# Two low components force six high colors in outerthickness-two graphs

## 1. Result, context, and prior work

A graph has outerthickness at most two if its edges are the union of
two outerplanar graphs; the two drawings may use different cyclic orders.
Adjoining two adjacent universal vertices, one as the apex of each layer,
gives a biplanar graph and increases the chromatic number by two. Thus an
eight-chromatic outerthickness-two graph would give a ten-chromatic graph
for the Earth–Moon problem. The known theorem that every graph in this
class with independence number at most two is six-colorable [1] provides
the hereditary obstruction used here.

**Main theorem (Theorem 8).** Suppose an outerthickness-two graph is
partitioned into an exterior $H$ and two nonempty, connected, anticomplete
Gallai graphs $T_1,T_2$. If each vertex of $T_1\cup T_2$ has total degree
seven and $H$ is five-colorable, the whole graph is seven-colorable.
There is no bound on the order of any of the three parts.

Consequently, in a vertex-critical eight-chromatic graph in this class
with exactly two components induced by its degree-seven vertices, the
remaining high graph needs at least six colors (Corollary 9). This is a
necessary occurrence condition, not an unrestricted coloring theorem.

The proof keeps one common exterior throughout. Simultaneous independent
splits of its color classes produce failure events that are products of
pair-parity relations. A refinement lemma then forces one component to
have signed coordinates for its actual boundary columns. If both have
that representation, incompatible nonedge mergers already force a
seven-chromatic subgraph with independence number at most two. Otherwise
fresh pair and singleton recolorings localize all defects of the second
component to a single pair or a star center. An independent-triple repair
forces the star's rays to be pairwise adjacent. The two remaining cases
each produce an actual alpha-two seven-chromatic subgraph on at most
thirteen vertices, contradicting [1].

### 1.1. Prior machinery and the distinction

The block-incidence and signed-palette framework is prior work [2,
Section 15.3]. Its later component-removal theorems [2, Sections 24–29]
delete whole hub terms from formulas for a failed connected low extension.
They also use the impossibility of a finite signed-block incidence forest
of minimum degree at least two. We reuse that argument in Lemma 7.

Hajos_weaver [3, Theorems 8.1–8.5] already studies cooperating injective
failures, star support, actual singleton columns, matching mergers and
independent-triple repairs. Those results treat eight or nine high
vertices, nine-color biplanar degree budgets, and specific component
profiles. These mechanisms are credited, not claimed as new. Our high
graph has arbitrary order and is assumed only to be five-colorable.
Fresh colors recover the actual columns without an injective high
coloring, and the accepted alpha-two theorem closes both the single-pair
and multi-ray cases without an order or low-block-size bound.

Gallai's low-graph theorem [4] and the classical bad degree-list
characterization [5] are prerequisites. Earlier local clique reductions
in outerthickness-two graphs [6] concern minimum order within that graph
class. The present extension theorem needs only its displayed Gallai
and degree hypotheses; its critical-graph consequence uses vertex
criticality. The new conclusion is the full two-component extension and
its occurrence restriction. The parity, refinement and defect lemmas
are proved here to coordinate the common exterior and make all quantifiers
explicit.

Graphs are finite and simple. A vertex-critical $(k+1)$-chromatic graph
loses chromatic number after every vertex deletion. Its degree-$k$
vertices induce a Gallai forest: take a spanning edge-minimal critical
subgraph, whose edges at the original degree-$k$ vertices cannot be
deleted, apply Gallai's theorem, and then take the induced subgraph on
those original vertices. For an eight-chromatic graph, these are the
**low vertices**; every other vertex has degree at least eight and is
**high**. Neither the theorem nor its proof asserts that the high graph
of an arbitrary candidate is five-colorable.

