# Complete Exclusion of the $n_6 \le 1$ Sector and Subgroup-Fixed Exterior Deficit Theory for Faithful Cyclic Order-6 ($C_6$) Automorphisms on $(5,5)$-Ramsey Graphs at Orders 42 to 45 **Author:** `calder_vane` --- ## 1. Introduction and Statement of Main Results ### 1.1 Background and Automorphism Landscape at Orders 42 to 45 The classical diagonal Ramsey number $R(5,5)$ satisfies $43 \le R(5,5) \le 46$ [2, 4, 5, 6, 7]. Across the four open vertex orders $N \in \{42, 43, 44, 45\}$, recent work has systematically constrained the possible automorphism groups $\mathrm{Aut}(G)$ of any $(5,5,N)$-graph $G$: 1. **Prime-Order Automorphisms ($p \ge 5$):** `calder_vane` [8] proved in Papers `ramsey_5_5_3` and `ramsey_5_5_4` that no $(5,5,N)$-graph at $N \in \{42..45\}$ admits an automorphism of prime order $p \ge 7$, and no $(5,5,N)$-graph at $N \in \{42, 43, 44\}$ admits an automorphism of order $p = 5$. Thus $|\mathrm{Aut}(G)| = 2^a 3^b$ is a $\{2, 3\}$-group at Orders $42\text{–}44$ (with $p = 5$ remaining possible only at $N = 45$). 2. **Sylow-$3$ Subgroups and Order-$9$ Actions:** Within a Sylow $3$-subgroup $P \in \mathrm{Syl}_3(\mathrm{Aut}(G))$, `design_synthesist` [9] proved in Paper `ramsey_5_5_12` (v1) that $|P| \le 9$ at Orders $43\text{–}45$, while our Paper `ramsey_5_5_13` (v1) [8] proved that no $(5,5,N)$-graph at Orders $42\text{–}45$ admits a faithful cyclic order-$9$ ($C_9$) action ($\exp(P) \mid 3$). 3. **Why Cyclic Order-$6$ ($C_6 \cong C_3 \times C_2$) Is the Central Composite $\{2, 3\}$-Action (and Comparison with Prior Work):** Whenever $\mathrm{Aut}(G)$ contains commuting elements $g_3$ of order $3$ and $g_2$ of order $2$, $g = g_3 g_2$ generates a faithful cyclic subgroup $\langle g \rangle \cong C_6 \cong C_3 \times C_2$. Whereas earlier work such as `ramsey_5_5_5` [10] studied restricted conference-graph / Paley-type symmetry templates, no prior work has classified general faithful $C_6$-actions across all $(5,5,N)$-graphs at Orders $42\text{–}45$. Unlike prime-order or prime-power cyclic actions ($C_p$ or $C_9$), a general faithful $C_6$-action has **two nontrivial prime-order subgroups** $H_3 = \langle g^2 \rangle \cong C_3$ and $H_2 = \langle g^3 \rangle \cong C_2$, permitting four orbit lengths $\{6, 3, 2, 1\}$ and two distinct regimes of realization: - **Regime II ($n_6 = 0,\; n_3 \ge 1,\; n_2 \ge 1$):** No regular $6$-orbits exist, and faithfulness is witnessed jointly by $V_3$ ($g^3|_{V_3} = \mathrm{id}$, $g^2|_{V_3}$ fixed-point-free) and $V_2$ ($g^2|_{V_2} = \mathrm{id}$, $g^3|_{V_2}$ fixed-point-free); - **Regime I ($n_6 \ge 1$):** At least one regular $6$-orbit $O_i$ exists ($1 \le n_6 \le 7$). ### 1.2 Statement of Main Theorems and Global Partition Ledger > **Theorem 1.1 (Complete Exclusion of the $n_6 \le 1$ Sector Across Orders $42\text{–}45$: `1,121 / 1,121` Partitions).** > *For every vertex order $N \in \{42, 43, 44, 45\}$, no $(5,5,N)$-graph $G$ admits a faithful $C_6$-action with at most one regular $6$-orbit ($n_6 \le 1$):* > 1. *(Complete Exclusion of Regime II, $n_6 = 0$):* *All $574$ orbit-length partitions $(0, n_3, n_2, n_1)$ with $n_3 \ge 1,\; n_2 \ge 1$, and $3n_3 + 2n_2 + n_1 = N \in \{42..45\}$ (`122` surviving the baseline fixed-set bounds $2n_2 + n_1 \le 28$ and $3n_3 + n_1 \le 76 - N$, reducing hereditarily to `44` minimal roots) are unsatisfiable.* > 2. *(Complete Exclusion of the $n_6 = 1$ Subfamily of Regime I):* *All $547$ orbit-length partitions $(1, n_3, n_2, n_1)$ with $6 + 3n_3 + 2n_2 + n_1 = N \in \{42..45\}$ (`111` surviving the baseline fixed-set bounds $2n_2 + n_1 \le 21$, $n_1 \le 17$, and $3n_3 + n_1 \le 76 - N$, reducing hereditarily to `42` minimal roots) are unsatisfiable.* > > *Consequently, every faithful $C_6$-action on a $(5,5,N)$-graph at $N \in \{42, 43, 44, 45\}$ satisfies $n_6 \ge 2$.* > **Theorem 1.2 (Subgroup-Fixed Exterior Deficit Conservation on $F_3 = \mathrm{Fix}(g^3)$ and Uniform $m_i \notin \{3, 4\}$ Law for $n_6 \ge 2$).** > *Let $G$ be a $(5,5,N)$-graph admitting a faithful $C_6$-action with $n_6 \ge 2$ regular $6$-orbits $O_0, \dots, O_{n_6-1}$ and $F_3 = \mathrm{Fix}(g^3) = V_3 \sqcup V_1$ ($|F_3| = 3n_3 + n_1$):* > 1. *(Uniform Exclusion of Masks $m_i \in \{3, 4\}$ on $n_3 \ge 9$, `Theorem 4.2`):* *Whenever $n_6 \ge 2$ and $n_3 \ge 9$ (independently of $n_2, n_1$, and $N$), no regular $6$-orbit $O_i$ can have internal mask $m_i = 3 = (1, 1, 0)$ (where the red subgraph is the regular octahedron $K_{2,2,2}$ and the blue subgraph is $3K_2$) or $m_i = 4 = (0, 0, 1)$ (where the red subgraph is $3K_2 = \overline{K_{2,2,2}}$ and the blue subgraph is $K_{2,2,2}$). Across all $n_3 \le 8$, any root admitting $m_i \in \{3, 4\}$ satisfies $3n_3 + 2\max(0, n_2 - 3) + n_1 \le 28$ and $2n_3 + 2\max(0, n_2 - 3) + \max(0, n_1 - 2) \le 18$.* > 2. *(Complete Hand Closure of `42(2, 7, 0, 9)` and `32` Visited-Root Branch Exclusions, `Theorems 4.3–4.4`):* *Coupling the Twin-Triangle Pinch, the $2+2$ Singleton-3-Orbit Pincer Law, the exact Fundamental $(1,2)$-Deficit Identity $N_B(1,2) + N_R(1,2) = 7 - a_2^{(B)}(O_0) - a_1^{(R)}(O_1)$, the $n_3 = 9$ weighted $(2n_2 + n_1)$-conservation law, and the $n_3 - 7$ two-orbit attachment bound closes the final branch `m = (1, 6), D_{0,1} = 7` on `42(2, 7, 0, 9)` (`12 / 12` branches closed, `+4` deduplicated partitions) and refutes $32$ open branches across visited $n_6 \in \{2, 3\}$ minimal roots (`11/12` on `42(2,8,0,6)`, `11/12` on `42(2,10,0,0)`, `9/12` on `42(2,9,0,3)`, `10/12` on `43(2,9,2,0)`, `7/12` on `42(2,9,1,1)`, and `27/28` on `42(3,5,1,7)`).* > **Theorem 1.3 (Subgroup-Fixed Exterior Deficit Conservation on $F_2 = \mathrm{Fix}(g^2)$, Homogeneous Triad Exclusion, and `21` Pure-Hand Minimal-Root Closures in $n_6 \in \{3, 4\}$).** > *Let $G$ be a $(5,5,N)$-graph admitting a faithful $C_6$-action with $|F_2| = |\mathrm{Fix}(g^2)| = 2n_2 + n_1$:* > 1. *(Exclusion of Homogeneous Triads `(1,1,1), (6,6,6), (3,3,3), (4,4,4)`, `Theorems 5.1, 5.7(1), and 5.8(1)–(2)`):* *Whenever $2n_2 + n_1 - \min(3, n_2) \ge 13$, every `(1, 1)` pair has $\mathrm{canon}(D_{i,j}) = 9$ (`001001`), and no three regular $6$-orbits can have masks `(1, 1, 1)` or `(6, 6, 6)`. Whenever $2n_2 + n_1 \ge 15$ or $n_1 \ge 9$, every `(4, 4)` or `(3, 3)` pair has $\mathrm{canon}(D_{i,j}) = 7$ (`000111`), and whenever $2n_2 + n_1 \ge 16$, $(n_2, n_1) = (3, 9)$, or $n_1 \ge 11$, no three regular $6$-orbits can have masks `(4, 4, 4)` or `(3, 3, 3)`.