\section{Rematching through the label-4 class}\label{sec:reservoir-comparison}
We next construct a matching by pairing retained $C$ vertices with
vertices of $X_4$. Their original $D$ mates are then available for a
separate matching using labels at most three. We call this the
\emph{reservoir comparison}; the auxiliary set of $D$ vertices can
later be enlarged.

Every retained $D$ has label 4 to every originally uncovered vertex.
Indeed, if its new mate is $A$, whose old mate is $B$, then
\[
 BB'(4),AD(1)\longrightarrow Du(\le3),AB(2)
\]
would reduce primary weight from 4 to at most 3. The matching size
is unchanged; $B'$ becomes uncovered in place of $u$. Similarly,
using $C'$ as the new mate of a retained $C$ and $D$ as the old mate
of $C'$, the switch
\[
 BB'(4),AD(1),CC'(1)\longrightarrow Cu(\le2),C'D(2),AB(2)
\]
proves that retained $C$ has outward labels at least 3.
The corresponding switches with an original top edge $TT'(4)$
give $DT=4$ and $CT\ge3$. Thus at most $e_{24}M^2$ of the
retained-$C$/top contacts fail to have label 3.


Choose a label-3 matching from retained $C$ to original top vertices,
using $tM$ vertices on each side, with $tM$ even after discarding at
most one edge. Thus we match mass $t$ of retained $C$ to top vertices at
label 3. The top matching supplies
$t\ge[r-e_{24}/(2w)]_+-o(1)$ when $r\le2w$.
Choose this matching first. On its selected old $D$ mates, take a
maximum matching of labels at most 3; suppose its covered mass is
$s$. The remaining $u_D=t-s$ vertices form a label-4 clique and
can be paired internally. Keep all original $AB$ edges and all
original $CD$ edges outside the selected target, and pair the
remaining top vertices internally. The latter is possible because
$X_4$ is a label-4 clique and an even number of its vertices remain.
All original label-1 and label-3 matching edges are unchanged. This
specifies an $M$-edge matching on exactly the original covered set;
write its normalized critical count as $C_T$.

A matched edge has at most $u_D+o(1)$ label-at-most-3 contacts from
its two endpoints to these unmatched vertices: two contacts with
distinct unmatched endpoints would give a three-edge augmenting path.
Only the unmatched $D$ have birth 4. Hence new critical label-3 $DD$
contacts number at most
\[
 u_Ds/2+u_D(q-t)
 =u_D(q-t/2)-u_D^2/2\le(q-t/2)^2/2.
\]
Every new label-2 $DD$ contact is charged to a nonblue mate $CC$
edge, since blue $CC$ together with $DD(2)$ would improve the two
old $CD(2)$ edges. Label-1 $DD$ contacts were already critical.
Thus the $DD$ contribution is at most $(q-t/2)^2/2+E_{CC}$.

The remaining direct contact count, with $a=E_{AD}$, $c=E_{CC}$,
$j=E_{AC}$, is
\begin{align*}
 C_T-C_0\le{}&t(\tfrac32q-r-z+2y+3h_3/r)
              -\tfrac78t^2+q^2/2\\
 &+2c+j+(2+t/r)a+e_{12}+e_{23}+e_{24}+e_{34}+o(1).
\end{align*}
Indeed $CD$ gains cost at most $tq+t(q-t)$; demoted top vertices
lose at least $tr-e_{24}$ contacts to retained $C$ and
$t(z-h_3/r)-e_{34}$ to good $P$. Bad old label-3 pairs cost at most
$2th_3/r$. The original half bounds supply the $X_1$ term and the
cross-deficit charges. For omitted $B$ contacts, two insertions into
the initial core force label 4 unless the omitted $A$ row misses
almost all initial retained $D$; those rows cost at most $(t/r)a$.
The separate $c$ charges for $C$ promotion and $DD(2)$ gains are
retained, since both may occur on the same mate pair.

The derivative of the displayed $t$ polynomial is at least
$-(5r/4+z)$ for $0\le t\le r\le q$. Substituting the top-matching
deficiency and \eqref{eq:original} gives
\begin{equation}\label{eq:reservoir-aug}
 \begin{split}
 C_T\le{}&B-b^2+\tfrac32qr-\tfrac{15}{8}r^2-rz+2ry+q^2/2
       +3h_3\\
 &-E_{AA}+2a+c-e_{13}-e_{14}
       +\frac{5r/4+z}{2w}e_{24}+o(1).
 \end{split}
\end{equation}
All selected $C,D$ retain their exact outward caps, including when
some $D$ edges have label less than 4.