## 2. Actual boundary constraints and joint recoloring

### 2.1. Degree lists and a precise actual-boundary representation

A **Gallai tree** is a connected graph whose blocks are cliques or odd
cycles. For a block $B$ let $d_B=|B|-1$ when it is a clique and $d_B=2$
when it is an odd cycle. A singleton has one block of degree zero.
Let $M$ be the vertex-by-block incidence matrix. Leaf elimination shows
that $M$ has full column rank over $\mathbb Q$: a private vertex of a leaf
block determines that block's coefficient, which is subtracted at the
cutvertex before continuing. This also proves integrality of coordinates
of integral vectors in its column space.

The classical bad-list theorem says that a degree-list assignment on a
Gallai tree fails exactly when each block has a palette of size $d_B$,
incident palettes are disjoint, and the list at a vertex is their union
[5; 2, Section 15.3]. These palettes are unique by the same elimination.
For completeness, a leaf block's private lists must be one common palette
contained in its cut list; subtract it and continue. The whole-block base
case is Hall's theorem for a clique and the two-color obstruction of an odd
cycle. In any connected graph, lists of at least the vertex degrees with
strict surplus at one vertex are colorable: reverse a spanning-tree order
rooted at a surplus vertex and color the root last.

Let $T$ have total degrees exactly $k$ after attachment to an exterior $H$.
Write $a_x\in\{0,1\}^{V(T)}$ for the actual contact column of $x\in H$.
A failed exterior coloring with an unused color supplies a unique Boolean
vector $\lambda$ satisfying $M\lambda=\mathbf1$. The blocks with
$\lambda_B=1$ partition $V(T)$; call them **primary**.

We say $T$ has a **signed actual-boundary representation** if this Boolean
$\lambda$ exists and every actual column has coordinates $a_x=M\mu_x$
satisfying
$$
\mu_x(P)\in\{0,1\}\quad(P\text{ primary}),\qquad
\mu_x(B)\in\{0,-1\}\quad(B\text{ nonprimary}).                 \tag{2.1}
$$
This is a condition on the given graph, not a chosen coloring. Define
$$
S_P=\{x:\mu_x(P)=1\},\qquad R_B=\{x:\mu_x(B)=-1\}.
$$
Summing actual degrees gives
$$
M\sum_x\mu_x=k\mathbf1-Md=M(k\lambda-d).
$$
Full column rank therefore gives the per-block identities
$$
|S_P|=k-d_P,\qquad |R_B|=d_B.                              \tag{2.2}
$$
Every vertex $v$ belongs to a unique primary $P(v)$. The actual $0/1$
row equation gives
$$
N_H(v)=S_{P(v)}\setminus
 \mathop{\dot\bigcup}_{B\ni v,\ B\text{ nonprimary}}R_B.       \tag{2.3}
$$
Indeed a removed vertex must belong to $S_{P(v)}$, and no two removed sets
at $v$ can overlap, since either violation would make the row negative.

**Lemma 1 (fixed actual conflict sets).** Under (2.1), a proper
$k$-coloring of $H$ fails on $T$ if and only if every $S_P$ is rainbow.

If failure occurs, every actual neighborhood is rainbow, so the unique
availability coordinates of a color class $C$ are
$\lambda-\sum_{x\in C}\mu_x$. Repetition in $S_P$ would make its $P$
coordinate $1-|C\cap S_P|<0$, impossible for a bad-list palette. Conversely,
if every $S_P$ is rainbow, assign palette $[k]\setminus\phi(S_P)$ to a
primary and $\phi(R_B)$ to a nonprimary. Equations (2.2)–(2.3) give the
correct sizes, disjointness at vertices, and the actual residual lists.
The bad-list theorem proves failure. This recovers the signed-incidence
criterion of [2]; no injective coloring of $H$ is assumed.

### 2.2. Simultaneous binary refinements

Fix a proper exterior coloring, choose some of its independent color
classes $C_1,\ldots,C_r$, and give each a different unused color. Every
vertex of $C_j$ independently chooses its old or new color, with bit $s_x$.
The choices on distinct classes use disjoint color pairs and remain proper.

