Ordinal Less than Successor Aleph

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

Then:


 * $y < \aleph_{x+1} \iff y < \aleph_x \lor y \sim \aleph_x$

Sufficient Condition
But $\left|{ y }\right|$ is a cardinal number, so it is either finite or an element of the infinite cardinal class.

It is a finite set iff $y \in \omega$ by Ordinal is Finite iff Natural Number.

If $y \in \omega$, then $y \in \aleph_x$ since $\aleph_x$ is an infinite set by Aleph is Infinite.

If it is an element of the infinite cardinal class, then $y = \aleph_z$ for some ordinal $z$ by the definition of the aleph mapping.

This means that:

Case 1: $z < x$
In the first case $z < x$:

Case 2: $z = x$
Therefore, $y < \aleph_x \lor y \sim \aleph_x$.

Necessary Condition
Suppose that $y < \aleph_x \lor y \sim \aleph_x$.

Case 1: $y < \aleph_x$
Suppose that $y < \aleph_x$.

It follows that $y < \aleph_{x+1}$ since $\aleph$ is strictly monotone by the definition of the aleph mapping.

Case 2: $y \sim \aleph_x$
Suppose that $y \sim \aleph_x$.

Then $\left|{ y }\right| = \aleph_x$ by Equivalent Sets have Equal Cardinal Numbers.

But $\aleph_x < \aleph_{x+1}$ since $\aleph$ is strictly monotone.

Therefore, $\left|{ y }\right| < \aleph_{x+1}$ and $y < \aleph_{x+1}$ by Ordinal in Aleph iff Cardinal in Aleph.