%
\label{thm:saturation-e3}
The self-enrichment ladder of Category~$\tau$
has exactly four layers:
\[
    E_0 \;\subsetneq\; E_1 \;\subsetneq\;
    E_2 \;\subsetneq\; E_3,
    \qquad
    E_4 = E_3.
\]
The first three inclusions are strict
(each layer contains genuinely new structure).
The fourth step collapses:
$[E_3^{\op}, E_3] \subseteq E_3$,
so $E_4 = E_3$.
The enrichment ladder is complete.

The proof assembles three components.

\medskip
\textbf{Step~1: Strictness.}
Each inclusion $E_k \subsetneq E_{k+1}$
for $k = 0, 1, 2$ is strict.
This was proved in Chapters~4--6:
$E_1$ contains split-complex-enriched Hom~objects
absent from~$E_0$;
$E_2$ contains discrete rational carriers
absent from~$E_1$;
$E_3$ contains self-referential codes
absent from~$E_2$.

\medskip
\textbf{Step~2: Collapse.}
The functor category $[E_3^{\op}, E_3]$
contains no objects of genuinely new structural type
(Proposition~\ref{prop:functor-collapse}).
The obstruction is the Ontic Closure Theorem (I.T01):
the five generators produce exactly four orbit channels,
and any fifth channel factors through
the $\omega$-absorber.
The Ladder Saturation Theorem (I.T02)
translates this orbit closure
into a statement about canonical injectivity:
pentation --- the operation that would populate
a fifth level --- lacks the structural scaffold
for type distinction.

\medskip
\textbf{Step~3: Identity.}
Since $E_4 = [E_3^{\op}, E_3] \subseteq E_3$
and $E_3 \subseteq E_4$
(every object of $E_3$ is in particular
a constant presheaf in $[E_3^{\op}, E_3]$),
we have $E_4 = E_3$.