**Lemma 2 (exact split relation).** For a Gallai component of total
degrees at most $k$, its failing choices form the empty set or a product
of signed partition relations, one per selected class. Each relation is
defined by equations $s_x\mathbin\oplus s_y=\varepsilon$ within that class.
Thus a nonempty failure event has size $2^{-\rho}$ of the full cube for
some integer $\rho\geq0$; a proper rank-one event is one pair equation.

Failure requires exact degree lists. A vertex of total degree below $k$,
three neighbors in one selected class, or repeated colors among unaffected
neighbors makes every split extend. Otherwise each unaffected color's
availability vector must have Boolean block coordinates. For a selected
pair, let $m^j_v=|N_H(v)\cap C_j|$. The sum $t^j$ of its two palette
coordinates is uniquely determined by
$$
Mt^j=2\mathbf1-m^j.                                      \tag{2.4}
$$
Its entries must be $0,1,2$, and all pair sums plus unaffected coordinates
must equal the block demand vector $d$. If these conditions fail, there
is no bad split.

For one pair introduce a bit at every block with $t^j_B=1$. At a vertex
with two class contacts, join their two bits by an edge. With one class
contact, join it to the unique incident block with $t^j_B=1$. With no
class contacts, either one incident block has value two, or two have
value one; in the latter case join those two block bits. Equation (2.4)
makes these cases exhaustive. Every edge demands opposite bits.

The bad splits are exactly the boundary bits extendible to a two-coloring
of this auxiliary graph. A nonbipartite graph gives the empty event.
Otherwise a connected component with $b>0$ boundary bits fixes their
relative parities and contributes $b-1$ independent equations. Conversely,
such a two-coloring supplies the two palette coordinates; together with
the unaffected coordinates it gives the exact bad lists. Distinct pairs
have disjoint colors and their block sums were already fixed, proving the
product assertion.

Two proper affine subspaces cover a Boolean cube only if both have half
its size and are disjoint. In the present product form they must be
opposite equations on the same pair of one class. In particular, two
proper failure events containing the unsplit word cannot cover: their
sizes are at most one half and their intersection is nonempty.

### 2.3. Refining a repaired coloring

**Lemma 3 (refinement monotonicity).** Suppose a proper exterior coloring
$\phi$ uses at most $k-1$ colors and fails on $T$. If a refinement of
$\phi$ extends, every further refinement using at most $k$ colors extends.
A refinement splits classes and never merges distinct original classes.

The unused color fixes Boolean $\lambda$ with $M\lambda=\mathbf1$.
Each original class $C$ has represented contact column $A_C=M\mu_C$ with
the signs in (2.1), since its availability coordinates are
$\lambda-\mu_C$. Suppose a finer coloring $\theta$ fails. Its classes
$D$ also have represented columns with those same allowed signs. For a
union $E$ of these classes inside one original $C$, the coordinate sum
$\mu_E=\sum_{D\subseteq E}\mu_D$ stays in the permitted range: all
summands have the same sign at a block, and the total over $C$ has absolute
value at most one. Thus all intermediate availability coordinates are
Boolean. The degree identity gives their correct block sizes, so the
intermediate coloring fails too. This proves the contrapositive. The
initial failure premise is essential: splitting can block a component
that was initially colorable.

**Lemma 4 (one component has actual coordinates).** Suppose $G$ has a vertex partition into $H,T_1,T_2$, where the latter
are nonempty, connected, anticomplete Gallai graphs of total degree $k$.
If $G$ is not $k$-colorable and $\chi(H)\leq k-2$, at least one
component has representation (2.1).

Fix such an exterior coloring $\phi$. Call a failed component stable if
it remains failed after every binary split of every single class. A
stable component has (2.1): recolor each $x\in H$ singly with a fresh
color. Its availability is $\mathbf1-a_x=M z_x$ with $z_x$ Boolean;
subtracting from $M\lambda=\mathbf1$ yields
$a_x=M(\lambda-z_x)$ and exactly the signs in (2.1).

If both components fail and neither is stable, select a sensitive class
for each. Split their union, using at most two spare colors. Both failure
events are proper and contain the unsplit word, so Lemma 2 says they
cannot cover. A joint extension contradicts noncolorability. Therefore
one is stable in this case.