\subsection{Missing edges wholly outside the retained set}
Here $q=m\le d$. The maximum-core choice includes the following
construction: first take a maximum blue matching on all
$C$, then match its old $D$ mates to $A$, deleting whole $CC$ edges
when an endpoint cannot be matched. The rectangular vertex-cover
bound loses at most $a/d$ $D$ vertices, and deletion costs at most
twice this many endpoints. If the full $CC$ matching covers $r_C$,
maximality of the chosen core therefore gives $r_C\le r+2a/d$.

Outside the actual core's $C$ vertices, a blue matching consequently
has at most $aM/d+o(M)$ edges. A graph with matching number $k$ has
at most $k|V|+O(|V|)$ edges, by the matching extremal bound. Its
missing $CC$ density outside the core is thus at least
\[
 c_{\rm out}\ge(q-r)^2/2-a(q-r)/d-o(1).
\]
Both positive $c$ charges in the reservoir comparison must touch a
selected $C$ index, so both exclude $c_{\rm out}$. Subtracting
$2c_{\rm out}$ from \eqref{eq:reservoir-aug} is valid. Since
$q-r\le q/5\le d/5$, the resulting count has missing-edge terms at
most $12a/5+c-E_{AA}$, and main polynomial
\begin{equation}\label{eq:AB-polynomial}
 f_T=\tfrac72qr-\tfrac{23}{8}r^2-rz+2ry-q^2/2.
\end{equation}
On $r+z\le6w$ and $r\le2w$, the positive $e_{24}$ coefficient
in \eqref{eq:reservoir-aug} is at most $13/4$.


\subsection{Averaging the choice of top vertices}
Select uniformly a subset $T_0$ of even mass $r$ from the original top
class of mass $2w$. Match retained $C$ to $T_0$ at label 3 as far as
possible; let the used mass be $t=r-\delta$, after harmless parity
rounding. The expected number of missing label-3 pairs in this balanced
bipartite graph is at most $re_{24}/(2w)$. A vertex cover for a matching
of size $tM$ forces at least $r\delta M^2$ missing pairs. Therefore
\[
 \mathbb E\delta\le e_{24}/(2w)+o(1).
\]
Now choose the maximum low-label matching on the corresponding $D$
vertices, as in the reservoir construction. Its count bounds hold for
every choice of $T_0$ and the subsequent $CT$ matching.

Every old $X_3$--top edge has label at least 3: a label at most 2
would, together with an arbitrary complementary edge, replace an old
label-3 and label-4 matching edge at strictly smaller primary cost.
Consequently the total label-3 density between these classes is
$2zw-e_{34}$. Uniform sampling, followed by deletion of at most
$\delta M$ unused top vertices, gives expected losses
\[
 L_{TC}\ge r^2-r\mathbb E\delta-\frac r{2w}e_{24}-o(1),
 \qquad
 L_{T3}\ge rz-2z\mathbb E\delta-\frac r{2w}e_{34}-o(1).
\]
This counts contacts with \emph{all} original label-3 pairs, so there
is no exceptional-pair allowance for these top-vertex losses. The remaining
$D$ promotion allowance is at most $2th_3/r\le2h_3$.

After these losses are separated, the $t$ polynomial is
$F(t)=t(3q/2+2y)-7t^2/8$. Since $F'(t)\ge-r/4$ for
$0\le t\le r\le q$, replacing $t$ by $r$ costs at most
$r\mathbb E\delta/4$. Combining all terms with the old count gives
the same main reservoir polynomial, but replaces its $3h_3$ by
$2h_3$. The $e_{24},e_{34}$ coefficients become respectively
\[
 \frac{9r/4+2z}{2w}-1\le\frac{21}{4},
 \qquad \frac r{2w}-1\le0.
\]
The first bound uses $r+z\le6w$ and $r\le2w$. Including the
outside-$CC$ correction, some actual matching therefore satisfies
\begin{equation}\label{eq:random-top}
 C_T\le B-b^2+f_T+2h_3+\frac{12}{5}a+c-E_{AA}
           -e_{13}-e_{14}+\frac{21}{4}e_{24}+o(1).
\end{equation}
The averaging selects among valid finite matchings. Its error terms
are uniform in the number of originally uncovered vertices.


