\section{\texorpdfstring{Rematching when the $CD$ family is larger}{Rematching when the CD family is larger}}\label{sec:larger-CD}
Suppose $q\ge d$, and put $h=q-d$, $a=E_{AD}$, $c=E_{CC}$,
and $j=E_{AC}$. We use the surplus $h$ of $CD$ pairs to construct
a matching different from the reservoir matching of
Section~\ref{sec:reservoir-comparison}. Throughout this section we
assume the following normalized parameter conditions:
\[
 3d/10\le h\le5d/4,\quad4d/5\le r\le d,\quad E\le9d^2/50,
\]
\[
 y\le13d/20,\quad y\le3z/8,\quad z\le3d,\quad
 y\le w,\quad d\le2w.
\]
Section~\ref{sec:certificates} shows that all instances in this
orientation outside these conditions already satisfy the asserted
upper bound. No additional condition $h\le r$ or $h\le d$ is used.

\subsection{Which omitted pairs can replace a retained pair}
Index the initial retained $AD$ links as $A_iD_i$, write $C_i$ for
the original mate of $D_i$, and let $\sigma$ be the fixed-point-free
involution defined by the retained blue edges $C_iC_{\sigma(i)}$.
An omitted $C_0D_0$ inserts into the initial core at index $i$ if
$A_iD_0$ and $C_0C_{\sigma(i)}$ are both blue: replace $A_iD_i$ and
$C_iC_{\sigma(i)}$ by those two edges, and exchange the retained
original pair $C_iD_i$ for $C_0D_0$. This gives another neutral core
of the same size. Let $U$ be the actual
set of noninsertable pairs, of mass $u$. Let $a_U$ count missing
$AD$ edges from initial core $A$ to $D_U$, and $c_U$ missing $CC$
edges between $C_U$ and initial core $C$. Failure at every index gives
\[
 ru\le a_U+c_U,
 \qquad u\le\frac{a+c}{r}\le9d/40<h.
\]
The other $CD$ indices are \emph{eligible}; call their set $V$, of mass
$q-u$. Its internal missing-edge counts $a_V$ (all missing $AD$ with eligible
$D$) and $c_V$ (missing $CC$ with both indices eligible) satisfy
\[
 a_U+a_V\le a,\qquad c_U+c_V\le c.
\]
Inserting any eligible pair and applying the outward switches proves
that its $D$ has outward label 4 and its $C$ has outward labels at
least 3. The initial retained $B$ set does not change, so its good
$PQ$ orientation still gives $QD=4$ and $QC\ge3$ for eligible pairs.

\subsection{Transporting the initial core deficit}
Let $I,J$ be the initial and rebuilt retained $B$ sets, of masses
$r,s$. For a subset $W$ of either actual core, let $H(W)$ count
its label-3 contacts to old $X_3$ and set $h(W)=|W|z-H(W)$.
We have $h(W)\ge-o(1)$. Indeed each core $B$ set is a label-4
clique and can be re-paired arbitrarily. Distinct label-3 neighbors
$B,B'$ of the two endpoints of an old $PQ(3)$ would allow
$BB'(4),PQ(3)\to BP(3),B'Q(3)$, reducing primary weight.
Thus all such contacts use one endpoint of $PQ$, or there are at
most two contacts. This proves the claim for every subset of size
at least two; a singleton costs only $O(M)$ unnormalized.
Additivity therefore gives
\[
 h(J)=h(I)-h(I\setminus J)+h(J\setminus I)
 \ge h_3-z(d-s)-o(1).
\]
If $\overline R(s)$ is \eqref{eq:rawcore} without its $h_3$ term,
and $\kappa=1-y/z$, then
\begin{equation}\label{eq:transport}
 C_s\le\overline R(s)-\kappa h_3+(z-y)(d-s)+o(1).
\end{equation}
The two cores may have different matchings and different good-pair
orientations; their union need not be a core.

\subsection{Completing the target without a repair loss}
Put $T=h-u>0$, so there are $d+T$ eligible indices in normalized
units. On these indices define a graph $H$ by
\[
 ij\in E(H)\quad\Longleftrightarrow\quad
 C_iC_j\text{ is blue and }D_iD_j\text{ has label 3}.
\]
A blue $CC$ pair cannot have its $DD$ mate at label at most 2,
because that would reduce the primary weight of two old $CD(2)$
edges. Let a maximum matching of $H$ cover mass $k$.
If $k\ge T$, use $T$ of its endpoints as the target, pairing $C$
blue and $D$ at 3 along that matching.

