\section{The Cubic Skeleton}\label{sec:cubic}

Represent a nonempty union of the seven boxes (6.1) by its Boolean mask in
the order $(E,P,Q,P1,Q1,F,G)$.  The mask $E$ is already closed at degree two.
For every other mask, the following formula characterizes cubic closure:

A homomorphism $A_3(S)\to H$ is a function on $S$ annihilating every
zero-sum multiset of length at most three.  Closure means that each such
function is the restriction of one character of the generated CRT subgroup.
Writing its coordinates as $(\eta,\alpha,\beta)$, a character has
\[
 2\eta=0,\qquad p\alpha=0,\qquad q\beta=0,                  \tag{7.0}
\]
and the seven restrictions
\[
\begin{array}{c|c@{\qquad}c|c}
E&\eta&P[a]&a\alpha\\
Q[b]&b\beta&P1[a]&\eta+a\alpha\\
Q1[b]&\eta+b\beta&F[a,b]&a\alpha+b\beta\\
G[a,b]&\eta+a\alpha+b\beta&&
\end{array}
\]
The local cubic equations already force the displayed linear forms on
$P,Q,F$.  Initially the parity-one boxes may carry arbitrary odd functions
$x(a),y(b),z(a,b)$.  The problem is exactly to decide which selected-box
patterns force these odd functions into the common character form (7.0).
\begin{equation}
\begin{aligned}
T_3={}&(F\land G)
\lor(P\land P1\land F)
\lor(Q\land Q1\land F)\\
&\lor(E\land P\land Q\land P1)
\lor(E\land P\land Q\land Q1)\\
&\lor(P\land\neg Q1\land\neg G)
\lor(Q\land\neg P1\land\neg G)\\
&\lor(F\land\neg P1\land\neg Q1)
\lor(P\land Q\land G\land\neg E)\\
&\lor(P1\land Q1\land F\land\neg E).
\end{aligned}
\tag{7.1}
\end{equation}

\begin{theorem}[Cubic iff theorem]\label{thm:cubic}
A nonempty seven-box mask other than $E$ closes at degree three if and only if
it satisfies (7.1).  Exactly 77 masks do so, and all have first closing degree
three.
\end{theorem}

\subsection{Exhaustive triple types}

A zero-sum triple contains zero or two parity-one points.  Coordinate
feasibility gives exactly the following nineteen box-count patterns:
\[
\begin{gathered}
 PPP,\ QQQ,\ FFF,\ PFF,\ QFF,\ PQF,\ FGG,\ Q1FG,\ P1FG,\\
 P1Q1F,\ QGG,\ QQ1Q1,\ QP1G,\ PGG,\ PQ1G,\ PP1P1,\\
 EFG,\ EQQ1,\ EPP1.
\end{gathered}                                                \tag{7.2}
\]
Up to exchanging $p$ and $q$, every mixed equation is consequently of one
of the ten forms
\begin{equation}
\begin{aligned}
&P+2P1,\quad P+2F,\quad P+Q+F,\quad P+2G,\quad F+2G,\\
&E+P+P1,\quad E+F+G,\quad Q+P1+G,\quad P1+F+G,
  \quad P1+Q1+F.
\end{aligned}
\tag{7.3}
\end{equation}
There is no parity-compatible twentieth pattern.

Here is the coordinate check behind exhaustiveness.  The parity coordinate of
a triple is zero only when it contains zero or two points from
$P1,Q1,G$.  With no parity-one point, each prime coordinate must either occur
three times in one active box or be shared among $P,Q,F$, giving the first six
patterns in (7.2).  With two parity-one points, their nonzero prime
coordinates determine which parity-zero box can cancel them.  If the two
points lie in the same box this gives $FGG,QGG,PGG,QQ1Q1,PP1P1$ and their
mixed-axis versions; if they lie in different boxes it gives the remaining
attachments $Q1FG,P1FG,P1Q1F,QP1G,PQ1G,EFG,EQQ1,EPP1$.
Every case is realized by nonzero coordinates.  Three parity-one points are
impossible before the prime coordinates are considered.

\subsection{Sufficiency}

We use five propagation facts, all valid with values in an arbitrary abelian
group.

\begin{lemma}[Cubic propagation]\label{lem:cubic-propagation}
The equations in (7.3) have the following consequences.
\begin{enumerate}
\item If $P,P1$ occur, the relation $P+2P1$ forces
$x(a)=\eta+a\alpha$ with $2\eta=0$.  The exchanged assertion holds.
\item $P+2F$ identifies the $P$ and $F$ $p$-slopes; the exchanged and
two-axis versions identify the corresponding slopes.
\item If $P,G$ occur, then $z(a,b)=a\alpha+v(b)$ with $v$ odd.  If $Q,G$
also occur, then $z(a,b)=\eta+a\alpha+b\beta$, $2\eta=0$.
\item If $F,G$ occur, then $z(a,b)=\eta+a\phi+b\psi$ with $2\eta=0$.
\item The rows $E+P+P1,E+Q+Q1,E+F+G$ identify the parity constant
with $E$.  The rows $Q+P1+G$ and $P+Q1+G$ give, respectively,
$z(a,b)=x(a)+b\beta$ and $z(a,b)=a\alpha+y(b)$.  Once
$x(a)=\eta+a\phi$, the row $P1+F+G$ gives
$z(a,b)=\eta+a\phi+b\psi$, and similarly after exchanging the primes.
Finally $P1+Q1+F$ forces
$x(a)-a\phi=-(y(b)-b\psi)=\eta$ with $2\eta=0$.
\end{enumerate}
\end{lemma}

\begin{proof}
For (1), $P+2P1$ gives
$x(a)+x(c)=(a+c)\alpha$.  If $h(a)=x(a)-a\alpha$, taking $c=a$ gives
$2h(a)=0$, while varying $c$ makes $h$ constant.  This includes $p=3$.