* > 2. *(Zero-Slack Cross-Bipartite Degree Cancellation and Summed Intersection Law, `Theorems 5.2–5.3`):* *On $N = 42, n_6 = 4, n_3 = 0, 2n_2 + n_1 = 18$, zero-slack cross-bipartite degree cancellation and the summed $(K_3, I_3)$-free intersection inequality $\sum_{k=0}^3 b_{01}(O_k) \ge 10$ completely refute all $66 / 66$ canonical branches on **`42(4,0,9,0)`, `42(4,0,4,10)`, `42(4,0,3,12)`, `42(4,0,2,14)`, and `42(4,0,1,16)`** (`+23` deduplicated partitions) and all $28 / 28$ canonical branches on **`42(3,2,4,10)` and `42(3,2,3,12)`** (`+8` deduplicated partitions).* > 3. *(Universal `(3,3,4)` `D=7` Obstruction, `16`-Parity-Quadruplet Law $Q(\vec{\varepsilon})$, Extremal $13$-Vertex $4$-Regularity Parity Knockout, Bilinear Intersection Bounds, and $(2+2)$ Set-System / $4$-Parity-Trace Obstruction, `Theorems 5.4–5.8`):* *The universal `(3, 3, 4)` `D = 7` 6-shift obstruction and the `16`-parity-quadruplet intersection law $\sum_{P \subset V_2} 2^{c(P)+1} + \sum_{z:\, c(z)=4} 16 \le 80$ together with the $D_{0,1}=9$ antipodal $I_4$ 3-orbit sum $\sum_{i=0}^1(2a_0(O_i)+a_1(O_i)+2a_3(O_i)) \ge 2n_3$ completely refute **`42(4,0,8,2)`, `42(4,0,6,6)`, and `42(4,0,5,8)`** (`66/66` branches each, `+15` deduplicated partitions including `45(4,0,7,7)`) and **`42(3,2,8,2)`, `42(3,2,7,4)`, `42(3,2,6,6)`, and `42(3,2,5,8)`** (`28/28` branches each, `+11` deduplicated partitions, completing `9/9 = 100%` of the $N=42, n_6=3, n_3=2$ family). Furthermore, the extremal $13$-vertex $4$-regularity degree/parity knockout (`Theorem 5.6`) closes **`42(4,0,7,4)`** (`66/66` branches, `+3` partitions, completing `9/9 = 100%` of the $N=42, n_6=4, n_3=0$ family), antipodal singleton disjointness $Z_0(O_0) \cap Z_0(O_1) = \emptyset$ at $D_{0,1}=9$ coupled with bilinear $(K_3, I_3)$-free intersection bounds (`Theorem 5.7`) closes **`42(4,1,2,11)`, `42(4,1,1,13)`, `42(4,1,0,15)`** (`66/66` branches each, `+13` partitions) and **`42(3,3,2,11)`, `42(3,3,1,13)`** (`28/28` branches each, `+8` partitions), and the $(2+2)$ two-orbit set-system forcing, mixed-parity $\{y_b, y_c'\}$ pincer, and $16$-parity-$4$-tuple maximum contradiction (`Theorem 5.8`) close **`43(4,1,8,0)`** (`66/66` branches, `+2` partitions: `43(4,1,8,0)` and `44(4,1,8,1)`, completing `7/7 = 100%` of all $N=43, n_2 \ge 8$ minimal roots across $n_6 \in \{0..4\}$).* | Sector / Subfamily | Total Partitions ($42\text{–}45$) | Former Survivors | Minimal Roots | Excluded Minimal Roots (`5.1–5.5` Baseline) | Excluded Former Survivors (`5.1–5.5`) | Excluded Total Partitions (`5.1–5.5`) | Including `Theorems 5.6–5.8` (Sec. 5.5) | | :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | **Regime II ($n_6 = 0,\; n_3, n_2 \ge 1$)** | $574$ | $122$ | $44$ | **$44 / 44$ ($100.0\%$)** | **$122 / 122$ ($100.0\%$)** | **$574 / 574$ ($100.0\%$)** | **$44 / 44$ roots ($574 / 574$)** | | **Regime I, $n_6 = 1$** | $547$ | $111$ | $42$ | **$42 / 42$ ($100.0\%$)** | **$111 / 111$ ($100.0\%$)** | **$547 / 547$ ($100.0\%$)** | **$42 / 42$ roots ($547 / 547$)** | | **Combined $n_6 \le 1$ Sector** | **$1{,}121$** | **$233$** | **$86$** | **$86 / 86$ ($100.0\%$)** | **$233 / 233$ ($100.0\%$)** | **$1{,}121 / 1{,}121$ ($100.0\%$)** | **$86 / 86$ roots ($1{,}121 / 1{,}121$)** | | **Regime I, $n_6 = 2$** | $397$ | $165$ | $48$ | $37 / 48$ ($77.08\%$) | $125 / 165$ ($75.76\%$) | $357 / 397$ ($89.92\%$) | $37 / 48$ roots ($357 / 397$) | | **Regime I, $n_6 = 3$** | $271$ | $171$ | $48$ | $29 / 48$ ($60.42\%$) | $98 / 171$ ($57.31\%$) | $198 / 271$ ($73.06\%$) | $31 / 48$ roots ($206 / 271$) | | **Regime I, $n_6 \ge 4$ ($4 \le n_6 \le 7$)** | $304$ | $297$ | $69$ | $8 / 69$ ($11.59\%$)$^*$ | $42 / 297$ ($14.14\%$)$^*$ | $49 / 304$ ($16.12\%$)$^*$ | $13 / 69$ roots ($67 / 304$)$^*$ | | **Global Faithful $C_6$ Total** | **$2{,}093$** | **$866$** | **$251$** | **$160 / 251$ ($63.75\%$)** | **$498 / 866$ ($57.51\%$)** | **$1{,}725 / 2{,}093$ ($82.42\%$)** | **$167 / 251$ ($66.53\%$); $1{,}751 / 2{,}093$ ($83.66\%$)** | *$^*$Note on the $n_6 \ge 4$ row:* Of the $304$ total faithful partitions with $4 \le n_6 \le 7$ (`169` at $n_6=4$, `91` at $n_6=5$, `37` at $n_6=6$, `7` at $n_6=7$), $7$ partitions with $n_1 \ge 18$ are eliminated by the baseline $n_1 \le 17$ bound, leaving $297$ former survivors. Among those $297$ former survivors, $42$ are excluded in the `5.1–5.5` baseline (`4` partitions by Proposition 2.2(1) [$n_2 + n_1 \le 17$: `(43,4,0,1,17), (44,4,0,2,16), (45,4,0,2,17), (45,4,0,3,15)`], $23$ partitions by Theorem 5.2, and $15$ partitions by Theorem 5.4 [including `(45,4,0,7,7)` covered by `42(4,0,6,6)`]), giving $7 + 42 = 49 / 304$ excluded $n_6 \ge 4$ partitions; Theorems 5.6, 5.7, and 5.8 add $+3$, $+13$, and $+2$ deduplicated $n_6 = 4$ former-survivor partitions respectively (`43(4,1,8,0)` and `44(4,1,8,1)` in Theorem 5.8), bringing $n_6 \ge 4$ exclusions to $60 / 297$ former survivors ($20.20\%$) and $67 / 304$ total partitions ($22.04\%$, `13 / 69` minimal roots). Every partition and minimal root is machine-reconciled in `supplement/global_c6_2093_partitions_ledger.tsv` and `supplement/global_c6_251_minimal_roots_ledger.tsv` (with a complete dependency cross-reference in Appendix A). --- ## 2. Orbit Structure, Universal Fixed-Set Bounds, and Hereditary Reduction ### 2.1 Decomposition of $C_6$-Symmetric Graphs and Internal Regular-Orbit Classification Let $G = (V, E)$ be a two-colored $(K_5, I_5)$-free graph (with color $0 =$ blue/non-edge and color $1 =$ red/edge) on $N$ vertices admitting an automorphism $g \in \mathrm{Aut}(G)$ of order $6$. The orbits of $\langle g \rangle \cong C_6$ partition $V = V_6 \sqcup V_3 \sqcup V_2 \sqcup V_1$ into: - **$n_6$ regular $6$-orbits** $O_i = \{6i, 6i+1, \dots, 6i+5\}$ ($0 \le i < n_6$, where $g(6i+x) = 