Order Sum of Well-Founded Orderings is Well-Founded Ordering

Theorem
Let $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ be ordered sets.

Let $\preccurlyeq_1$ and $\preccurlyeq_2$ be well-founded.

Then the order sum $\struct {S_1, \preccurlyeq_1} \oplus \struct {S_2, \preccurlyeq_2}$ of $\struct {S_1, \preccurlyeq_1}$ and $\struct {S_2, \preccurlyeq_2}$ is also well-founded.

Proof
Let $\struct {S, \preccurlyeq} := \struct {S_1, \preccurlyeq_1} \oplus \struct {S_2, \preccurlyeq_2}$.

Let $T \subseteq S$ such that $T \ne \O$.

Let $T = T_1 \sqcup T_2$ where $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$.

Let $T_1 \ne \O$.

Then:
 * $\exists x \in T_1: \forall y \in T_1: y \preccurlyeq x \implies y = x$

That is, $T_1$ has a minimal element $x$ of $T_1$.

Thus we have:
 * $\forall y \in T_1: \tuple {y, 0} \preccurlyeq \tuple {x, 0} \implies \tuple {y, 0} = \tuple {x, 0}$

by definition of order sum.

Also by definition of order sum:
 * $\forall \tuple {z, 1} \in T: \tuple {x, 0} \preccurlyeq \tuple {z, 1}$

and so vacuously:
 * $\forall \tuple {z, 1} \in T: \tuple {z, 1} \preccurlyeq \tuple {x, 0} \implies \tuple {z, 1} = {x, 0}$

and so $\tuple {x, 0}$ is seen to be a minimal element of $T$.

So if $T_1 \ne \O$ it follows that $T$ has a minimal element.

Let $T_1 = \O$.

Then $T$ consists completely of elements of the form $\tuple {t, 1}$ where $t \in T_2$.

So, let $x \in T_2$ be a minimal element $T_2$.

Then:
 * $\forall y \in T_2: y \preccurlyeq x \implies y = x$

Thus we have:
 * $\forall y \in T_2: \tuple {y, 1} \preccurlyeq \tuple {x, 1} \implies \tuple {y, 1} = \tuple {x, 1}$

by definition of order sum.

So $\tuple {x, 1}$ is seen to be a minimal element of $T$.

So if $T_1 = \O$ it follows that $T$ has a minimal element.

In both cases it is seen that an arbitrary non-empty subset of $T$ has a minimal element.

Hence the result by definition of well-founded.