Membership Rank Inequality

Theorem
Let $S$ and $T$ be sets.

Let $\operatorname{rank} \left({ S }\right)$ denote the rank of $S$.

Then:


 * $S \in T \implies \operatorname{rank} \left({S}\right) < \operatorname{rank} \left({ T }\right)$

Proof
By Ordinal Equal to Rank:
 * $T \in V\left({ \operatorname{rank} \left({ T }\right) + 1 }\right)$

By the definition of rank:
 * $T \subseteq V\left({ \operatorname{rank} \left({ T }\right) }\right)$

Since $S \in T$:
 * $S \in V\left({ \operatorname{rank} \left({ T }\right) }\right)$

By Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy:
 * $\operatorname{rank} \left({ T }\right) \not \subseteq \operatorname{rank} \left({ S }\right)$

Therefore by Ordinal Membership is Trichotomy and Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
 * $\operatorname{rank} \left({ S }\right) < \operatorname{rank} \left({ T }\right)$