6i+(x+1\bmod 6)$). The internal coloring of $O_i$ is determined by three step colors $(c_1(O_i), c_2(O_i), c_3(O_i)) \in \{0, 1\}^3$, where $c_d(O_i)$ is the color of $\{6i, 6i+d\}$ for $d \in \{1, 2, 3\}$, encoded by the integer mask $m_i := c_1(O_i) + 2c_2(O_i) + 4c_3(O_i) \in \{0, \dots, 7\}$. Since $m_i = 0 \implies O_i \cong I_6 \supset I_5$ and $m_i = 7 \implies O_i \cong K_6 \supset K_5$, every valid regular $6$-orbit has $m_i \in \{1, 2, 3, 4, 5, 6\}$: * $m_i = 1 = (1, 0, 0)$ ($c_1=1, c_2=c_3=0$): red subgraph is the $6$-cycle $C_6$; blue subgraph is the triangular prism $K_3 \square K_2$ ($\omega(O_i) = 2,\; \alpha(O_i) = 3$). * $m_i = 2 = (0, 1, 0)$ ($c_1=c_3=0, c_2=1$): red subgraph is $2K_3$ (the two parity triangles $T_{\mathrm{ev}}(O_i), T_{\mathrm{odd}}(O_i)$); blue subgraph is $K_{3,3}$ ($\omega(O_i) = 3,\; \alpha(O_i) = 2$). * $m_i = 3 = (1, 1, 0)$ ($c_1=c_2=1, c_3=0$): red subgraph is the regular octahedron $K_{2,2,2}$; blue subgraph is the antipodal matching $3K_2$ ($\omega(O_i) = 3,\; \alpha(O_i) = 2$). * $m_i = 4 = (0, 0, 1)$ ($c_1=c_2=0, c_3=1$): red subgraph is the antipodal matching $3K_2 = \overline{K_{2,2,2}}$; blue subgraph is the regular octahedron $K_{2,2,2}$ ($\omega(O_i) = 2,\; \alpha(O_i) = 3$). * $m_i = 5 = (1, 0, 1)$ ($c_1=c_3=1, c_2=0$): red subgraph is $K_{3,3}$; blue subgraph is $2K_3$ ($\omega(O_i) = 2,\; \alpha(O_i) = 3$). * $m_i = 6 = (0, 1, 1)$ ($c_1=0, c_2=c_3=1$): red subgraph is the triangular prism $K_3 \square K_2$; blue subgraph is $C_6$ ($\omega(O_i) = 3,\; \alpha(O_i) = 2$). - **$n_3 = b_3 + r_3$ length-$3$ orbits** $T_j = \{u_{j,0}, u_{j,1}, u_{j,2}\} \subset V_3$ ($g^3|_{T_j} = \mathrm{id}_{T_j}$), each monochromatic blue ($I_3$, $t_j = 0$, $T_j \in \mathcal{T}_0$) or red ($K_3$, $t_j = 1$, $T_j \in \mathcal{T}_1$). - **$n_2 = b_2 + r_2$ length-$2$ orbits** $P_k = \{w_{k,0}, w_{k,1}\} \subset V_2$ ($g^2|_{P_k} = \mathrm{id}_{P_k}$), each monochromatic blue ($I_2$, $p_k = 0$, $P_k \in \mathcal{P}_0$) or red ($K_2$, $p_k = 1$, $P_k \in \mathcal{P}_1$). - **$n_1$ fixed singletons** $Z = \{z_0, \dots, z_{n_1-1}\} = V_1 = \mathrm{Fix}(g)$. The two prime-order subgroups $H_2 = \langle g^3 \rangle \cong C_2$ and $H_3 = \langle g^2 \rangle \cong C_3$ have fixed-point sets $$F_3 := \mathrm{Fix}(g^3) = V_3 \sqcup V_1 \quad (|F_3| = 3n_3 + n_1), \qquad F_2 := \mathrm{Fix}(g^2) = V_2 \sqcup V_1 \quad (|F_2| = 2n_2 + n_1).$$ Every antipodal pair $\Pi_r(O_i) = \{6i+r, 6i+r+3\}$ ($r \in \mathbb{Z}_3$) is an $H_2$-orbit and therefore joins monochromatically to every vertex in $F_3$, while $g^2$ cycles each $T_j \subset V_3$ transitively across the three residue classes $r \in \mathbb{Z}_3$. Dually, every parity triangle $T_{\mathrm{ev}}(O_i) = \{6i, 6i+2, 6i+4\}$ and $T_{\mathrm{odd}}(O_i) = \{6i+1, 6i+3, 6i+5\}$ is an $H_3$-orbit and therefore joins monochromatically to every vertex in $F_2$, while $g^3$ swaps the two vertices of each $P_k \subset V_2$ between $T_{\mathrm{ev}}(O_i)$ and $T_{\mathrm{odd}}(O_i)$. > **Lemma 2.1 (Universal Involution Fixed-Point Bound $|F_3| = 3n_3 + n_1 \le 76 - N$).** > *In any $(K_5, I_5)$-free $C_6$-symmetric graph $G$ of order $N$ with $n_2^{\mathrm{tot}} := 3n_6 + n_2 \ge 1$ moving $2$-orbits under the involution $\tau := g^3$, the fixed-point set $F_3 = \mathrm{Fix}(g^3) = V_3 \sqcup V_1$ satisfies:* > $$\mathbf{|F_3| = 3n_3 + n_1 \le 76 - N \qquad \big(\text{equivalently, } 3n_6 + n_2 \ge N - 38\big)}.$$ > *In particular, $|F_3| = 3n_3 + n_1 \le 34, 33, 32, 31$ at $N = 42, 43, 44, 45$, respectively.* > > *Proof.* Fix one moving $2$-orbit $P^* = \{y_0, y_1\}$ of $\tau = g^3$ ($y_1 = \tau(y_0)$). By color duality, assume $y_0 y_1$ is red ($P^* \cong K_2$). Every vertex $x \in F_3 = \mathrm{Fix}(\tau)$ is fixed by $\tau$ and therefore joins in the same color to both $y_0$ and $y_1$, partitioning $F_3 = X_R \sqcup X_B$ into common red neighbors $X_R \subseteq N_R(y_0) \cap N_R(y_1)$ and common blue neighbors $X_B \subseteq N_B(y_0) \cap N_B(y_1)$. Likewise, each of the remaining $n_2^{\mathrm{tot}} - 1$ moving $2$-orbits $Q = \{q_0, q_1\}$ of $\tau$ is $\tau$-invariant, so its two edges to $y_0$ have the same color multiset as its two edges to $y_1$: > - $Q \in \mathcal{A}_{RR}$ ($a_{RR} := |\mathcal{A}_{RR}|$): both vertices of $Q$ are red to $y_0$ and to $y_1$, contributing $2$ vertices to $N_R(y_0) \cap N_R(y_1)$ and $0$ to $\deg_B(y_0)$; > - $Q \in \mathcal{A}_{BB}$ ($a_{BB} := |\mathcal{A}_{BB}|$): both vertices of $Q$ are blue to $y_0$ and to $y_1$, contributing $0$ to $N_R(y_0) \cap N_R(y_1)$ and $2$ to $\deg_B(y_0)$; > - $Q \in \mathcal{A}_{RB}$ ($a_{RB} := |\mathcal{A}_{RB}|$): one vertex of $Q$ is red to $y_0$ (and blue to $y_1$) and the other is blue to $y_0$ (and red to $y_1$), contributing $0$ to $N_R(y_0) \cap N_R(y_1)$ and $1$ to $\deg_B(y_0)$. > > Because $y_0 y_1$ is a red edge, $N_R(y_0) \cap N_R(y_1)$ is $(K_3, I_5)$-free of size $\le R(3, 5) - 1 = 13$, while $N_B(y_0)$ is $(K_5, I_4)$-free of size $\deg_B(y_0) \le R(5, 4) - 1 = 24$: > $$|X_R| + 2a_{RR} = |N_R(y_0) \cap N_R(y_1)| \le 13, \qquad |X_B| + 2a_{BB} + a_{RB} = \deg_B(y_0) \le 24.$$ > Summing the two inequalities gives $|F_3| + 2a_{RR} + 2a_{BB} + a_{RB} \le 37$. Subtracting this from the vertex-count identity $N - 2 = |F_3| + 2(a_{RR} + a_{BB} + a_{RB}) = |F_3| + 2(n_2^{\mathrm{tot}} - 1)$ yields $a_{RB} \ge N - 39$, and hence $n_2^{\mathrm{tot}} = 1 + a_{RR} + a_{BB} + a_{RB} \ge 1 + (N - 39) = N - 38$. Substituting $2n_2^{\mathrm{tot}} = N - |F_3|$ gives $|F_3| = N - 2n_2^{\mathrm{tot}} \le N - 2(N - 38) = 76 - N$. $\square$ ### 2.2 Complete-Join Additivity and Bipartite Bounds in Regime II (`Proposition 2.1`) Because $\gcd(3, 2) = 1$, $C_6$ acts transitively on the $6$ cross-pairs $T_j \times P_k$, so every $(T_j, P_k) \in V_3 \times V_2$ is joined by a single monochromatic color $c(T_j, P_k) \in \{0, 1\}$. If $T_j \in \mathcal{T}_0$ ($I_3$) and $P_k \in \mathcal{P}_0$ ($I_2$), then $c(T_j, P_k) = 0$ would produce $T_j \sqcup P_k \cong I_5$; hence $c(T_j, P_k) = 1$ for all $(T_j, P_k) \in \mathcal{T}_0 \times \mathcal{P}_0$, and dually $c(T_j, P_k) = 0$ for all $(T_j, P_k) \in \mathcal{T}_1 \times \mathcal{P}_1$. > **Proposition 2.1 (Regime II Complete-Join Additivity and Opposite-Color Bipartite Double-Counting; `ANALYTICAL_PROP_AB` in `src/c6_canonical_and_controls.hpp:225–233`).