Limit Ordinals Preserved Under Ordinal Multiplication

Theorem
Let $x$ and $y$ be ordinals.

Let $x$ be non-empty.

Let $y$ be a limit ordinal.

It follows that the ordinal product $\left({x \times y}\right)$ is a limit ordinal.

Proof
$y$ is a limit ordinal and thus is nonzero, by definition.

$x$ and $y$ are both nonzero.

So by Ordinals have No Zero Divisors:
 * $\left({ x \times y }\right) \ne 0$

So by definition of limit ordinal:
 * $\left({ x \times y }\right) \in K_{II} \lor \exists z \in \operatorname{On}: \left({ x \times y }\right) = z^+$

Suppose that $\left({ x \times y }\right) = z^+$ for some ordinal $z$.

It follows that:

But by Subset is Right Compatible with Ordinal Addition:
 * $\left({x \times w}\right) + 1 \subseteq \left({x \times w}\right) + x$

Therefore:

But by Successor in Limit Ordinal:
 * $w^+ \in y$

Therefore:
 * $z^+ \in \left({ x \times y }\right)$

contradicting the fact that $z^+ = \left({x \times y}\right)$.

Thus:
 * $z^+ \ne \left({x \times y}\right)$

and:
 * $\left({x \times y}\right) \in K_{II}$