Successor Mapping is Slowly Progressing

Theorem
Let $V$ be a basic universe.

Let $s: V \to V$ denote the successor mapping on $V$:
 * $\forall x \in V: \map s x := x \cup \set x$

Then $s$ is a slowly progressing mapping.

Proof
From Successor Mapping is Progressing we have that $s$ is a progressing mapping.

Then we have that:
 * $\set x \notin x$

Thus:
 * $\card {x \cup \set x} = \card x + 1$

Hence the result.