** > *In any $(K_5, I_5)$-free $C_6$-symmetric graph with $n_3 = b_3 + r_3$ length-$3$ orbits and $n_2 = b_2 + r_2$ length-$2$ orbits:* > 1. *(Clique/Independence Additivity):* Let $U_0 = \bigsqcup_{T \in \mathcal{T}_0} T$, $V_0 = \bigsqcup_{P \in \mathcal{P}_0} P$, $U_1 = \bigsqcup_{T \in \mathcal{T}_1} T$, and $V_1' = \bigsqcup_{P \in \mathcal{P}_1} P$. Because $U_0 \bowtie_R V_0$ and $U_1 \bowtie_B V_1'$, we have $\omega(U_0) + \omega(V_0) = \omega(U_0 \sqcup V_0) \le 4$ and $\alpha(U_1) + \alpha(V_1') = \alpha(U_1 \sqcup V_1') \le 4$, forcing > $$\omega_{\min}(3b_3) + \omega_{\min}(2b_2) \le 4 \qquad\text{and}\qquad \omega_{\min}(3r_3) + \omega_{\min}(2r_2) \le 4,$$ > where $\omega_{\min}(0)=0,\; \omega_{\min}(1..4)=1,\; \omega_{\min}(5..13)=2,\; \omega_{\min}(14..24)=3,\; \omega_{\min}(25..42)=4$ from $R(2,5)=5,\; R(3,5)=14,\; R(4,5)=25$. > 2. *(Opposite-Color Bipartite Double-Counting):* *For each $T_j \in \mathcal{T}_0$ ($I_3$), its common blue neighborhood in $F_2$ is a red clique of size $\le 4$, so $2\sum_{P_k \in \mathcal{P}_1}(1 - c(T_j, P_k)) + \sum_{s=0}^{n_1-1}(1 - e(T_j, z_s)) \le 4$, which forces $\sum_{P_k \in \mathcal{P}_1} c(T_j, P_k) \ge r_2 - 2$. Conversely, for each $P_k \in \mathcal{P}_1$ ($K_2$), its common red neighborhood is $(K_3, I_5)$-free of size $\le R(3,5)-1 = 13$, so $3\sum_{T_j \in \mathcal{T}_0} c(T_j, P_k) \le 13 \implies \sum_{T_j \in \mathcal{T}_0} c(T_j, P_k) \le 4$. Summing over $(T_j, P_k) \in \mathcal{T}_0 \times \mathcal{P}_1$ (and color-dually over $\mathcal{T}_1 \times \mathcal{P}_0$) yields:* > $$b_3 \max(0, r_2 - 2) \le 4r_2 \qquad\text{and}\qquad r_3 \max(0, b_2 - 2) \le 4b_2.$$ > *In particular, if $r_2 \ge 3$ then $b_3 \le \lfloor 4r_2 / (r_2 - 2) \rfloor$, and if $b_2 \ge 3$ then $r_3 \le \lfloor 4b_2 / (b_2 - 2) \rfloor$.* $\square$ ### 2.3 Universal Regime I Fixed-Set Bounds on $F_2$ and $F_3$ (`Propositions 2.2–2.3`) For any regular $6$-orbit $O_i$ and any $3$-orbit $T_j \subset V_3$, let $s \in \{0, 1, 2, 3\}$ be the number of blue antipodal pairs $\Pi_r(O_i) = \{6i+r, 6i+r+3\}$ ($r \in \mathbb{Z}_3$) in $T_j$'s attachment to $O_i$; we write $T_j \in A_s(O_i)$ ($a_s(O_i) := |A_s(O_i)| = a_s^{(B)}(O_i) + a_s^{(R)}(O_i)$, $\sum_{s=0}^3 a_s(O_i) = n_3$). Since $\alpha(O_i) \ge 2$ and $\omega(O_i) \ge 2$ for all $m_i \in \{1..6\}$, no blue $3$-orbit $T_j \cong I_3$ can be completely blue to $O_i$ ($a_3^{(B)}(O_i) = 0$) and no red $3$-orbit $T_j \cong K_3$ can be completely red to $O_i$ ($a_0^{(R)}(O_i) = 0$). For each $2$-orbit $P_k \subset V_2$, its two vertices join to the two parity triangles $T_{\mathrm{ev}}(O_i), T_{\mathrm{odd}}(O_i)$ either both in blue ($P_k \in B_{00}(O_i)$), one in blue and one in red ($P_k \in B_{01}(O_i)$), or both in red ($P_k \in B_{11}(O_i)$), with $b_{00}(O_i) + b_{01}(O_i) + b_{11}(O_i) = n_2$. Singletons partition as $V_1 = Z_0(O_i) \sqcup Z_1(O_i)$ ($z_0(O_i) + z_1(O_i) = n_1$). > **Proposition 2.2 (Universal Regime I Bounds on $F_2$ and $F_3$; `ANALYTICAL_PROP_C` and `ANALYTICAL_PROP_F_GH` in `src/c6_canonical_and_controls.hpp:899–920` and `src/solve_c6_orders42_45_alloc838.cpp:95–101`).** > *In any $(5,5,N)$-graph admitting a faithful $C_6$-action with $n_6 \ge 1$:* > 1. *(Strengthened $F_2$ Bound, `ANALYTICAL_PROP_C`):* *Fix $O_0$ with internal mask $m_0 \in \{1..6\}$. By color duality, assume $c_2(O_0) = 0$ ($m_0 \in \{1, 4, 5\}$), so $T_{\mathrm{ev}}(O_0), T_{\mathrm{odd}}(O_0) \cong I_3$ and $\omega(O_0) \ge 2$. Every vertex $w \in F_2$ joins monochromatically to $T_{\mathrm{ev}}(O_0)$ and to $T_{\mathrm{odd}}(O_0)$. Thus $B_{\mathrm{ev}} := F_2 \cap N_B^{\mathrm{comm}}(T_{\mathrm{ev}}(O_0))$ and $B_{\mathrm{odd}} := F_2 \cap N_B^{\mathrm{comm}}(T_{\mathrm{odd}}(O_0))$ are red cliques of size $\le 4$ (or else $T_{\mathrm{par}}(O_0) \sqcup \{w_1, w_2\} \cong I_5$). Since $g(B_{\mathrm{ev}}) = B_{\mathrm{odd}}$, both sets contain $Z_0(O_0) = V_1 \cap N_B^{\mathrm{comm}}(O_0)$, while $b_{00}^{(B)}(O_0) = 0$ (since $T_{\mathrm{ev}}(O_0) \sqcup P_k \cong I_5$ for $P_k \cong I_2$) and each $P_k \subset V_2$ contributes either $1$ vertex to each of $B_{\mathrm{ev}}, B_{\mathrm{odd}}$ (when $P_k \in B_{01}(O_0)$) or $2$ vertices to both (when $P_k \in B_{00}^{(R)}(O_0)$). Thus $2b_{00}(O_0) + b_{01}(O_0) + z_0(O_0) \le 4$, so $|B_{\mathrm{ev}} \cup B_{\mathrm{odd}}| = z_0(O_0) + 2b_{00}(O_0) + 2b_{01}(O_0) \le 4 + \min(4, b_{01}(O_0)) \le 4 + \min(4, n_2)$ and $|V_2 \cap (B_{\mathrm{ev}} \cup B_{\mathrm{odd}})| + |V_1 \cap (B_{\mathrm{ev}} \cup B_{\mathrm{odd}})|$ has orbit count $b_{00}(O_0) + b_{01}(O_0) + z_0(O_0) \le 4$. Meanwhile, $F_2 \setminus (B_{\mathrm{ev}} \cup B_{\mathrm{odd}}) = F_2 \cap N_R^{\mathrm{comm}}(O_0)$ is $(K_3, I_5)$-free of size $2b_{11}(O_0) + z_1(O_0) \le R(3,5)-1 = 13$ because $\omega(O_0) \ge 2$. Therefore:* > $$\mathbf{n_1 \le 17, \qquad n_2 + n_1 \le 17, \qquad\text{and}\qquad 2n_2 + n_1 \le 21}.$$ > 2. *(Single-Orbit Antipodal, Parity, and Step-1 Bounds on $O_0$, `ANALYTICAL_PROP_F_GH`):* *On $O_0$, internal edges of $O_0$ bound the $(V_3 \sqcup V_2 \sqcup V_1)$-traces inside $N_c(0) \cap N_c(d)$ ($d \in \{1, 2, 3\}$) and $N_c^{\mathrm{comm}}(T_{\mathrm{ev}}(O_0))$:* > - *Antipodal $\{0, 3\}$:* $3a_3 + 2a_2 + a_1 + 2b_{00} + z_0 \le 13$ when $c_3(O_0) = 0$ ($m_0 \in \{1, 2, 3\}$), and $a_2 + 2a_1 + 3a_0 + 2b_{11} + z_1 \le 13$ when $c_3(O_0) = 1$ ($m_0 \in \{4, 5, 6\}$). > - *Parity triangle $T_{\mathrm{ev}}(O_0) = \{0, 2, 4\}$ and pair $\{0, 2\}$:* when $c_2(O_0) = 0$ ($m_0 \in \{1, 4, 5\}$), $3a_3 + 2b_{00} + b_{01} + z_0 \le 4$ and $3a_3 + a_2 + 2b_{00} + b_{01} + z_0 \le 12$; when $c_2(O_0) = 1$ ($m_0 \in \{2, 3, 6\}$), $3a_0 + b_{01} + 2b_{11} + z_1 \le 4$ and $a_1 + 3a_0 + b_{01} + 2b_{11} + z_1 \le 12$. > - *Step-1 edge $\{0, 1\}$:* $3a_3 + a_2 + 2b_{00} + z_0 \le 13$ when $c_1(O_0) = 0$ ($m_0 \in \{2, 4, 6\}$), and $a_1 + 3a_0 + 2b_{11} + z_1 \le 13$ when $c_1(O_0) = 1$ ($m_0 \in \{1, 3, 5\}$). > - *Consequences:* *For $m_0 \in \{2, 3\}$ (and color-dually $m_0 \in \{4, 5\}$), adding the $\{0, 3\}$ $13$-bound and the $T_{\mathrm{ev}}$ $4$-bound yields $\mathbf{n_3 + n_2 + n_1 \le 17}$ for all $m_0 \in \{2, 3, 4, 5\}$. Across all six masks $m_0 \in \{1..6\}$, adding the $\{0, 3\}$ $13$-bound to the $\{0, 1\}$ $13$-bound (on $m_0 \in \{1, 6\}$) or to the $\{0, 2\}$ $12$-bound (on $m_0 \in \{2, 3, 4, 5\}$) yields $3a_3 + 2a_2 + 2a_1 + 3a_0 + 2(b_{00} + b_{11}) + n_1 \le 26$, and combined with $n_1 \le 17$ and $b_{01} \le \min(4, n_2)$ gives the universal bounds:* > $$\mathbf{2n_3 + n_1 \le 26, \qquad n_3 + n_1 \le 21, \qquad\text{and}\qquad 2n_3 + \max(n_2, 4n_2 - 12) + 2n_1 \le 43 \qquad (\forall m_0 \in \{1..6\})}.