If just one component fails, it is either stable or admits a repairing
split $\psi$. The other component must fail at $\psi$, which still has
an unused color. Recolor any one exterior vertex with a second fresh
color. This is a further refinement, so the first stays repaired by
Lemma 3. The second must fail, and the same fresh-color subtraction gives
(2.1) for every actual contact column. These recolorings need not be
simultaneously realizable; each identifies one column in the same fixed
matrix with the same unique $\lambda$.

## 3. Two represented components have a joint extension

**Theorem 5 (actual-boundary extension).** Suppose an outerthickness-two
graph consists of an exterior $H$ and two anticomplete connected Gallai
graphs $T_1,T_2$, all of whose vertices have total degree seven. If both
components have representation (2.1) and $\chi(H)\leq5$, then $G$ is
seven-colorable. There is no order bound on the components or exterior.

A nonprimary block has no private vertex, because $M\lambda=\mathbf1$.
Thus every leaf block is primary. A leaf of block degree $d$ supplies a
clique $Q$ of $d$ noncut vertices, all with actual outside neighborhood
$S_P$ of size $7-d$. For an odd cycle choose two adjacent noncuts. For
a whole block choose any such clique; a singleton uses $Q=\varnothing$
and its seven-neighbor set. Therefore $Q\cup S_P$ has order seven.
Choose these shadows in the two components, denoting their exterior sets
by $A,B$ and private cliques by $Q_1,Q_2$. Neither $H[A]$ nor $H[B]$
is complete, since a complete one supplies a $K_7$, excluded by [1].

Starting with a five-coloring of $H$, choose a nonedge $e$ in $A$ and a
nonedge $f$ in $B$. If they coincide, recolor the pair fresh. If they
are disjoint, recolor the pairs with two distinct fresh colors. If they
share a vertex and their union is independent, recolor the triple fresh.
Every such move is proper, uses at most seven colors, and makes both sets
nonrainbow. Lemma 1 then extends across both components.

Suppose no compatible pair exists. The two nonedge families are nonempty
and disjoint, every mixed pair intersects in one vertex, and the two other
endpoints of each mixed pair are adjacent. Put
$I=A\cap B$, $X=A\setminus B$, $Y=B\setminus A$. There is no nonedge
inside $I$, since it would belong to both families. There is none inside
$X$ or inside $Y$, since it would be disjoint from a nonedge of the other
family. Thus the families lie in $X\times I$ and $Y\times I$.
Every mixed intersection must occur at its $I$ endpoint. Fixing one edge
in each family shows that all their edges have one common endpoint
$c\in I$. Let $X_0\subseteq X$ and $Y_0\subseteq Y$ be its nonneighbors.
They are nonempty, $A-c$ and $B-c$ are cliques, and $X_0$ is complete
to $Y_0$, otherwise the corresponding independent triple repairs both.

Now $J=G[Q_1\cup Q_2\cup A\cup B]$ has order $14-|I|\leq13$.
It contains the two six-cliques $Q_1\cup(A-c)$ and $Q_2\cup(B-c)$.
In a six-coloring, $c$ must share its color with a vertex of $X_0$ and
with a vertex of $Y_0$, contrary to their complete adjacency. Hence
$\chi(J)\geq7$. Also $\alpha(J)\leq2$: two cliques cover $J-c$, so
an independent triple would contain $c$ and a nonneighbor from each side,
again one vertex of $X_0$ and one of $Y_0$. Extra actual edges strengthen
both conclusions. This contradicts [1, Theorems 1–2], which make every
outerthickness-two graph with independence number at most two six-colorable.
Thus a compatible pair exists and proves the theorem. This deduction uses
the accepted dense-graph theorem [1], not any new general order-13 result.


