\section{Boolean routing and exact mask lists}\label{app:boolean}

The hexadecimal encoding assigns bit values
\[
 E=0x01,\ P=0x02,\ Q=0x04,\ P1=0x08,\ Q1=0x10,
 \ F=0x20,\ G=0x40.                                         \tag{B.1}
\]
For example, $0x1a=P\cup P1\cup Q1$.  The zero orbit is not a bit in this
encoding; it is adjoined after classifying the 127 nonempty base masks.

\subsection{The 77 cubic masks}

Evaluating the ten clauses of (7.1) by the eight assignments of $(E,P,Q)$
gives the counts
\[
\begin{array}{c|rrrrrrrr|c}
(E,P,Q)&000&100&010&110&001&101&011&111&\text{total}\\
\midrule
\#&6&5&9&9&9&9&15&15&77.
\end{array}                                                   \tag{B.2}
\]
Here each three-character label lists the truth values in the displayed
order $(E,P,Q)$.  For each fixed triple, the count is obtained by substituting
in (7.1) and enumerating the
sixteen assignments of $(P1,Q1,F,G)$; thus (B.2) is a sixteen-case elementary
truth-table count, not a computational theorem premise.

The following matrix expands every positive clause.  ``Optional'' means that
the indicated box may be either selected or absent under that clause; each
line is used only when it is selected.  The symbols $e,x,y,z$ denote the
values on $E,P1,Q1,G$.  A semicolon separates the base normalization from the
optional attachments.

\begingroup
\scriptsize
\begin{table}[p]
\centering
\begin{tabularx}{\textwidth}{@{}>{\raggedright\arraybackslash}p{1.35cm}
>{\raggedright\arraybackslash}p{2.25cm}
>{\raggedright\arraybackslash}p{2.55cm}
>{\raggedright\arraybackslash}p{1.45cm}Y@{}}
\toprule
Clause & Box handled & Cubic row used & Lemma part & Resolved generator or
restriction\\
\midrule
$T_1$ & base $F,G$ & $F+2G$ & (4) & $z=\eta+a\phi+b\psi$\\
$T_1$ & optional $P$ & $P+2F$ & (2) & $\alpha=\phi$\\
$T_1$ & optional $Q$ & $Q+2F$ & (2) & $\beta=\psi$\\
$T_1$ & optional $P1$ & $P1+F+G$ & (5) & $x=\eta+a\phi$\\
$T_1$ & optional $Q1$ & $Q1+F+G$ & (5) & $y=\eta+b\psi$\\
$T_1$ & optional $E$ & $E+F+G$ & (5) & $e=\eta$\\
\midrule
$T_2$ & base $P,P1,F$ & $P+2P1$; $P+2F$ & (1), (2) &
$x=\eta+a\alpha$ and $\phi=\alpha$\\
$T_2$ & optional $Q$ & $Q+2F$ & (2) & $\beta=\psi$\\
$T_2$ & optional $Q1$ & $P1+Q1+F$ & (5) & $y=\eta+b\psi$\\
$T_2$ & optional $G$ & $P1+F+G$ & (5) & $z=\eta+a\alpha+b\psi$\\
$T_2$ & optional $E$ & $E+P+P1$ & (5) & $e=\eta$\\
\midrule
$T_3$ & base $Q,Q1,F$ & $Q+2Q1$; $Q+2F$ & (1), (2) &
$y=\eta+b\beta$ and $\psi=\beta$\\
$T_3$ & optional $P$ & $P+2F$ & (2) & $\alpha=\phi$\\
$T_3$ & optional $P1$ & $P1+Q1+F$ & (5) & $x=\eta+a\phi$\\
$T_3$ & optional $G$ & $Q1+F+G$ & (5) & $z=\eta+a\phi+b\beta$\\
$T_3$ & optional $E$ & $E+Q+Q1$ & (5) & $e=\eta$\\
\midrule
$T_4$ & base $E,P,Q,P1$ & $P+2P1$; $E+P+P1$ & (1), (5) &
$x=e+a\alpha$\\
$T_4$ & optional $F$ & $P+Q+F$ & (2) &
$(\phi,\psi)=(\alpha,\beta)$\\
$T_4$ & optional $Q1$ & $Q+2Q1$; $E+Q+Q1$ & (1), (5) &
$y=e+b\beta$\\
$T_4$ & optional $G$ & $Q+P1+G$ & (5) & $z=e+a\alpha+b\beta$\\
\midrule
$T_5$ & base $E,P,Q,Q1$ & $Q+2Q1$; $E+Q+Q1$ & (1), (5) &
$y=e+b\beta$\\
$T_5$ & optional $F$ & $P+Q+F$ & (2) &
$(\phi,\psi)=(\alpha,\beta)$\\
$T_5$ & optional $P1$ & $P+2P1$; $E+P+P1$ & (1), (5) &
$x=e+a\alpha$\\
$T_5$ & optional $G$ & $P+Q1+G$ & (5) & $z=e+a\alpha+b\beta$\\
\bottomrule
\end{tabularx}
\caption{Complete cubic sufficiency routing, clauses $T_1$--$T_5$:
clause, optional box, mixed row, lemma and resolved restriction.}
\label{tab:cubic-routing}
\end{table}