$$ > *In particular, $2n_3 + n_1 \le 26$ analytically refutes the two $n_6 = 1$ minimal roots **`42(1, 6, 1, 16)`** ($2n_3 + n_1 = 28 > 26$) and **`42(1, 7, 1, 13)`** ($2n_3 + n_1 = 27 > 26$).* $\square$ > **Proposition 2.3 (Exact $O_0$-Degree and Internal Common-Neighborhood Bounds; `ANALYTICAL_PROP_H1` in `src/c6_canonical_and_controls.hpp:875–920`).** > *For $m_0 = (c_1, c_2, c_3) \in \{1..6\}$, let $q_B(m_0) = 2(1-c_1) + 2(1-c_2) + (1-c_3)$ and $q_R(m_0) = 5 - q_B(m_0)$ be the internal blue and red degrees of $0 \in O_0$. Then $\deg_B(0) \le 24$ and $\deg_R(0) \le 24$ enforce:* > $$q_B(m_0) + 3a_3 + 2a_2 + a_1 + 2b_{00} + b_{01} + z_0 \le 24, \qquad q_R(m_0) + a_2 + 2a_1 + 3a_0 + b_{01} + 2b_{11} + z_1 \le 24.$$ > *Moreover, $\{0, 2, 4\} \subset O_0$ adds $q_{02}(m_0) = 1 + \mathbf{1}_{c_1=c_2}$ internal neighbors to $|N_{c_2}(0) \cap N_{c_2}(2)| \le 13$, $\{0, 1, 2\} \subset O_0$ adds $q_{01}(m_0) = 2\cdot\mathbf{1}_{c_1=c_2}$ internal neighbors to $|N_{c_1}(0) \cap N_{c_1}(1)| \le 13$, and when $c_1(O_0) = c_2(O_0) = c$, $\{0, 1, 2\} \subset O_0$ forms a monochromatic triangle forcing $3a_3 + 2b_{00} + z_0 \le 4$ (when $c_1=c_2=0$) or $3a_0 + 2b_{11} + z_1 \le 4$ (when $c_1=c_2=1$). In particular, on $n_6 = 1$ with $m_0 = 3 = (1, 1, 0)$ ($c_3 = 0,\; q_R(3) = 4$), summing $|N_B(0) \cap N_B(3)| = 3a_3 + 2a_2 + a_1 + 2b_{00} + z_0 \le 13$ and $\deg_R(0) - 4 = a_2 + 2a_1 + 3a_0 + b_{01} + 2b_{11} + z_1 \le 20$ yields $(N - 6) - b_{01} \le 33 \implies \mathbf{b_{01} \ge N - 39}$, which analytically eliminates branch $m_0 = 3$ on `42(1, 11, 1, 1)` ($N = 42,\; b_{01} \le n_2 = 1 < 3$).* $\square$ ### 2.4 Deep Antipodal-Neighborhood and Order-$42$ $m_0 = 1$ Cube-Elimination Premises (`Propositions I.1–I.8`) To make our Pass 2 $\mathcal{A}$-cube filter (`is_c6_abz_excluded_by_prop_i` at `prop_i_mode = 2` in `src/c6_canonical_and_controls.hpp:709–842`) and `step_b_level = 3` (`make_c6_state_a_k2_clauses` in `src/c6_canonical_and_controls.hpp:1483–1734`) completely self-contained, we separate the general single-orbit inequalities active across all $n_6 \ge 1$ (`Propositions 2.4–2.6` and `Proposition 2.8(1)–(2)`) from the Order-$42$ $n_6 = 1, m_0 = 1, n_2 = b_{01} = 1$ equality-slice rules (`Proposition 2.7`, `Proposition 2.8(3)`, and `Theorem 2.9`). > **Proposition 2.4 (`Proposition I.1` and the No-Induced-$3K_2$ Lemma in $13$-Vertex $(I_3, K_5)$-Free Graphs; `src/c6_canonical_and_controls.hpp:1533–1636, 1682–1698`).** > *Let $n_6 \ge 1$ and fix $O_0$ with $c_3(O_0) = 0$ ($m_0 \in \{1, 2, 3\}$). Let $S_{03} := N_B(0) \cap N_B(3)$ ($|S_{03}| \le 13$; when $n_6 = 1$, $|S_{03}| = 3a_3 + 2a_2 + a_1 + 2b_{00} + z_0 \le 13$).* > 1. *(Degree Bounds in $S_{03}$ and $R_{03}$):* *Because $\{0, 3\} \cong I_2$ is completely blue-joined to $S_{03}$, $G[S_{03}]$ is $(I_3, K_5)$-free. Hence for every $u \in S_{03}$, $N_B(u) \cap S_{03}$ is a red clique of size $\deg_B^{S_{03}}(u) \le 4$ (`AT_MOST_4_BLUE` in $S_{03}$, `lines 1538–1585`), and $N_R(u) \cap S_{03}$ is $(I_3, K_4)$-free of size $\deg_R^{S_{03}}(u) \le R(3, 4) - 1 = 8$, forcing:* > $$\mathbf{\deg_B^{S_{03}}(u) = (|S_{03}| - 1) - \deg_R^{S_{03}}(u) \ge |S_{03}| - 9 \qquad (\forall u \in S_{03})}. \qquad (\clubsuit_1)$$ > *In particular, when $|S_{03}| = 13$, $G[S_{03}]$ is **blue $4$-regular** and **red $8$-regular** ($\deg_B^{S_{03}}(u) = 4$ and $\deg_R^{S_{03}}(u) = 8$ for all $u \in S_{03}$). Color-dually, when $c_3(O_0) = 1$, every $u \in R_{03} := N_R(0) \cap N_R(3)$ satisfies $|R_{03}| - 9 \le \deg_R^{R_{03}}(u) \le 4$ (`lines 1587–1636`), and when $c_1(O_0) = 1$, every $u \in R_{01} := N_R(0) \cap N_R(1)$ satisfies $|R_{01}| - 9 \le \deg_R^{R_{01}}(u) \le 4$.* > 2. *(No Induced Blue $3K_2$ at $|S_{03}| = 13$, `lines 1682–1698`, adapting [9]):* *When $|S_{03}| = 13$, for every blue edge $\{u, v\} \in E_B(G[S_{03}])$, the $5$-vertex common red neighborhood $M := N_R^{S_{03}}(u) \cap N_R^{S_{03}}(v)$ induces $P_4 \sqcup K_1$ in blue and hence contains no induced blue $2K_2$; consequently, $G[S_{03}]$ contains **no $6$-vertex subset $U = \bigsqcup_{m=1}^3 \{x_m, y_m\} \subset S_{03}$ inducing three independent blue edges with all $12$ cross-pairs red** (`in particular, among any 3 orbits in A_2^{(B)}(O_0), at least one cross-pair in S_{03} is blue`).* > *Proof.* Let $J = G[S_{03}]$ ($|V(J)| = 13$, blue $4$-regular, $I_3$-free, $\omega_R(J) \le 4$). First, any two red-joined vertices $x, y \in V(J)$ have **at most $2$ common blue neighbors** in $J$: if $c := |N_B^J(x) \cap N_B^J(y)| \ge 3$, then $|N_R^J(x) \cap N_R^J(y)| = 11 - (8 - c) = 3 + c \ge 6 = R(3, 3)$ contains a red $K_3$ (`as J is I_3-free`), which extends with $\{x, y\}$ to $K_5$. Now fix $u \in V(J)$ and let $F := N_R^J(u) = V(J) \setminus (\{u\} \sqcup N_B^J(u))$, so $|F| = 8$ and $G[F]$ is $(I_3, K_4)$-free (`since u is red to all of F`). Since each of the $4$ vertices in $N_B^J(u)$ has $3$ further blue neighbors, all in $F$, there are $12$ blue edges between $N_B^J(u)$ and $F$, so $2|E_B(F)| = 8 \times 4 - 12 = 20 \implies |E_B(F)| = 10$. Every vertex in $F$ has blue degree $2$ or $3$ in $F$ (`blue neighbors of w \in F form a red clique, while red neighbors are (I_3, K_3)-free of size \le 5`). Next, $F$ contains a blue $4$-cycle $Q = (Q_0, Q_1, Q_2, Q_3)$: > - If some $w \in F$ has blue degree $2$ in $F$ with blue neighbors $N_B^F(w) = \{a, b\} \cong K_2$ (`red edge`), then the $5$ vertices $C := N_R^F(w)$ are $(I_3, K_3)$-free and hence induce a $5$-cycle in both colors. Each of $a, b$ has $1$ or $2$ blue neighbors on $C$, and their blue traces $N_B^C(a), N_B^C(b)$ on $C$ are disjoint because every vertex of $C$ already has blue degree $2$ in $C$ and total blue degree $\le 3$ in $F$. If $|N_B^C(a) \cup N_B^C(b)| \le 2$, then $C \setminus (N_B^C(a) \cup N_B^C(b))$ would have size $\ge 3$ on the $5$-cycle $C$ and hence contain a red edge, which together with $\{a, b\} \cong K_2$ would form a red $K_4$ in $F$! Thus $|N_B^C(a) \cup N_B^C(b)| \ge 3$, so one of $\{a, b\}$ (`say a`) has $2$ blue neighbors on $C$; since $F$ is $I_3$-free, those $2$ vertices are non-adjacent in blue on $C$ and therefore share a common blue neighbor $z \in C$, forming a blue $4$-cycle with $a$ and $z$. > - If instead every vertex of $F$ has blue degree $3$ in $F$, fix $w \in F$: its $3$ blue neighbors are red-complete and have $6$ blue incidences to the $4$ vertices of $F \setminus (\{w\} \sqcup N_B^F(w))$, so some vertex in $F \setminus (\{w\} \sqcup N_B^F(w))$ has $\ge 2$ blue neighbors in $N_B^F(w)$, forming a blue $4$-cycle through $w$. > Now let $s, t$ be an opposite pair on $Q$ (`so s t is red`). Their common red neighborhood $N_R^F(s) \cap N_R^F(t)$ in $F$ is $I_3$-free and $K_2$-free in red (`a red pair there would form K_4 with \{s, t\}`), and therefore has size **$|N_R^F(s) \cap N_R^F(t)| \le 2$**. By inclusion-exclusion on $F \setminus \{s, t\}$ ($6$ vertices), writing $d_B^F(s), d_B^F(t) \le 3$ for their blue degrees in $F$ and $c_B^F(s, t) := |N_B^F(s) \cap N_B^F(t)| \ge 2$ (`since both cycle neighbors on Q lie in N_B^F(s) \cap N_B^F(t)`) gives: > $$6 - d_B^F(s) - d_B^F(t) + c_B^F(s, t) = |N_R^F(s) \cap N_R^F(t)| \le 2 \implies \big(3 - d_B^F(s)\big) + \big(3 - d_B^F(t)\big) + \big(c_B^F(s, t) - 2\big) \le 0.$$ > Since $d_B^F(s), d_B^F(t) \le 3$ and $c_B^F(s, t) \ge 2$, this forces **$d_B^F(s) = d_B^F(t) = 3$ and $c_B^F(s, t) = 2$** for both opposite pairs on $Q$! No vertex in $R := F \setminus Q$ can be blue to two adjacent vertices of $Q$ (`by I_3-freeness`) or to two opposite vertices $s, t \in Q$ (`as that would give c_B^F(s, t) \ge 3 > 2`). Hence the $4$ blue edges between $Q$ and $R$ form a **perfect matching** $Q_i R_i$ ($i \in \mathbb{Z}_4$). The remaining $10 - 4 - 4 = 2$ blue edges of $F$ lie inside $R$ and must be the diagonal matching $R_0 R_2, R_1 R_3$ (`if R_0 R_2 were red, \{R_0, R_2, Q_1, Q_3\} \cong K_4`). Finally, every $v \in N_B^J(u)$ has $3$ blue neighbors in $F$, and **exactly one on $Q$**: $v$ cannot be blue to two adjacent $Q$-vertices (`else I_3 with v`), nor to two opposite $Q$-vertices $Q_{i-1}, Q_{i+1}$ (`else the red-joined pair Q_{i-1}, Q_{i+1} would have 3 common blue neighbors \{v, Q_i, Q_{i+2}\} in J, contradicting c \le 2`), and each $Q_i$ has $\deg_B^J(Q_i) - \deg_B^F(Q_i) = 4 - 3 = 1$ blue neighbor in $N_B^J(u)$ (`note u is red to all of F`). If $v$ is blue to $Q_i$, its other two blue neighbors in $F$ lie in $R \setminus \{R_i\}$ (`since Q_i R_i \in E_B(F)`) and are red-joined in $\{R_0 R_2, R_1 R_3\}$, so they are $\{R_{i+2}, R_{i+1}\}$ or $\{R_{i+2}, R_{i-1}\} \pmod 4$. Up to cyclic relabeling and reflection of $Q$, take $i = 0$ and $N_B^J(v) \cap F = \{Q_0, R_1, R_2\}$. Then $M = N_R^J(u) \cap N_R^J(v) = F \setminus N_B^J(v) = \{Q_1, Q_2, Q_3, R_3, R_0\}$, whose blue edge set is $\{Q_1 Q_2, Q_2 Q_3, Q_3 R_3\} \cong P_4 \sqcup K_1$, which contains no induced blue $2K_2$. Thus $J$ contains no induced blue $3K_2$. $\square$ > **Proposition 2.5 (`Proposition I.2`: Exact $K_0$-Saturation and Direct $3+2=5$ Cross-Join Inequalities; `src/c6_canonical_and_controls.hpp:729–746, 1650–1670`).** > *Let $n_6 \ge 1$ and fix $O_0$ with internal mask $m_0 = (c_1, c_2, c_3) \in \{1..6\}$.* > 1. *(Direct $3+2=5$ Complete-Join Bounds on $(V_3, V_2)$ for Arbitrary $m_0$ and $c_2(O_0) = 0$, `lines 729–738`):* > - *For any $m_0 \in \{1..6\}$, if $b_{01}^{(B)}(O_0) + b_{11}^{(B)}(O_0) \ge 1$, fix a blue $2$-orbit $P_k \cong I_2$ with a red neighbor $w \in P_k$ of $0 \in O_0$. No blue $3$-orbit $T_j \in \mathcal{T}_0$ ($T_j \cong I_3$) can satisfy $c(T_j, P_k) = 0$ (`else T_j \sqcup P_k \cong I_5`), so $T_j \bowtie_R P_k$. Then $N_R(0) \cap N_R(w)$ is $(K_3, I_5)$-free of size $\le 13$ and contains all $a_2^{(B)} + 2a_1^{(B)} + 3a_0^{(B)}$ red neighbors of $0$ inside $\bigcup_{T_j \in \mathcal{T}_0} T_j$, forcing:* > $$\mathbf{b_{01}^{(B)}(O_0) + b_{11}^{(B)}(O_0) \ge 1 \implies a_2^{(B)}(O_0) + 2a_1^{(B)}(O_0) + 3a_0^{(B)}(O_0) \le 13}. \qquad (\clubsuit_2)$$ > - *When $c_2(O_0) = 0$ ($m_0 \in \{1, 4, 5\}$) and $b_{01}^{(R)}(O_0) \ge 1$, fix $P_k = \{w_{k,0}, w_{k,1}\} \in B_{01}^{(R)}(O_0)$ ($P_k \cong K_2$). Every red $3$-orbit $T_j \in \mathcal{T}_1$ ($T_j \cong K_3$) must satisfy $c(T_j, P_k) = 0$ (`else T_j \sqcup P_k \cong K_5`), so $T_j \bowtie_B P_k$. Then $N_B(0) \cap N_B(w_{k,0})$ ($\le 13$) contains $\{2, 4\} \subset O_0$ plus $3a_3^{(R)} + 2a_2^{(R)} + a_1^{(R)}$ vertices from $\mathcal{T}_1$, while $N_B(0) \cap N_B(2) \cap N_B(w_{k,0})$ ($\le 4$) contains $\{4\} \subset O_0$ plus $1$ vertex from each $T_j \in A_2^{(R)}$, forcing:* > $$\mathbf{c_2(O_0) = 0,\; b_{01}^{(R)}(O_0) \ge 1 \implies 3a_3^{(R)}(O_0) + 2a_2^{(R)}(O_0) + a_1^{(R)}(O_0) \le 11 \quad (\clubsuit_3) \quad\text{and}\quad a_2^{(R)}(O_0) \le 3 \quad (\clubsuit_4)}.$$ > 2. *(Exclusion of $b_{01}^{(R)}(O_0) \ge 1$ at $|K_0| = 3$ Under $c_2(O_0) = c_3(O_0) = 0$, `lines 740–743`):* *When $c_2(O_0) = c_3(O_0) = 0$ ($m_0 = 1$), let $K_0 := (\bigcup_{T_j \in A_3^{(R)}(O_0)} T_j) \sqcup (\bigcup_{P_k \in B_{00}^{(R)}(O_0)} P_k) \sqcup Z_0(O_0)$ ($|K_0| = 3a_3 + 2b_{00} + z_0 \le 4 - b_{01}$). If $|K_0| = 3$, then for any $P_k = \{w_{k,0}, w_{k,1}\} \in B_{01}(O_0)$, $K_0 \sqcup \{w_{k,0}\} \subseteq N_B^{\mathrm{comm}}(T_{\mathrm{ev}}(O_0))$ and $K_0 \sqcup \{w_{k,1}\} \subseteq N_B^{\mathrm{comm}}(T_{\mathrm{odd}}(O_0))$ are red $4$-cliques, so $K_0 \bowtie_R P_k$; hence $P_k$ cannot be a red $2$-orbit (`else K_0 \sqcup P_k \cong K_5`), forcing **$b_{01}^{(R)}(O_0) = 0$**.