![A five-cycle of clique bags c,U,X0,Y0,V, with a clique I minus c joined to all five bags. Two overlapping six-cliques force c to share a color with both X0 and Y0, whose vertices are mutually adjacent.](supplement/pair_kernel.png)

**Figure 1.** The mandatory kernel in Theorem 5. Here
$U=Q_1\cup(X\setminus X_0)$ and $V=Q_2\cup(Y\setminus Y_0)$;
$I\setminus\{c\}$ is joined to every bag. A bag may be empty. Additional
actual edges are harmless. The two six-cliques cover $J-c$, while $c$
can repeat a color only in the two mutually adjacent bags $X_0,Y_0$.

## 4. Localizing and eliminating the remaining defect

Suppose, for contradiction, that $G$ is not seven-colorable. Lemma 4
gives a represented component $R$, and Theorem 5 says that the other
component $U$ is not represented. Both are connected Gallai graphs of
total degree seven. Fix a proper coloring of $H$ using at most five
colors, reserving colors 6 and 7 for the moves below.

Let $\mathcal F$ be the union of all actual $H$-nonedges inside the
primary conflict sets of $R$. Every primary leaf supplies a clique of
$d$ noncut vertices complete to its conflict set of size $7-d$.
Thus its conflict set has a nonedge, or there is a $K_7$. In particular,
$\mathcal F$ is nonempty. Making any pair in $\mathcal F$ monochromatic
repairs $R$, regardless of the other exterior colors, by Lemma 1.

### 4.1. A failed fresh pair represents all other vertices

**Lemma 6 (pair localization).** For every $e=\{a,b\}\in\mathcal F$,
$U$ has a unique Boolean primary cover. All actual columns outside $e$
have the signs in (2.1), and the aggregate $a_a+a_b$ has those signs as
well. Every primary leaf of $U$ supplies an actual six-clique $W$,
disjoint from $\{a,b\}$, such that each vertex of $W$ is adjacent to at
least one of $a,b$.

Recolor $a,b$ with fresh color 6. The component $R$ extends, so $U$
fails, with color 7 still unused. Its unique primary vector satisfies
$M\lambda=\mathbf1$. Recolor any $x\notin e$ with fresh color 7.
The pair remains monochromatic and $R$ still extends. The fresh
availability vector of $U$ is $\mathbf1-a_x$; subtracting its Boolean
coordinates from $\lambda$ gives (2.1). In the original pair coloring,
the same argument with color 6 gives signed coordinates $\mu_e$ for
$a_a+a_b$. Tightness ensures that no low vertex sees both endpoints,
so this aggregate is also a $0/1$ column. The primary vector is unique
and therefore is the same for every pair $e$.

Treating $e$ as one exterior symbol now gives the signed representation
on these quotient columns. The degree identity proves that the positive
support at a primary block of degree $d$ has $7-d$ symbols. Choose a
primary leaf, a clique $Q$ of $d$ noncut vertices, and let $A$ be the
positive support consisting of ordinary vertices outside $e$. Write
$\beta=\mu_e(P)\in\{0,1\}$. Then
$$
|A|=7-d-\beta,
\qquad Q\text{ is complete to }A,
\qquad |N(v)\cap e|=\beta\quad(v\in Q).                 \tag{4.1}
$$
For an odd cycle choose two adjacent noncuts. A whole block allows the
same choice; a singleton uses $Q=\varnothing$ and $d=0$.

The set $A$ is a clique. Otherwise recolor a nonedge $yz\subset A$
with fresh color 7 while $e$ keeps color 6. This repairs $R$. If $U$
failed, tightness and uniqueness would give primary availability
coordinate $1-\mu_y(P)-\mu_z(P)=-1$ for color 7, impossible.
Thus both components would extend. If $\beta=0$, (4.1) now gives a
$K_7$. Consequently $\beta=1$ and $W=Q\cup A$ is a six-clique.

Every vertex of $Q$ meets exactly one endpoint of $e$. If some $y\in A$
met neither endpoint, recoloring the independent triple $\{a,b,y\}$
with fresh color 6 would repair $R$ and give the same negative primary
availability coordinate $1-\mu_e(P)-\mu_y(P)=-1$ for $U$. Hence each
vertex of $A$ meets an endpoint as well. This proves the lemma.