\subsection{\texorpdfstring{Separating the two labels in the $DD$ count}{Separating the two labels in the DD count}}
We can bound together two types of newly critical $DD$ edges: those
of label 2, and those of label 3 whose other endpoint is outside the
target. When either is charged to its missing $CC$ mate, the same
$DD$ edge cannot contribute to both types. We therefore count their
shared set of missing $CC$ mates only once.
Let $U_{\rm out}$ count blue pairs between the initial retained $C$
set and its omitted complement. For an actual target $I$ of mass
$t\le r$, let $c_I$ count nonblue $CC$ pairs touching $I$, and let
$u$ be the unmatched target-$D$ mass in the low-label matching.

Every newly critical label-2 $DD$ pair has a nonblue $CC$ mate.
Charge these pairs together with outside-target label-3 $DD$ pairs
whose $CC$ mates are nonblue. Their combined density is at most
$c_I$, because a $DD$ pair cannot have both labels 2 and 3.
Unmatched-to-matched target contacts contribute at most $u(t-u)/2$.
The remaining outside-target contacts have blue $CC$ mates and
contribute at most $u(r-t)+U_{\rm out}$. Hence the whole $DD$ gain is
\[
 c_I+u(r-t/2)-u^2/2+U_{\rm out}
 \le c_I+(r-t/2)^2/2+U_{\rm out}+o(1).
\]
The corresponding bound used in \eqref{eq:random-top} was
$c_I+(q-t/2)^2/2$. For every $t\le r$,
\[
 \frac{(r-t/2)^2-(q-t/2)^2}{2}
 =-\frac{(q-r)(q+r-t)}2\le-\frac{q(q-r)}2.
\]
This inequality holds before averaging, so it incurs no additional
$e_{24}$ or sampling error. The separate $C$-promotion charge is
unchanged, as is the correction for pairs entirely outside the core.

The blue resource $U_{\rm out}$ is itself controlled. Extend the
retained $CC$ matching to a maximum one covering mass $R$, with
$r\le R\le r+2a/d+o(1)$, and let $c_R$ count missing pairs inside its
covered set. The Tutte--Berge calculation in
\eqref{eq:TB-core} gives
\[
 \operatorname{blue}(\text{unmatched},\text{covered})
 \le\frac{4(q-R)}R c_R+o(1),\qquad
 \left(5-\frac{4q}R\right)c_R\le c-\frac{q^2-R^2}2+o(1).
\]
Contacts with the newly covered vertices cost at most $r(R-r)$.
Consequently, for $r>4q/5$, put
\[
 K=\frac{q^2-r^2}2,\quad
 L=\frac{4(q-r)}{5r-4q},\quad
 \beta=\frac{2r}{d}+\frac{2a}{d^2}.
\]
Then
\begin{equation}\label{eq:blue-out-resource}
 U_{\rm out}\le\frac{2r}{d}a+L(c-K+\beta a)+o(1).
\end{equation}
Here $K(r)-K(R)\le2ra/d+2a^2/d^2\le\beta a$;
all expressions being multiplied by $L\ge0$ are nonnegative up to
$o(1)$. The same conclusion holds at $r=q$ with $L=0$.

Before rounding its coefficient to $12/5$, the random-top comparison
has $a$ coefficient $2+2(q-r)/d$. Combining it with
\eqref{eq:blue-out-resource} therefore gives
\begin{equation}\label{eq:disjoint-reservoir}
 \begin{split}
 C_T\le{}&B-b^2+f_T-\frac{q(q-r)}2-LK+2h_3\\
 &+\left(2+\frac{2q}{d}+L\beta\right)a+(1+L)c-E_{AA}
 -e_{13}-e_{14}+\frac{21}{4}e_{24}+o(1).
 \end{split}
\end{equation}
On a ratio box $h/q\ge v_0$, $r/q\ge\rho_0>4/5$, one may replace
$L$ by $4(1-\rho_0)/(5\rho_0-4)$ and
$\beta$ by $2/(1+v_0)+(9/25)/(1+v_0)^2$.
This uses $a\le9q^2/50$. Replacing $L$ by an upper bound is valid
for its entire nonnegative resource $c-K+\beta a$.