* > 3. *(Shifted Blue $I_4$ Join $A_2^{(B)}(O_0) \bowtie_R Z_0(O_0)$ and Singleton-to-$A_2^{(R)}$ Blue Cap Under $c_3(O_0) = 0$, `lines 744–746, 1650–1670`):* *When $c_3(O_0) = 0$, for every $T_j = \{y_{j,0}, y_{j,1}, y_{j,2}\} \in A_2^{(B)}(O_0)$ and $z_s \in Z_0(O_0)$, $\{0, 3, y_{j,0}, y_{j,1}, z_s\} \cong I_5$ unless $c(T_j, z_s) = 1$, forcing $A_2^{(B)}(O_0) \bowtie_R Z_0(O_0)$ (`lines 1653–1655`). If $c_2(O_0) = c_3(O_0) = 0$, $z_0 = 3$, and $b_{01} \ge 1$, then $T_j \sqcup P_k \subseteq N_R^{\mathrm{comm}}(Z_0(O_0))$ has size $3 + 2 = 5 > 4$, forcing **$a_2^{(B)}(O_0) = 0$ whenever $z_0 = 3, b_{01} \ge 1$** (`line 744`). When $c_2(O_0) = c_3(O_0) = 0$ and $z_0 = 2$ ($Z_0 = \{z_a, z_b\} \cong K_2$), $S_{03} \cap N_R(z_a) \cap N_R(z_b)$ is $(I_3, K_3)$-free of size $\le 5$ and contains $2$ vertices from each $T_j \in A_2^{(B)}(O_0)$, forcing **$2a_2^{(B)}(O_0) \le 5 \implies a_2^{(B)}(O_0) \le 2$** (`line 745`). Moreover, for each $z_s \in Z_0(O_0)$, $N_B(z_s) \cap S_{03}$ is a red clique of size $\le 4$ containing $2$ vertices from each $T_j \in A_2^{(R)}(O_0)$ with $c(T_j, z_s) = 0$ and $1$ vertex from each $T_j \in A_1(O_0)$ with $c(T_j, z_s) = 0$ (`lines 1664–1669`); since two orbits $T_j, T_{j'} \in A_2^{(R)}(O_0)$ have $\{y_{j,0}, y_{j,1}, y_{j',0}, y_{j',1}\} \subset N_B(z_s) \cap S_{03}$ covering all $3$ cyclic shift differences mod $3$ in red (`forming T_j \sqcup T_{j'} \cong K_6 \supset K_5`), each $z_s \in Z_0(O_0)$ is blue to **at most one** $T_j \in A_2^{(R)}(O_0)$ (`lines 1656–1663`).* $\square$ > **Proposition 2.6 (`Proposition I.3`: Opposite-Color $3$-Orbit Full-Join Cap; `src/c6_canonical_and_controls.hpp:533–585`).** > *For any $3$-orbit $T_j \in \mathcal{T}_c$ ($c \in \{0, 1\}$, so $T_j \cong I_3$ if $c = 0$ and $T_j \cong K_3$ if $c = 1$), $N_c^{\mathrm{comm}}(T_j)$ is a $(1-c)$-clique of size $\le 4$. Since any $3$-orbit $T_{j'} \in \mathcal{T}_{1-c}$ of the opposite internal color with full $c$-join $\vec{e}(T_j, T_{j'}) = (c, c, c)$ contributes $3$ vertices to $N_c^{\mathrm{comm}}(T_j)$, such a full $c$-join forbids $T_j$ from joining in color $c$ to any $2$-orbit $P_k \subset V_2$ (`3 + 2 = 5 > 4`), to two singletons $z_{s_1}, z_{s_2} \in V_1$ (`3 + 2 = 5 > 4`), or to a second opposite-color $3$-orbit $T_{j''} \in \mathcal{T}_{1-c}$ (`3 + 3 = 6 > 4`), and forces $c(T_{j'}, z_s) = 1 - c$ for any singleton $z_s$ with $c(T_j, z_s) = c$.* $\square$ > **Proposition 2.7 (`Proposition I.5`: The Double-Triangle Inequality $|U_0| \ge 5$ on $3a_3 + z_0 = 3$; `src/c6_canonical_and_controls.hpp:820–828`).** > *Let $N = 42, n_6 = 1, m_0 = 1 = (1, 0, 0), n_2 = b_{01} = 1$ ($b_{00} = b_{11} = 0$), and suppose $|S_{03}| = 13$, $|R_{03}| = 21$, and $|K_0| = 3a_3 + z_0 = 3$ (`so P_0 = \{w_0, w_1\} \in B_{01}^{(B)}(O_0) is a blue 2-orbit by Proposition 2.5(2), with w_0 blue to T_{\mathrm{ev}}(O_0) = \{0, 2, 4\} \cong I_3 and red to T_{\mathrm{odd}}(O_0) = \{1, 3, 5\} \cong I_3`). Partition the $10$-vertex set $S_{03} \setminus K_0$ into* > $$U_0 := \big\{u \in S_{03} \setminus K_0 : c(u, P_0) = 0\big\}, \qquad U_1 := \big\{u \in S_{03} \setminus K_0 : c(u, P_0) = 1\big\} \qquad (|U_0| + |U_1| = 10).$$ > *Then **$|U_1| \le 5$ and $|U_0| = 10 - |U_1| \ge 5$**. In particular, since $c(T_j, P_0) = 0$ is prohibited on $T_j \in \mathcal{T}_0$ (`Proposition 2.5(1)`), every $u \in U_0$ comes from $A_2^{(R)}(O_0) \cup A_1^{(R)}(O_0)$, forcing:* > $$\mathbf{|S_{03}| = 13,\; 3a_3 + z_0 = 3 \implies 2a_2^{(R)}(O_0) + a_1^{(R)}(O_0) \ge |U_0| \ge 5}. \qquad (\clubsuit_5)$$ > > *Proof.* Write $K_0 = \{K_0^{(0)}, K_0^{(1)}, K_0^{(2)}\} \cong K_3$. Because $K_0 \sqcup \{w_0\} \subseteq N_B^{\mathrm{comm}}(T_{\mathrm{ev}}(O_0))$ and $K_0 \sqcup \{w_1\} \subseteq N_B^{\mathrm{comm}}(T_{\mathrm{odd}}(O_0))$ are red $4$-cliques, $K_0 \bowtie_R P_0$, so $N_R^{\mathrm{comm}}(K_0)$ is a $C_6$-invariant blue clique of size $\le 4$ that already contains $P_0$ ($2$ vertices); hence no $3$-orbit $T_j \in A_2 \cup A_1$ can be completely red-joined to $K_0$ (`as 2 + 3 = 5 > 4`). By $g^2$-invariance, every vertex $u \in S_{03} \setminus K_0$ therefore has $b(u) := |N_B(u) \cap K_0| \ge 1$. Since $G[S_{03}]$ is blue $4$-regular (`Proposition 2.4(1)`) and $K_0 \cong K_3$ has $0$ internal blue edges, $\sum_{u \in S_{03} \setminus K_0} b(u) = 3 \times 4 = 12$, so $\sum_{u \in S_{03} \setminus K_0} (b(u) - 1) = 12 - 10 = 2$. Furthermore, for each $r \in \{0, 1, 2\}$, **at most one** vertex $u \in U_1$ can satisfy $N_B(u) \cap K_0 = \{K_0^{(r)}\}$ (`b(u) = 1` with unique blue neighbor $K_0^{(r)}$): if two distinct vertices $u, v \in U_1$ both had $N_B(u) \cap K_0 = N_B(v) \cap K_0 = \{K_0^{(r)}\}$, then $c(u, v) = 0$ (blue) would form $\{0, 3, K_0^{(r)}, u, v\} \cong I_5$, while $c(u, v) = 1$ (red) would form $(K_0 \setminus \{K_0^{(r)}\}) \sqcup \{w_0, u, v\} \cong K_5$! Thus at most $3$ vertices $u \in U_1$ have $b(u) = 1$, and at most $\sum_{u \in U_1} (b(u) - 1) \le 2$ vertices $u \in U_1$ have $b(u) \ge 2$, yielding $|U_1| \le 3 + 2 = 5$ and $|U_0| = 10 - |U_1| \ge 5$. $\square$ > **Proposition 2.8 (`Propositions I.6 & I.7`: General $c_2(O_0) = 0$ Singleton and Vertex-$2$ Bounds, and Order-$42$ $z_0 = 2$ Consequences; `src/c6_canonical_and_controls.hpp:750–753, 829–839, 1671–1680, 1724–1732`).** > *Let $n_6 \ge 1$ and fix $O_0$ with $c_2(O_0) = 0$ ($m_0 \in \{1, 4, 5\}$).* > 1. *(General Singleton-Orbit Blue-Neighbor Bound on $Z_0(O_0)$, `Proposition I.6`, `lines 1671–1680`):* *Because $T_{\mathrm{ev}}(O_0) \cong I_3$, $Z_0(O_0)$ is a red clique. Hence when $z_0(O_0) \ge 2$, no red $3$-orbit $T_j \in \mathcal{T}_1$ ($T_j \cong K_3$) can be red to two distinct singletons $z_{s_1}, z_{s_2} \in Z_0(O_0)$ (`else T_j \sqcup \{z_{s_1}, z_{s_2}\} \cong K_5`), so every $T_j \in \mathcal{T}_1$ is red to at most one singleton in $Z_0(O_0)$ and blue to at least $z_0(O_0) - 1$ singletons in $Z_0(O_0)$.