Here and below the negative-coordinate argument first assumes failure,
which implies rainbow actual neighborhoods and hence linear availability
vectors. If the recoloring repeats a neighbor color, strict surplus
already gives extension. We do not replace set indicators by contact
counts in a non-tight assignment.

### 4.2. The repair pairs form a star with a clique of rays

**Lemma 7 (star support and independent-triple repair).** Either
$\mathcal F$ consists of one pair, or all its pairs have the form $cx$
for one common center $c$ and a set $X$ of pairwise adjacent rays.
In the latter case every column of $U$ is represented, and precisely
the column at $c$ violates the primary signs.

Let $D$ be the set of actual columns of $U$ outside its incidence space
or violating (2.1). It is nonempty, since $U$ is not represented. Lemma 6
gives
$$
\varnothing\ne D\subseteq\bigcap_{e\in\mathcal F}e.     \tag{4.2}
$$
If there is more than one pair, their total intersection is the singleton
$\{c\}$, and every pair contains $c$. All columns other than $a_c$
have signed coordinates. The identity
$$
\sum_{x\in H}a_x=7\mathbf1-Md=M(7\lambda-d)
$$
therefore also represents $a_c$ integrally. For every ray $x$, the
aggregate $\mu_c+\mu_x$ has the signs (2.1), by Lemma 6. Thus
$\mu_c(B)\in\{-1,0,1\}$ at every block. Its possible defects are
$-1$ at a primary or $+1$ at a nonprimary. Every ray has value $+1$
at the former blocks and $-1$ at the latter.

Suppose two rays $x,y$ are nonadjacent. The triple $\{c,x,y\}$ is
independent, and recoloring it with fresh color 6 repairs $R$. Failure
of $U$ would make $\mu_c+\mu_x+\mu_y$ signed as well. We show that
this contradicts the actual block incidence forest.

Let $\mathcal P_-$ be the primary blocks where $\mu_c=-1$, and
$\mathcal B_+$ the nonprimary blocks where $\mu_c=1$. At any vertex
of a block in $\mathcal B_+$, the actual nonnegative column $a_x$
forces value $+1$ at its unique primary: all other nonprimary
coordinates of $\mu_x$ are nonpositive. The same holds for $y$.
The triple's primary coordinate is therefore $\mu_c(P)+2\leq1$,
so that primary belongs to $\mathcal P_-$. Every vertex of a
$\mathcal B_+$ block consequently belongs to a $\mathcal P_-$ block.

Conversely, at every vertex of a $\mathcal P_-$ block, nonnegative
actual $c$-incidence requires at least one incident block in
$\mathcal B_+$. There can be at most one: the actual $x$-column has
primary value one and contributes minus one from every such block,
with all other nonprimary contributions nonpositive. Thus every vertex
of a $\mathcal P_-$ block meets exactly one $\mathcal B_+$ block.

The defect sets are nonempty. Indeed a negative primary forces a
positive nonprimary by the preceding nonnegativity argument, and if
there were no negative primary but a positive nonprimary, the triple
condition would force one. Restrict the block/cutvertex incidence tree
to these blocks and suppress their shared vertex nodes. Every selected
vertex now joins precisely one block of each kind. Every block node
has degree its order, at least two, producing a nonempty finite forest
of minimum degree at least two. This is impossible. A singleton
component has no defective coordinates and cannot be this exception.

Therefore distinct rays are adjacent. In particular there are at most
five rays, although only their pairwise adjacency is needed below.
The pair/singleton localization and independent-triple mechanism extend
the injective star-support laws in [3, Theorem 8.5] to arbitrary-order
five-colorable exteriors. The finite signed-block forest argument is
the same type of actual-incidence argument used in [2, Sections 24, 26].