If $k<T$, retain all its $k$ endpoints. Among its unmatched indices,
whose mass is $d+(T-k)$, a blue $CC$ matching covers at least $T-k$.
Otherwise the unmatched vertices of a maximum blue matching would
have mass at least $d$, forming a blue-independent set and requiring
at least $d^2/2-o(1)$ missing pairs, contrary to $c\le9d^2/50$.
Choose $T-k$ such endpoints and add them to the target. Their $DD$
mate edges have label 4, since unmatched vertices of $H$ are
independent in $H$. Thus the entire target has a blue $CC$ perfect
matching, with $D$ birth 3 on the first part and 4 on the rest.
No random pairing or target repair is necessary.

In the second case only the last $T-k$ target $D$ vertices can
create new label-3 $DD$ contacts. Each matched edge of $H$ has at
most $T-k+o(1)$ total $H$ contacts to this unmatched target: two
contacts at distinct unmatched endpoints would augment the maximum
matching by one edge. Hence the new $H$ contacts number at most
\[
 \frac{k(T-k)}2\le\frac{T^2}{8}.
\]
Any other eligible label-3 $DD$ contact is charged to a missing $CC$
mate edge. Label-2 $DD$ contacts also have missing $CC$ mates, and
these two label categories are disjoint. Their combined allowance
is therefore at most $c_V$, not twice $c_V$. Contacts to noneligible $D$
are bounded separately by $uT$.

\subsection{Counting the new matching and allocating missing edges}
Rebuild a separate neutral core on the eligible complement of the
target, which has mass exactly $d$. Let its size be $s$. Use its
neutral replacement, the $CC$ and $DD$ edges just constructed on the
target, and all unchanged original matching edges elsewhere. The
chosen sets are disjoint, so this is an $M$-edge matching; call its
normalized critical count $C_D$. Comparing
the target matching to this core gives
\begin{align}
 C_D\le{}&\overline R(s)-\kappa h_3+(z-y)(d-s)
       +T(q-2s)+uT-\tfrac38T^2 \notag\\
 &+j+(1+T/r)a_V+2c_V+e_{12}+e_{23}
       +(2T/r)h_3+o(1).\label{eq:full-raw}
\end{align}
Here target $D$--$C$ contacts cost at most $Tq$, demoted $C$ removes
$2Ts+T^2/2$ blue contacts, and the preceding matching argument
returns at most $T^2/8$ label-3 $DD$ contacts. The $X_1$ and good
$PQ$ half bounds give $e_{12},e_{23}$; bad pairs cost $2Th_3/r$.
The outside-$B$ allowance $(T/r)a_V$ uses missing $AD$ into the
initial retained $D$, all of which are eligible.

The missing-edge terms can be allocated more precisely. Write $a_J$ for missing $AD$
into complement $D$, and $c_J$ for missing $CC$ entirely inside the
complement. The raw $+a_V$ in \eqref{eq:full-raw} is actually supported
on target $D$ columns, disjoint from $a_J$. Both raw $c$ charges are
supported on pairs incident to target indices, disjoint from $c_J$.
The two $c$ charges may overlap each other, so retain coefficient 2
on their common resource.

Let $v=h/d$ and set
\[
 P_0=\frac{10}{9}(53/10-v),\qquad
 P(T)=P_0+\frac{20T}{9d}.
\]
The packing derivative, including \eqref{eq:transport}, is bounded by
\[
 (6d-x-z+3y)_++z-y+2T
 =\max(4d-h+2y,z-y)+2T\le\frac9{10}dP(T).
\]
The last inequality uses $y\le13d/20$, $z\le3d$, and $v\le5/4$.
The balanced case of \eqref{eq:split-packing} gives
$s^2\ge d^2-2(a_J+c_J)-o(1)$ and $s\ge4d/5-o(1)$.
Dividing $d^2-s^2$ by $d+s$ therefore gives
$d-s\le10(a_J+c_J)/(9d)+o(1)$.
Since $P(T)\ge9/2>2$, disjoint allocation proves
\[
 P(T)a_J+a_{\rm target}\le P(T)a_V,\qquad
 P(T)c_J+2c_{\rm incident}\le P(T)c_V.
\]
Thus after replacing $s$ by $d$, the two eligible defect coefficients
are only $P(T)+T/r$ and $P(T)$.