\subsection{Including the exceptional label-3 pairs}
Write $b_3$ for the mass of old label-3 pairs outside the good subset;
thus $b_3\le h_3/r+o(1)$. For a selected $CT$ target of mass $t$,
form the graph of labels at most 3 on the target $D$ vertices together
with all $2b_3$ endpoints of these exceptional pairs. Start with their
old label-3 matching edges and augment to a maximum matching.
Every exceptional endpoint stays covered, so the unmatched set lies
entirely in $D$. Write $u$ for its mass and pair it internally at
label 4. Exceptional vertices now have birth at most 3; all their
other count changes are nonpositive. The existing outward caps
therefore remain valid.

The low-label matching covers mass $t+2b_3-u$. The three-edge
augmentation obstruction bounds its contacts with unmatched $D$ by
$u(t+2b_3-u)/2+o(1)$. In particular this jointly bounds the new
internal $DD(3)$ contacts and the new label-3 contacts with exceptional
vertices. Adding the same outside-target and disjoint missing-$CC$
charges as before gives
\[
 c_I+u(r-t/2+b_3)-u^2/2+U_{\rm out}
 \le c_I+(r-t/2+b_3)^2/2+U_{\rm out}+o(1).
\]
There is no separate $2tb_3$ allowance.

Before top-demotion losses, the target-dependent polynomial is
\[
 g(t)=t(2q+2y)-t^2+(r-t/2+b_3)^2/2,
 \qquad g'(t)\ge-r/4-b_3/2.
\]
The same random sampling therefore adds at most $b_3e_{24}/(4w)$
to the former deficiency error. At $t=r$, the square is
$r^2/8+rb_3/2+b_3^2/2$. Consequently
\eqref{eq:disjoint-reservoir} strengthens to
\begin{equation}\label{eq:joint-bad-reservoir}
 \begin{split}
 C_T\le{}&B-b^2+f_T-\frac{q(q-r)}2-LK
       +\frac{h_3}{2}+\frac{h_3^2}{2r^2}\\
 &+\left(2+\frac{2q}{d}+L\beta\right)a+(1+L)c-E_{AA}
 -e_{13}-e_{14}+\frac{13}{2}e_{24}+o(1).
 \end{split}
\end{equation}
The last coefficient follows from $b_3\le z$ and $r+z\le6w$:
$(9r/4+2z+b_3/2)/(2w)-1\le13/2$.
This inequality holds throughout its stated parameter domain. The exact certificates in
Section~\ref{sec:certificates} use its complete defect terms.

On the additional low-$c$ domain, \eqref{eq:dense-tutte-out} improves
\eqref{eq:joint-bad-reservoir} by replacing $L$ with
$4(q-r)/(7r-4q)$. On a box with $h/q\ge v_0$ and
$r/q\in[\rho_0,\rho_1]$, we use
\[
 L\le\frac{4(1-\rho_0)}{7\rho_0-4},\qquad
 \beta\le\frac{2\rho_1}{1+v_0}+\frac{9/25}{(1+v_0)^2}.
\]
As before, the coefficient $L$ multiplies the complete nonnegative
resource $c-K+\beta a$; its negative $-LK$ term is retained.

For the final bound we use maximality of the joint $AD$ and $CC$
choice in the fixed core.
Lemma~\ref{lem:joint-packing-resource}, applied before substituting for
$U_{\rm out}$, gives the sharper joint-reservoir inequality
\begin{equation}\label{eq:joint-max-reservoir}
\begin{split}
 C_T\le{}&B-b^2+f_T-\frac{q(q-r)}2-LK
       +\frac{h_3}{2}+\frac{h_3^2}{2r^2}\\
 &+\left(2+\frac{2q-r}{d}+L\frac qd\right)a+(1+L)c-E_{AA}
 -e_{13}-e_{14}+\frac{13}{2}e_{24}-I_3+o(1),\\
 &L=\frac{4(q-r)}{5r-4q},\qquad K=\frac{q^2-r^2}{2}.
\end{split}
\end{equation}
The favorable quadratic packing term $-a^2/(2d^2)$ was discarded.
The same substitution is used for the deterministic profile reservoir.
It is valid throughout $r>4q/5$ and requires no additional dense-$C$
case distinction.