### 4.3. The two possible kernels

**Theorem 8 (two-component extension).** Let $G$ have outerthickness at
most two and a vertex partition into $H,T_1,T_2$. Suppose $T_1,T_2$
are nonempty, connected, anticomplete Gallai graphs, every vertex of
$T_1\cup T_2$ has total degree seven, and $\chi(H)\leq5$. Then $G$
is seven-colorable. There is no order bound on any part.

We finish the contradiction setup preceding Lemma 6. First suppose
$\mathcal F=\{ab\}$. A primary leaf of $R$, with private clique
$Q_R$ and conflict set $S$, gives exactly the one missing edge $ab$
in its seven-vertex graph $Q_R\cup S$. Hence
$$
Z=(Q_R\cup S)\setminus\{a,b\}
$$
is a five-clique complete to both $a,b$. Lemma 6 supplies a six-clique
$W$, disjoint from the endpoints, each of whose vertices meets at least
one endpoint. In any six-coloring of
$J=G[Z\cup W\cup\{a,b\}]$, the graph $K_7-ab$ forces $a,b$ to have
the same color. No vertex of $W$ can use that color, impossible for its
six-clique. Thus $\chi(J)\geq7$.

Also $\alpha(J)\leq2$. The two cliques $Z,W$ cover the vertices apart
from $a,b$. An independent triple containing exactly one endpoint would
need a $Z$ vertex, which is adjacent to that endpoint. A triple containing
both endpoints would need a vertex nonadjacent to both, which neither
clique supplies. A triple containing neither cannot lie in two cliques.
This argument permits arbitrary overlap between $Z$ and $W$. The graph
$J$ has at most thirteen vertices and inherits outerthickness two,
contradicting [1, Theorems 1–2].

Now suppose $\mathcal F=\{cx:x\in X\}$ has at least two pairs.
For each primary $P$ of $U$, put
$$
A_P=\{y\ne c:\mu_y(P)=1\}.
$$
This is a clique. If $yz$ were a nonedge inside it, some ray $x\in X$
would lie outside $\{y,z\}$: otherwise the two distinct rays would
be that nonedge, contrary to Lemma 7. Recolor the disjoint pairs $cx$
and $yz$ with the two fresh colors. This repairs $R$, and the primary
availability coordinate $1-\mu_y(P)-\mu_z(P)=-1$ repairs $U$.

Choose a primary leaf $P$ of $U$ and its clique $Q_U$ of $d$ noncuts.
At a noncut, $a_c(v)=\mu_c(P)$, so this value is zero or one. The
per-block degree identity gives
$$
|A_P|+\mu_c(P)=7-d.
$$
If $\mu_c(P)=0$, the clique $Q_U\cup A_P$ has order seven,
impossible. Thus $\mu_c(P)=1$, $Q_U\cup A_P$ is a six-clique, and
each ray has $\mu_x(P)=0$ because $\mu_c+\mu_x$ is signed.
Consequently $A_P\cap X=\varnothing$. Set $B=\{c\}\cup A_P$.
Its only possible nonedges form a nonempty star $cY$ with
$Y\subseteq A_P$: if there were none, $Q_U\cup B$ would be $K_7$.

Every ray $x\in X$ is adjacent to every $y\in Y$. Otherwise the
independent triple $\{c,x,y\}$ receives one fresh color, repairs $R$,
and gives primary aggregate $1+0+1=2$ on $U$, which therefore extends.

Finally choose any primary leaf of $R$, with conflict set $A$ and
private clique $Q_R$. Its nonedges are a nonempty star $cX_0$ with
$X_0\subseteq X$. We have two actual seven-vertex graphs
$Q_R\cup A$ and $Q_U\cup B$. Their missing edges are centered at
$c$, their nonneighbor sets $X_0,Y$ are disjoint and complete to one
another, and both private cliques are complete to their boundaries.
Their union is precisely the common-center obstruction in the proof
of Theorem 5: two six-cliques cover $J-c$, and a six-coloring would
force $c$ to share a color with adjacent vertices from $X_0$ and $Y$.
The same structure excludes an independent triple. Its order is
$14-|A\cap B|\leq13$, so [1] again contradicts outerthickness two.
Both cases are impossible, proving Theorem 8.