For (2), $P+2F$ gives $2a(\phi-\alpha)=0$.  The difference is $p$-torsion
and $p$ is odd, hence $\phi=\alpha$.  The other cases are identical.

For (3), fixing one coordinate in $P+2G$ and subtracting two instances makes
the $p$-difference independent of the $q$-base point, producing
$a\alpha+v(b)$; opposite pairs make $v$ odd.  At $p=3$ the two apparent
sheets are identified by that simultaneous oddness.  Applying the exchanged
equation and comparing the two forms gives the affine expression and the
two-torsion constant.

For (4), subtract the $F$ slopes and set
$h(a,b)=z(a,b)-a\phi-b\psi$.  Relations $F+2G$ make $h$ constant on the
punctured-rectangle adjacency graph and pairs impose $2h=0$.  The graph is
connected for $p\ge5$.  For $p=3$, simultaneous oddness reduces opposite
first coordinates and a third nonzero $q$-coordinate completes the
comparison.

For (5), the three rows containing $E$ compare its value directly with the
constant term of the other parity-one box.  Substitution in $Q+P1+G$ and
$P+Q1+G$ gives the two displayed attachment formulas.  Substitution in
$P1+F+G$ gives the affine form on $G$.  In $P1+Q1+F$, separating the $p$
and $q$ variables makes $x(a)-a\phi$ and $-(y(b)-b\psi)$ one constant;
inverse pairs make it two-torsion.
\end{proof}

For each of the ten clauses in (7.1), Appendix~\ref{app:boolean} gives one
uniform routing scheme with five explicit fields: the optional selected box,
the mixed row used, the part of Lemma~\ref{lem:cubic-propagation} applied, and
the resulting normal form.  The table includes the cases in which no mixed
row is needed because a selected primary coordinate or the sole parity box is
an independent character factor.  Thus every bit not fixed by a clause is
accounted for.  Reading any row of that matrix shows that every selected box
carries the restriction of one common character
$(\eta,\alpha,\beta)$; hence the obstruction quotient vanishes.  This proves
the forward implication of Theorem~\ref{thm:cubic} without a finite mask
enumeration.

\subsection{Necessity}

The Boolean complement of (7.1), together with the special mask $E$, is the
union of the ten terms
\[
\begin{array}{ll}
N_1=G\neg F\neg P,&N_2=G\neg F\neg Q,\\
N_3=P1\neg F\neg P,&N_4=Q1\neg F\neg Q,\\
N_5=P1\neg G\neg P\neg Q1,&N_6=Q1\neg G\neg P1\neg Q,\\
N_7=\neg E\neg F\neg P\neg Q,&
N_8=EG\neg F\neg P1\neg Q1,\\
N_9=EQ1\neg G\neg P\neg Q,&
N_{10}=P1Q1\neg E\neg F\neg G.
\end{array}                                                   \tag{7.4}
\]
Conjunction signs are suppressed only in this display.

For $N_1,N_3$, take a nonzero odd function
$\sigma_p:C_p^\times\to\Z$ on the selected $P1$ and $G$ boxes.  Every
available dangerous triple either needs the missing $F$ or $P$, or pairs
opposite $p$-arguments; the function therefore kills every cubic row.
$N_2,N_4$ use the exchanged witness.  For $N_5,N_6$, use an odd function on
the single surviving parity-one axis.  For $N_7$, use any nonzero odd
function on a selected box among $P1,Q1,G$: three parity-one points cannot be
zero-sum.  For $N_8,N_9$, take target $C_2$, value one on $E$, and zero
elsewhere.  For $N_{10}$, take target $C_2$, value one on $P1$, and zero
elsewhere.  The missing-box hypotheses prevent every one-point attachment,
while two occurrences cancel in $C_2$.

One may choose the integer witness concretely: select a representative pair
$\{a,-a\}$, put $\sigma_p(a)=1$, $\sigma_p(-a)=-1$, and set all other values
to zero.  For $N_1,N_3$, the only surviving triple equations involving its
support pair opposite $p$-arguments, so the two values cancel.  Because the
generated CRT subgroup is finite, every homomorphism from it to $\Z$ is zero;
the nonzero witness cannot be an ambient character.  This avoids any appeal to
dimension counting.

For the $C_2$ witnesses, an ambient parity character gives the same value on
every selected parity-one box.  The $N_8,N_9,N_{10}$ assignments deliberately
give different values to two selected parity-one boxes, while the absent boxes
remove every cubic relation that would compare them.  Hence these witnesses
are also noncharacters for a structural reason, not merely because a finite
matrix has positive nullity.

The integer-valued witnesses cannot be characters of a finite group, and the
$C_2$ witnesses disagree with the unique ambient parity character on the
selected boxes.  Each is therefore a genuine cubic defect.  Appendix
\ref{app:boolean} expands the coverage table, proving the reverse implication
without a machine-generated Boolean premise.

Finally, degree two contains only inverse-pair rows inside each box.
Table~\ref{tab:local} shows that $E$ is the only nonempty base mask already
closed at two.  Direct Boolean reduction of the ten disjointly audited
clauses gives 77 positive masks; Appendix~\ref{app:boolean} includes an
elementary count.  Thus all 77 close first at three.

Adjoining the optional zero orbit doubles them, so 154 of the 255 supports
have depth three.