Cardinal Inequality implies Ordinal Inequality

Theorem
Let $T$ be a set.

Let $\left|{T}\right|$ denote the cardinal number of $T$.

Let $x$ be an ordinal.

Then:


 * $x < \left|{T}\right| \iff \left|{x}\right| < \left|{T}\right|$

Sufficient Condition
By Cardinal Number Less than Ordinal, it follows that $\left|{x}\right| \le x$.

So if $x < \left|{T}\right|$, then $\left|{x}\right| < \left|{T}\right|$.

Necessary Condition
Suppose $\left|{T}\right| \le x$.

By Subset of Ordinal implies Cardinal Inequality:
 * $\left|{\left({\left|{T}\right|}\right)}\right| \le \left|{x}\right|$.

Therefore, by Cardinal of Cardinal Equal to Cardinal:
 * $\left|{T}\right| \le \left|{x}\right|$

By the Rule of Transposition:
 * $\neg \left|{T}\right| \le \left|{x}\right| \implies \neg \left|{T}\right| \le x$

By Ordinal Membership is Trichotomy:
 * $x < \left|{T}\right| \iff \left|{x}\right| < \left|{T}\right|$