be arbitrary.

As every pair of elements of $B$ is comparable under $\RR$:
 * $\tuple {\map F a, \map F b} \in \RR$ or $\tuple {\map F b, \map F a} \in \RR$

Hence from $(1)$:
 * $a \preccurlyeq b$ or $b \preccurlyeq a$

Hence every pair of elements of $A$ is comparable under $\preccurlyeq$.

Thus $\preccurlyeq$ is a total ordering on $A$.

Let $C$ be an arbitrary non-empty subclass of $A$.

Let $C'$ be the class of all $\map F x$ such that $x \in C$.

We have that $C'$ is a subclass of $B$.

Then as $B$ is well-ordered by $\RR$, it follows that $C'$ has a smallest element.

This smallest element is $\map F b$ for some $b \in C$.

Let $x \in C$.

Then:
 * $\map F x \in C'$

and so:
 * $\tuple {\map F b, \map F x} \in \RR$

Hence by $(1)$:
 * $b \preccurlyeq x$

As $x$ is arbitrary:
 * $\forall x \in C: b \preccurlyeq x$

and so $b$ is the smallest element of $C$ $\preccurlyeq$.

It follows by definition that $\preccurlyeq$ is a well-ordering.

Thus $A$ is well-ordered by $\preccurlyeq$.

Hence the result.