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, it follows that $T \in V\left({ \operatorname{rank} \left({ T }\right) + 1 }\right)$.

By the definition of rank, it follows that $T \subseteq V\left({ \operatorname{rank} \left({ T }\right) }\right)$.

Since $S \in T$, it follows that $S \in V\left({ \operatorname{rank} \left({ T }\right) }\right)$.

By Ordinal Subset of Rank, it follows that $\operatorname{rank} \left({ T }\right) \not \subseteq \operatorname{rank} \left({ S }\right)$.

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