Membership Rank Inequality

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

Let $\map {\operatorname{rank} } S$ denote the rank of $S$.

Then:


 * $S \in T \implies \map {\operatorname{rank} } S < \map {\operatorname{rank} } T$

Proof
By Ordinal Equal to Rank:
 * $T \in \map V {\map {\operatorname{rank} } T + 1}$

By the definition of rank:
 * $T \subseteq \map V {\map {\operatorname{rank} } T}$

Since $S \in T$:
 * $S \in \map V {\map {\operatorname{rank} } T}$

By Ordinal is Subset of Rank of Small Class iff Not in Von Neumann Hierarchy:
 * $\map {\operatorname{rank} } T \nsubseteq \map {\operatorname{rank} } S$

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