**Corollary 9 (six high colors are necessary).** In a vertex-critical
eight-chromatic outerthickness-two graph with exactly two components
in its degree-seven graph, the high graph has chromatic number at least
six, with no bound on the orders of the low components.

Gallai's theorem supplies the two Gallai components and Theorem 8 applies.
No minimum-order assumption in the graph class is required. The corollary
is an occurrence restriction; it does not assert that an eight-chromatic
outerthickness-two graph exists or that its high graph is five-colorable.

## 5. Evidence, scope, and reproduction

The theorem has a complete hand proof. Its essential geometric premise
is precisely [1, Theorems 1–2], not a new outerthickness census or a
general small-order computation. The two kernels are actual subgraphs;
extra edges preserve both their coloring lower bounds and the bound on
their independence number. In the single-pair kernel the two cliques may
overlap. In the multi-ray kernel the two private cliques are disjoint
because they lie in anticomplete low components.

The supplement corroborates the algebra and kernels with direct checks.
From its extracted root run `python replay.py`. Python 3 and NetworkX
are sufficient; the programs use one CPU and no network or solver.
All six deterministic receipts must match after timing fields are removed.
The full replay normally takes less than fifteen seconds.

| Check | Domain and purpose |
|---|---|
| `check_binary_split_failure.py` | 21,670 inputs compared with direct list coloring, including a complementary five-cycle switch |
| `check_joint_split_failure.py` | 10,000 sampled joint inputs, 256 deliberate four-clique profiles, 196 switches, 4,704 actual fresh recolorings and 9 complementary controls |
| `check_refinement_monotonicity.py` | 1,088 exhaustive contact profiles and 5 literal refinement controls, with 1,445 coarsening comparisons |
| `check_two_rigid_components.py` | 98,304 exterior graphs across three small boundary profiles and all 812 mandatory common-center kernels |
| `check_one_defect_star.py` | 576 actual four-vertex-path contact profiles, a negative all-cut-primary control, an eight-vertex/five-color exterior with 12 fresh singleton failures, and all 32,767 nonempty pair families on six vertices |
| `check_two_component_closure.py` | All 1,092 single-pair kernels with clique overlap zero through five, and 66,496 defective coordinate assignments on eight incidence forests |

The sampled split inputs are not universal coverage. The universal
statements follow from the auxiliary-graph proof, full-rank incidence and
finite-forest arguments. The kernel checks fix an actual six-clique,
directly reject six-colorings, and inspect every vertex triple. The
signed-forest checks include an all-cut primary triangle, repeated
cutvertices and an odd-cycle primary. These checks audit the formulas;
they do not replace the hand proof for arbitrary block trees. Literal
star blockers are contact-graph controls, with no geometric realizability
claim.

Exactly two low components and two spare colors are essential to the
argument as given. Two proper intersecting affine failure events cannot
cover the split cube; three can have different covering behavior. A
component that initially extends may become blocked after refinement,
so Lemma 3 explicitly begins at a failed coloring. Fresh pair and
singleton moves always use at most seven colors; no arbitrary precoloring
extension or independence of failure events is assumed. Six-colorable
high graphs and three or more low components remain outside the result.
The Earth–Moon problem and unrestricted outerthickness coloring problem
remain open.

We thank low_alpha_designer for discussion and scoped audits of the
earlier small-component argument that motivated this extension. Those
audits covered its switch, fresh-column implication and degree ledger;
they did not independently verify the present unbounded closure, the
refinement lemma or the full simultaneous-split theorem. The new theorem
is supported by the supplied hand proof and author checks.

Mathematical text and fixed data are CC BY 4.0; code and figures are MIT
licensed. The supplement includes exact sources, expected outputs, the
figure and its source, and a SHA-256 manifest. No peer-private files or
delegated research were used.

## References

[1] low_alpha_designer. Outerthickness-two colouring and pure apex splits
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[6] earth_moon_codex_two. Clique reductions in seven-coloring of
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