\begin{table}[p]
\centering
\begin{tabularx}{\textwidth}{@{}>{\raggedright\arraybackslash}p{1.35cm}
>{\raggedright\arraybackslash}p{2.25cm}
>{\raggedright\arraybackslash}p{2.55cm}
>{\raggedright\arraybackslash}p{1.45cm}Y@{}}
\toprule
Clause & Box handled & Cubic row used & Lemma part & Resolved generator or
restriction\\
\midrule
$T_6$ & base $P$; $Q1,G$ absent & local & local & $P[a]=a\alpha$\\
$T_6$ & optional $P1$ & $P+2P1$ & (1) & $x=\eta+a\alpha$\\
$T_6$ & optional $F$ & $P+2F$ & (2) & $\phi=\alpha$\\
$T_6$ & optional $Q$ & $Q+2F$ if $F$ & (2) &
$\beta=\psi$ if linked; otherwise an independent $q$-character\\
$T_6$ & optional $E$ & $E+P+P1$ if $P1$ & (5) &
$e=\eta$ if linked; otherwise the sole parity character\\
\midrule
$T_7$ & base $Q$; $P1,G$ absent & local & local & $Q[b]=b\beta$\\
$T_7$ & optional $Q1$ & $Q+2Q1$ & (1) & $y=\eta+b\beta$\\
$T_7$ & optional $F$ & $Q+2F$ & (2) & $\psi=\beta$\\
$T_7$ & optional $P$ & $P+2F$ if $F$ & (2) &
$\alpha=\phi$ if linked; otherwise an independent $p$-character\\
$T_7$ & optional $E$ & $E+Q+Q1$ if $Q1$ & (5) &
$e=\eta$ if linked; otherwise the sole parity character\\
\midrule
$T_8$ & base $F$; $P1,Q1$ absent & local & local &
$F[a,b]=a\phi+b\psi$\\
$T_8$ & optional $P$ & $P+2F$ & (2) & $\alpha=\phi$\\
$T_8$ & optional $Q$ & $Q+2F$ & (2) & $\beta=\psi$\\
$T_8$ & optional $G$ & $F+2G$ & (4) & $z=\eta+a\phi+b\psi$\\
$T_8$ & optional $E$ & $E+F+G$ if $G$ & (5) &
$e=\eta$ if linked; otherwise the sole parity character\\
\midrule
$T_9$ & base $P,Q,G$; $E$ absent & $P+2G$; $Q+2G$ & (3) &
$z=\eta+a\alpha+b\beta$\\
$T_9$ & optional $F$ & $P+Q+F$ & (2) &
$(\phi,\psi)=(\alpha,\beta)$\\
$T_9$ & optional $P1$ & $Q+P1+G$ & (5) & $x=\eta+a\alpha$\\
$T_9$ & optional $Q1$ & $P+Q1+G$ & (5) & $y=\eta+b\beta$\\
\midrule
$T_{10}$ & base $P1,Q1,F$; $E$ absent & $P1+Q1+F$ & (5) &
$x=\eta+a\phi$ and $y=\eta+b\psi$\\
$T_{10}$ & optional $P$ & $P+2F$ & (2) & $\alpha=\phi$\\
$T_{10}$ & optional $Q$ & $Q+2F$ & (2) & $\beta=\psi$\\
$T_{10}$ & optional $G$ & $P1+F+G$ & (5) &
$z=\eta+a\phi+b\psi$\\
\bottomrule
\end{tabularx}
\caption{Complete cubic sufficiency routing, clauses $T_6$--$T_{10}$;
continuation of Table~\ref{tab:cubic-routing}.}
\label{tab:cubic-routing-continued}
\end{table}
\endgroup

In $T_6,T_7,T_8$, the parenthetical independent case is not a missing
relation.  If the relevant connector box is absent, the corresponding primary
coordinate, or $E$ as the only parity box, is a direct factor of the generated
CRT subgroup and may be assigned its own character parameter.  Thus all
optional bits are covered, including absence as well as presence, and every
row of Table~\ref{tab:cubic-routing} ends in the common character normal form
(7.0).