\subsection{Accounting for the noneligible pairs}
Normalize lengths by $d$ and defects by $d^2$. Write
$v=h/d$, $\rho=r/d$, $\sigma=u/d$, and $l=T/d=v-\sigma$.
The main target polynomial is
\[
 f(l)=l(2v-1)-\frac{11}{8}l^2,
 \qquad \frac{df}{d\sigma}=1-2v+\frac{11}{4}l\le31/16.
\]
The eligible coefficients $P_0+20l/9+l/\rho$ and $P_0+20l/9$
both decrease with $\sigma$. Their minimum is at least $P_0\ge9/2$.
Increasing $\sigma$ removes at least $\rho\,d\sigma$ eligible
missing-edge resource; its count cost is at least $18/5$ per unit.
All target $h_3$ coefficients also decrease. The resource envelope
therefore has derivative at most
\[
 31/16-18/5=-133/80<0.
\]
This holds at resource-allocation kinks by comparing optimal
allocations at two nearby values of $\sigma$. Consequently the
actual count is bounded by the formula evaluated at $u=0$, $T=h$.
This is a comparison of count bounds; the actual construction still
uses $T=h-u$.

Put $P_h=\frac{10}{9}(53/10+h/d)$. After discarding favorable cross
terms on the stated domain, the resulting inequality is
\begin{equation}\label{eq:full-target}
 \begin{split}
 C_D\le{}&B-b^2+d(x+z-3y)-3d^2-dh+\tfrac58h^2\\
 &+(P_h+h/r)a+P_hc+j+e_{12}+\tfrac38e_{23}
       +(2h/r-\tfrac58)h_3+o(1).
 \end{split}
\end{equation}
The certificate in Section~\ref{sec:certificates} uses these
coefficients. The allocation above is essential: the target and
complementary packing use disjoint parts of the same missing-edge
count.


\subsection{\texorpdfstring{Using $C$ vertices with several retained blue neighbors}{Using C vertices with several retained blue neighbors}}
To bound contacts with omitted $B$ vertices, call an original $CD$ index
rich if its $C$ vertex has at least four blue neighbors in the initial
retained $C$ set, and write $g$ for the rich mass. If an omitted
$A_0B_0$ has two blue $A_0D$ links to rich indices, then
$B_0D_*=4$ for every eligible $D_*$. To see this, insert $C_*D_*$
into the initial core. Choose a distinct rich neighbor of $A_0$ and
insert $A_0B_0$ through it, retaining the target slot. Four blue
neighbors suffice to avoid the removed vertex and the target's mate.
Re-pair the resulting $B$ clique through $B_0$ and the old mate $B_i$
of the target's $A_i$. The switch
$B_0B_i(4),A_iD_*(1)\to B_0D_*(\le3),A_iB_i(2)$
would reduce primary weight, a contradiction.
A row with only one rich neighbor has at most one exceptional target
contact. Charging zero- and one-neighbor rows to their missing rich
$AD$ links bounds the outside-$B$ term by $(T/g)a_{\rm out}+o(1)$.

If the nonrich mass is $b_C$, with $b_0$ of it initially retained,
then $c\ge rb_C-b_0^2/2-o(1)$. The bound $c\le9d^2/50$ first
implies $b_C<r+o(1)$, and then
\[
 b_C\le r-\sqrt{r^2-2c}+o(1)\le\frac{5c}{4r}+o(1),
 \qquad g\ge q-9d/32-o(1).
\]
The last estimate uses $r\ge4d/5$. The resource $a_{\rm out}$ uses
$A$ rows outside the initial core, whereas $a_U$ uses initial rows;
these resources are disjoint. Thus the preceding noneligible-pair
argument remains valid, with outside coefficient $T/g$.
More precisely, before substituting $r\ge4d/5$, the same argument
gives $g\ge q-9d^2/(40r)-o(1)$. On a box with
$v=h/d\in[v_0,v_1]$ and $r/d\ge\rho_0$, the $a$ coefficient in
\eqref{eq:full-target} can consequently be replaced by
\[
 A_D=\frac{10}{9}(53/10+v_1)
       +\frac{v_1}{1+v_1-9/(40\rho_0)},
 \qquad \rho_0\le r/d.
\]
Also retain the exact deficit term
$-(1-y/z)h_3+2hh_3/r$, rather than its constant envelope.
These coefficients and the unrounded $h_3$ term are the ones used
for this orientation in Section~\ref{sec:certificates}.