\subsection{Edge counts for the two auxiliary comparisons}\label{sec:auxiliary-counts}
We now construct and count the two matchings used in
Section~\ref{sec:auxiliary-matchings}. The first extends a blue
matching from retained $C$ to good $P$ vertices; the second replaces
an $AD$ edge together with a good $PQ$ edge. Both use the label
restriction \eqref{eq:Q-no-blue-X2}.

For the $CP$ construction, retain all actual $AD$ and $BB$ core edges.
Start with its blue $CC$ matching in the blue graph on retained $C$
and all good $P$, then augment to a maximum matching. Every retained
$C$ remains covered. If its additionally covered $P$ mass is $v$,
the blue matching covers $r+v$ vertices. Replace the old $CC$ edges
and these $vM$ original $PQ$ edges by this blue matching and an
internal label-4 matching on their $Q$ mates. The finite integer
$vM$ is even, since $rM$ and $rM+vM$ are even. All unselected $PQ$
edges remain. Edge counts agree:
$(r+v)M/2+vM/2=rM/2+vM$. Primary weight also agrees, since the new
$QQ$ edges have total weight $2vM$, the same as the removed label-3
edges. Thus this is an actual primary-neutral matching with the
same covered vertices; in particular it has no outward critical edges.

In the $CP$ construction only selected $P$ births decrease from 3
to 1 and their $Q$ mates increase from 3 to 4. Retained $A,C,D$
give net change at most $-2rv$ plus their portion of $e_{23}$:
the three $P$ losses total $3rv$ up to this deficit, and only the
$Q$--$C$ label-3 gain, at most $rv$, can offset them.
An insertable omitted $AB$ pair has $QB=4$ and $PB\ge3$ and costs
only its cross deficit. Noninsertable rows have mass at most $a/r$,
so they cost at most $2va/r$. An eligible omitted $CD$ pair costs
at most $v$; noneligible pairs have mass at most $(a+c)/r$ and
cost at most one further contact each. These terms total
$v(q-r)+v(a+c)/r$.

Inside $X_3$, selected $Q$ can gain at most $vZ$ contacts with good
$P$ and $2v(z-Z)$ with exceptional endpoints. Selected $P$ loses at
least $v^2/2-I_3$ internal contacts, using $E_{PP}\le I_3$.
Thus this part is at most
$vz-v^2/2+I_3+v(z-Z)$.
The $X_1$ contribution is at most $e_{13}$ by comparing its old
half bound with the at most two contacts from each selected $Q$.
All uses of $e_{23}$ here are on disjoint old pair rectangles.
Their sum gives
\begin{equation}\label{eq:CPcomparison}
 C_{CP}-C_r\le v(z+q-3r)-v^2/2+e_{13}+e_{23}+I_3
       +(v/r)h_3+(3v/r)a+(v/r)c+o(1).
\end{equation}

For $AD/PQ$, let $\pi=r/Z$ and first perform the formal birth change
on every assigned group, before repairing incompatible assignments.
Only retained $A,D$ promote from 1 to 2; selected $P,Q$ demote
from 3 to 2. Promotions cost at most $4ry$ against $X_1$ and
$5r^2/2$ inside the retained core: the latter consists of $AA,AC,AD$.
The expected $PA(2)$ loss is $\pi W$. The $PC(2),PD(2)$ losses
are at least $2r^2-\pi U_{PC}$, except for their cross deficits and
blue $PD$ edges. A blue $PD$ edge forces its current $AD$ partner
$A$ to have $AQ>2$, since otherwise $AD(1),PQ(3)$ could be replaced
at smaller primary cost. These failures are charged to the same
$AB$--$PQ$ cross deficit used below.

The $QA(2)$ loss is at least $rd-a$ up to its cross deficit: the
preliminary exclusion rules out blue $QA$, every insertable row
has $QB=4,PB\ge3$, and noninsertable rows have mass at most $a/r$.
The combined two uses of the $AB$ cross deficit and the one use
of the $CD$ cross deficit cost at most $2\pi e_{23}$.
Selected $PP(2)$ loses at least
$r^2/2-\pi^2(U_{PP}+I_3)+o(1)$ in expectation. All other contact
changes are nonpositive. Summation gives the formal bound
\[
 r(4y-d)-\pi W+\pi U_{PC}+\pi^2(U_{PP}+I_3)
       +a+2\pi e_{23}+o(1).
\]

The expected failed assignment mass is at most $e_{23}/Z$.
Restoring a failed group decreases its $A,D$ births and raises only
its $P,Q$ births from 2 to 3. The latter creates only label-2
contacts. The old per-pair half bounds give at most $2x+2y$
contacts to $X_1\cup X_2$, and good-$Q$ low-label exclusion leaves
at most $r$ further contacts to selected $P$ endpoints.
Thus repair costs at most $(2x+2y+r)e_{23}/Z$, proving the stated
coefficient $(2x+2y+3r)/Z$. Every repaired assignment is an actual
primary-neutral matching. The formal assignment is also outward-safe:
$A,D$ never exceed their original birth 2, and $P,Q$ only demote.
Consequently no error depends on the number of uncovered vertices.