\section{The Quartic Skeleton}\label{sec:quartic}

We next work modulo the local degree-four forms (6.2).  Fourteen uniform
zero-sum connector types, listed in Appendix~\ref{app:connectors}, attach the
remaining local parameters.  The specialization
\[
 P1[a]+G[-a,b]+G[c,d]+G[-c,-b-d]=0                          \tag{8.1}
\]
and its $p\leftrightarrow q$ exchange are load-bearing.  This is the second
parameter choice of type X10, not a fifteenth connector: it supplies both the
slope edge and the parity/sign edge.

The connector types perform three overlapping jobs:
\[
\begin{array}{c|l}
\text{job}&\text{types}\\
\midrule
\text{axis-to-rectangle slope attachment}&X2,X3,X6\\
\text{rectangle and cross-box comparison}&X1,X4,X7,X9,X10B,X13,X14\\
\text{parity anchoring}&X5,X8,X10A,X10B,X11,X12.
\end{array}                                                   \tag{8.0}
\]
Some types necessarily occur in two rows.  In particular, merging X10A and
X10B merely because they have the same box counts would erase (8.1).
Appendix~\ref{app:connectors} supplies uniform nonvanishing parameter choices,
including the $p=3,q=5$ endpoints.

We repeatedly use the following elementary isolation principle.  If a
connector image is $r+bs$, where $s$ has odd prime order, two admissible
values of $b$ whose difference is a unit isolate $s$ and then $r$.  A
difference $2s$ also suffices by Bezout.  Thus primary components can be
separated symbolically, without sampling the primes.

\begin{lemma}[Primary isolation]\label{lem:isolation}
Let $s$ have order dividing an odd prime $r$.  If $x+b_1s=0$ and
$x+b_2s=0$, with $b_1-b_2$ a unit modulo $r$, then $s=x=0$.  The conclusion
also holds when $b_1-b_2=2$.
\end{lemma}

\begin{proof}
Subtraction gives $(b_1-b_2)s=0$.  Multiplication by a unit is an
automorphism of the $r$-primary group, so $s=0$ and substitution gives $x=0$.
Two is a unit because $r$ is odd.  The argument may be applied after primary
projection and therefore does not assume that a splitting was already chosen.
\end{proof}

\begin{theorem}[Quartic closure theorem]\label{thm:quartic}
Among the masks not closed by degree three, the exact degree-four failure sets
are
\[
\begin{array}{ll}
p,q\ge5:&0x18,\ 0x1a,\ 0x1c,\\
p=3:&0x08,\ 0x0c,\ 0x18,\ 0x1a,\ 0x1c,\
       0x40,\ 0x44,\ 0x48,\ 0x4c.
\end{array}                                                   \tag{8.2}
\]
Consequently 46 generic and 40 prime-three residual masks close first at
degree four.
\end{theorem}

\subsection{Generic branch}

Suppose $p,q\ge5$.  All local parameters split into their $2$-, $p$-, and
$q$-primary layers.  On the $p$-layer, the vertices $P,P1,F,G$ have the six
edges
\[
\begin{array}{lll}
P-P1:\mathrm{X2},&P-F:\mathrm{X6},&P-G:\mathrm{X3},\\
P1-F:\mathrm{X13},&P1-G:\mathrm{X10B},&F-G:\mathrm{X1}.
\end{array}                                                   \tag{8.3}
\]
This is a complete graph.  The exchanged connectors give the complete graph
on $Q,Q1,F,G$ for the $q$-layer.  Therefore each selected prime layer has
exactly one ambient coordinate.

An edge in (8.3) means equality of the named local slope parameters.  X1, for
example, gives twice the $F-G$ difference; Lemma~\ref{lem:isolation} removes
the factor two.  X10B gives the $P1-G$ difference, while X10A by itself only
anchors a parity combination.  Since the graph is complete, every nonempty
induced subgraph is connected, including supports containing only two of the
four slope vertices.