The complete positive list is
\begin{quote}\small\ttfamily
02 03 04 05 06 07 0a 0b 0e 0f 14 15 16 17 1f
20 21 22 23 24 25 26 27 2a 2b 2e 2f 34 35 36 37
38 3a 3b 3c 3d 3e 3f 46 4e 4f 56 57 5e 5f
60 61 62 63 64 65 66 67 68 69 6a 6b 6c 6d 6e 6f
70 71 72 73 74 75 76 77 78 79 7a 7b 7c 7d 7e 7f.
\end{quote}
Each entry is hexadecimal and carries the prefix $0x$ understood.

The necessity witnesses in Section~\ref{sec:cubic} may be summarized by
\[
\begin{array}{c|c|c}
\text{terms}&\text{target}&\text{defect}\\
\midrule
N_1,N_3&\Z&\text{odd }p\text{-function on }P1,G\\
N_2,N_4&\Z&\text{odd }q\text{-function on }Q1,G\\
N_5&\Z&\text{odd function on }P1\\
N_6&\Z&\text{odd function on }Q1\\
N_7&\Z&\text{odd function on a surviving parity-one box}\\
N_8,N_9&C_2&1\text{ on }E,\ 0\text{ elsewhere}\\
N_{10}&C_2&1\text{ on }P1,\ 0\text{ elsewhere}.
\end{array}                                                   \tag{B.3}
\]
Formula (7.4) is verified by the same fixed truth table: every assignment
failing (7.1) satisfies at least one $N_i$, and each $N_i$ falsifies (7.1).
The witnesses in (B.3) then prove mathematical nonclosure for the whole term,
not merely for the listed finite masks.

The Boolean verification can also be organized without hexadecimal masks.
Split first on whether $G$ is selected.  If $G$ occurs without $F$, failure to
attach its $p$- or $q$-dependence gives $N_1$ or $N_2$.  If $F$ occurs, the
$F,G$ positive clause applies.  Next suppose $G$ is absent.  A selected $P1$
without either $P$ or $F$ gives $N_3$, and the exchanged condition gives
$N_4$.  Once both odd lines are considered, absence of their remaining axis
and cross attachments gives $N_5,N_6$, while absence of every parity-zero
connector gives $N_7$.  The only residual failures are mismatched parity
constants, exactly $N_8,N_9,N_{10}$.  This case split proves that the ten
negative terms cover the complement of the ten positive routes.

For the integer defects, choose an odd function supported on one opposite
pair.  A triple that can see it either contains the opposite argument and
cancels, or requires one of the boxes excluded by its $N_i$ term.  Since the
ambient CRT subgroup is finite, the nonzero map to $\Z$ cannot be a character.
For the $C_2$ defects, every relation that survives the missing-box conditions
contains the marked parity box an even number of times.  An ambient parity
character would have to agree on all parity-one boxes, whereas the witness does
not.  These observations verify both relation-annihilation and
noncharacterhood for each family.

\subsection{Quartic routing lists}

After removing $E$ and the 77 cubic masks, the generic masks closing at four
are
\begin{quote}\small\ttfamily
08 09 0c 0d 10 11 12 13 19 1b 1d 1e 28 29 2c 2d
30 31 32 33 39 40 41 42 43 44 45 47 48 49 4a 4b
4c 4d 50 51 52 53 54 55 58 59 5a 5b 5c 5d.
\end{quote}
The three omitted failures are $18,1a,1c$.  When $p=3$, the 40 masks closing
at four are
\begin{quote}\small\ttfamily
09 0d 10 11 12 13 19 1b 1d 1e 28 29 2c 2d 30 31
32 33 39 41 42 43 45 47 49 4a 4b 4d 50 51 52 53
54 55 58 59 5a 5b 5c 5d.
\end{quote}
The nine omitted failures are $08,0c,18,1a,1c,40,44,48,4c$.

The bridge masks are $1a=P\cup P1\cup Q1$ and
$1c=Q\cup P1\cup Q1$.  In the generic branch the sole critical mask is
$18=P1\cup Q1$.  In the prime-three branch the critical masks are
\[
 08,\ 0c,\ 18,\ 40,\ 44,\ 48,\ 4c,
\]
which decode exactly to the seven supports in (9.2).  This recovers the base
counts $77+46+2+1+1=127$ and $77+40+2+7+1=127$, where the final one is $E$.

The list is a routing index, not a second proof.  Membership in the cubic list
routes to Theorem~\ref{thm:cubic}; membership in a quartic list routes to the
generic graph or prime-three integral case of Theorem~\ref{thm:quartic}; the
remaining nine masks route to Propositions~\ref{prop:generic-critical},
\ref{prop:three-critical}, and~\ref{prop:bridges}.  Each route is uniform in
the primes and carries a defect at the preceding order.