The parity vertices are $E,P1,Q1,G$.  Edges from $E$ to the other three and
from each odd line to $G$ are available whenever their endpoint boxes occur.
The remaining edge $P1-Q1$ comes from X7 if $F$ occurs, or from X14 if both
$P,Q$ occur.  Disconnection is possible only if $P1,Q1$ occur,
$E,G,F$ do not, and $P,Q$ are not both present.  In the bit order (6.1) these
are exactly $0x18,0x1a,0x1c$.  This proves the first line of (8.2).

These masks decode to $P1\cup Q1$, $P\cup P1\cup Q1$, and
$Q\cup P1\cup Q1$.  The graph already identifies the surviving class as a
diagonal parity discrepancy: it is the generic depth-six obstruction on the
first mask and the depth-five obstruction on the latter two.

\subsection{The prime-three branch}

Let $p=3$.  The $C_q$ vertices still form a complete graph.  The parity/three
part has six generators
\[
 (e,p,u,c,f,g)=(E,P,P1,Q1,F,G),
\]
with $2e=3p=2c=3f=0$ and $u,g$ free.  Their common images in $C_6$ are
$(3,4,1,3,4,1)$.  The connector types yield the following twelve rows:
\[
\begin{array}{c|c}
FG:f+2g & PP1:p+2u\\
PG:p+2g & PF:p+2f\\
P1F:2u-2f & P1G:u-g\\
EG:e+3g & EP1:e+3u\\
EQ1:e+c & Q1G:c+3g\\
P1Q1F:u+c-f & PP1Q1:p-u+c.
\end{array}                                                   \tag{8.4}
\]
Every row maps to zero in $C_6$, and the isolation principle separates an
attached $C_q$ component.

The target map is
\[
 e,p,u,c,f,g\longmapsto3,4,1,3,4,1\quad\text{in }C_6.        \tag{8.5}
\]
Each row of (8.4) visibly lies in its kernel.  The free elements $u,g$ are
integral lifts of a sign coordinate.  An anchor turns such a lift into one
element of order six; it is not legitimate to split it in advance as an
independent $C_2$ constant and $C_3$ slope.  The integral presentation is
exactly what keeps that distinction visible.

If neither free generator occurs, ordinary connectors give the ambient
target.  If $g$ occurs without $u$, one of $E,P,F,Q1$ anchors it to order six;
without an anchor the failures are $0x40,0x44$.  If both occur, (8.1) gives
$u=g$, and the same anchors close the common generator; the failures are
$0x48,0x4c$.  If $u$ occurs without $g$, one of $E,P,F$ anchors it.  Without
an anchor the failures are $0x08,0x0c,0x18,0x1c$.  When anchored, $Q1$ is
attached by the exchanged X12, X7, or X14 except for $0x1a$, where an extra
$C_2$ survives.  These are precisely the nine failures in (8.2).

The division is exhaustive because $u,g$ are the only free generators.  If
both are absent, all remaining generators have order two or three and the
ordinary connectors identify them with (8.5).  If exactly one is present,
the listed anchor set contains every row in (8.4) with a nonzero coefficient
on it.  If both occur, X10B first reduces them to one common generator and the
same anchor analysis applies.  The exceptional $Q1$ attachment on $0x1a$
leaves exactly the diagonal $C_2$ killed by the fifth-order bridge.

Theorem~\ref{thm:cubic} accounts for 77 masks.  Removing those, $E$, and the
failure sets from the 127 base masks leaves 46 generic and 40 prime-three
masks.  The local forms show they were not closed earlier, so their first
closing degree is four.  This proves Theorem~\ref{thm:quartic}.

\begin{remark}
The hexadecimal masks are only compact names.  Appendix~\ref{app:boolean}
decodes the bit order and lists the routed families.  The proof above is
uniform in $p,q$ and uses no sampled Smith normal form.
\end{remark}

The finite atlases at $n=30,42,70,110$ were used for discovery and regression
only.  The proof of Theorem~\ref{thm:quartic} consists of the parameterized
rows in Appendix~\ref{app:connectors}, Lemma~\ref{lem:isolation}, the complete
primary graphs, and the fixed integral presentation (8.4).  None of those
steps interpolates from a numerical prime sample.