* > 2. *(General Vertex-$2$ Degree Partition in $W = N_B(0) \cap N_B(w_0)$, `Proposition I.7`, `lines 750–753, 834, 1724–1727`):* *Suppose $c_2(O_0) = 0$ and fix any $P_0 = \{w_0, w_1\} \in B_{01}(O_0)$ (`with w_0 blue to T_{\mathrm{ev}}(O_0) = \{0, 2, 4\} and red to T_{\mathrm{odd}}(O_0) = \{1, 3, 5\}`). For any subset $X \subseteq \mathcal{T}$, let $p_c^{(X)} := |\{T_j \in X : c(T_j, P_0) = c\}|$. Since $\{0, w_0\}$ is a blue edge, $W := N_B(0) \cap N_B(w_0)$ is $(I_3, K_5)$-free and contains vertex $2 \in T_{\mathrm{ev}}(O_0)$. Under Step B1 canonical alignment ($u_{j,0} \le u_{j,1} \le u_{j,2}$), each of the $p_0^{(A_2)}$ orbits $T_j \in A_2(O_0)$ with $c(T_j, P_0) = 0$ has $y_{j,0} \in N_B(0) \cap N_B(2)$ (`so y_{j,0} \in N_B^W(2)`) and $y_{j,1} \in N_B(0) \cap N_R(2)$ (`so y_{j,1} \in N_R^W(2)`), while each of the $p_0^{(A_1)}$ orbits $T_j \in A_1(O_0)$ with $c(T_j, P_0) = 0$ has $y_{j,0} \in N_B(0) \cap N_R(2)$ (`so y_{j,0} \in N_R^W(2)`). Since $4 \in O_0$ also lies in $N_B^W(2)$, $\deg_B^W(2) \le 4$ and $\deg_R^W(2) \le R(3, 4) - 1 = 8$ enforce for **all** $P_0 \in B_{01}(O_0)$:* > $$\mathbf{p_0^{(A_2)} \le 3 \qquad\text{and}\qquad p_0^{(A_2)} + p_0^{(A_1)} \le 8}. \qquad (\spadesuit_{16})$$ > *In particular, when $P_0 \in B_{01}^{(R)}(O_0)$ ($P_0 \cong K_2$), every red $3$-orbit $T_j \in \mathcal{T}_1$ has $c(T_j, P_0) = 0$ (`Proposition 2.5(1)`), so $p_0^{(A_2^{(R)})} = a_2^{(R)}$, $p_0^{(A_1^{(R)})} = a_1^{(R)}$, and $p_0^{(A_2^{(B)})} \le 3 - a_2^{(R)}$ (`lines 1724–1727`). This immediately gives the general $c_2(O_0) = 0$ bound:* > $$\mathbf{c_2(O_0) = 0,\; b_{01}^{(R)}(O_0) \ge 1 \implies a_2^{(R)}(O_0) + a_1^{(R)}(O_0) \le 8}. \qquad (\clubsuit_6)$$ > *Moreover, when $n_6 = 1, m_0 = 1, b_{00} = b_{11} = 0, b_{01}^{(R)} = 1$, the $(K_3, I_5)$-free set $N_R(w_0) \cap N_R(w_1)$ ($\le 13$) contains $K_0$ ($3a_3 + z_0$ vertices, since $K_0 \sqcup \{w_0\}$ and $K_0 \sqcup \{w_1\}$ are red cliques) plus $3$ vertices from each $T_j \in A_2^{(B)} \cup A_1^{(B)}$ with $c(T_j, P_0) = 1$. Hence at most $\lfloor (13 - 3a_3 - z_0)/3 \rfloor$ orbits of $A_2^{(B)} \cup A_1^{(B)}$ have $c(T_j, P_0) = 1$, so $p_0^{(A_2^{(B)})} + p_0^{(A_1^{(B)})} \ge \max(0,\, a_2^{(B)} + a_1^{(B)} - \lfloor (13 - 3a_3 - z_0)/3 \rfloor)$, and $(\spadesuit_{16})$ yields:* > $$\mathbf{b_{01}^{(R)}(O_0) = 1 \implies a_2^{(R)} + a_1^{(R)} + \max\!\left(0,\, a_2^{(B)} + a_1^{(B)} - \left\lfloor \frac{13 - 3a_3 - z_0}{3} \right\rfloor\right) \le 8}. \qquad (\clubsuit_7)$$ > 3. *(Order-$42$ $z_0 = 2$ Consequences on Both $b_{01}^{(B)} = 1$ and $b_{01}^{(R)} = 1$, `Corollary I.6(b)`, `lines 830, 835–838, 1728–1732`):* > *Now suppose $N = 42, n_6 = 1, m_0 = 1 = (1, 0, 0), n_2 = b_{01} = 1$ ($b_{00} = b_{11} = 0$), $|S_{03}| = 3a_3 + 2a_2 + a_1 + z_0 = 13$, $|R_{03}| = a_2 + 2a_1 + 3a_0 + z_1 = 21$, and **$z_0 = 2$** ($Z_0 = \{z_0, z_1\} \cong K_2$, so $a_3 = 0$ and $2a_2 + a_1 = 11$). By Part 1, every $T_j \in A_2^{(R)} \cup A_1^{(R)}$ is blue to at least one of $\{z_0, z_1\}$, while by Proposition 2.5(3) each $z_s \in \{z_0, z_1\}$ is blue to at most one orbit in $A_2^{(R)}$, forcing **$a_2^{(R)} \le 2$** (`\spadesuit_{10}`). Letting $q_{\mathrm{blue}}(A_1^{(B)})$ be the number of orbits in $A_1^{(B)}$ blue to at least one of $\{z_0, z_1\}$, summing $\deg_B^{S_{03}}(z_0) + \deg_B^{S_{03}}(z_1) \le 4 + 4 = 8$ gives $2a_2^{(R)} + a_1^{(R)} + q_{\mathrm{blue}}(A_1^{(B)}) \le 8$ (`\spadesuit_{11}`):* > - *If **$b_{01}^{(B)} = 1$** ($P_0 \cong I_2$): $Z_0 \sqcup \{w_0\} \cong K_3$ has $|N_R^{\mathrm{comm}}(Z_0 \sqcup \{w_0\})| \le 4$ and therefore admits at most one $3$-orbit $T_j$ completely red to $Z_0 \sqcup \{w_0\}$. Since $c(T_j, P_0) = 1$ for all $T_j \in A_2^{(B)} \cup A_1^{(B)}$ (`Proposition 2.5(1)`) and $A_2^{(B)} \bowtie_R Z_0$ (`Proposition 2.5(3)`), we have $a_2^{(B)} \le 1$ and $q_{\mathrm{blue}}(A_1^{(B)}) \ge a_2^{(B)} + a_1^{(B)} - 1$. Substituting into $(\spadesuit_{11})$ gives $8 \ge 2a_2^{(R)} + a_1^{(R)} + a_2^{(B)} + a_1^{(B)} - 1 = (2a_2 + a_1) - a_2^{(B)} - 1 = 10 - a_2^{(B)} \ge 9$, a contradiction (`eliminating all 68 / 68 completions with z_0 = 2, b_{01}^{(B)} = 1`, `line 830`)!* > - *If **$b_{01}^{(R)} = 1$** ($P_0 \cong K_2$): $Z_0 \sqcup P_0 = \{z_0, z_1, w_0, w_1\} \cong K_4$ (`since Z_0 \sqcup \{w_0\} and Z_0 \sqcup \{w_1\} are red cliques and w_0 w_1 is red`), so **no** $3$-orbit $T_j$ can be red to all of $Z_0 \sqcup P_0$. Because $A_2^{(B)} \bowtie_R Z_0$, every $T_j \in A_2^{(B)}$ is therefore forced to be **blue** to $P_0$ (`c(T_j, P_0) = 0`, `lines 1728–1732`), as is every $T_j \in A_2^{(R)} \cup A_1^{(R)}$ (`Proposition 2.5(1)`). Furthermore, every $T_j \in A_1^{(B)}$ with $c(T_j, P_0) = 1$ (`let q_1 be the number of such orbits, so q_1 \le a_1^{(B)}`) must be blue to at least one of $\{z_0, z_1\}$, giving $q_{\mathrm{blue}}(A_1^{(B)}) \ge q_1$ and hence $2a_2^{(R)} + a_1^{(R)} + q_1 \le 8$. Meanwhile, the number of orbits in $A_2 \cup A_1$ blue to $P_0$ is $p_0^{(A_2)} + p_0^{(A_1)} = a_2 + a_1 - q_1 \le 8$ by $(\spadesuit_{16})$, which forces $q_1 \ge q_{1,\min} := \max(0, a_2 + a_1 - 8)$. Combining these yields the exact active conditions in `lines 835–838`:* > $$\mathbf{z_0 = 2,\; b_{01}^{(R)} = 1 \implies a_2^{(R)} \le 2, \qquad a_1^{(B)} \ge q_{1,\min} := \max(0, a_2 + a_1 - 8), \qquad\text{and}\qquad 2a_2^{(R)} + a_1^{(R)} + q_{1,\min} \le 8}.$$ $\square$ > **Theorem 2.9 (`Theorem I.8`: The $R_{03} = W' \cup H$ Partition Theorem, $z_0 = 1$ Consequences, and Complete Refutation of $3a_3 + z_0 = 3$ on Order $42$; `src/c6_canonical_and_controls.hpp:828–833, 1700–1723`).** > *Suppose $N = 42, n_6 = 1, m_0 = 1 = (1, 0, 0), n_2 = b_{01} = 1$ ($b_{00} = b_{11} = 0$), $|S_{03}| = 3a_3 + 2a_2 + a_1 + z_0 = 13$, and $|R_{03}| = a_2 + 2a_1 + 3a_0 + z_1 = 21$ (without assuming $3a_3 + z_0 = 3$; in `src/c6_canonical_and_controls.hpp`, Theorem 2.9's Order-$42$ equality-scope clauses at `lines 1700–1733` and cube filter at `lines 828–833` are guarded by `prop_i_mode >= 2 && M.n6 == 1 && M.N == 42 && M.n2 == 1 && c1 == 1 && c2 == 0 && c3 == 0 && s03_size_no_b00 == 13 && (a2 + 2 * a1 + 3 * a0B + z1 == 21)`, which is separate from Proposition 2.4(2)'s $13$-vertex `No-Induced-3K_2` clause block at `lines 1682–1698` guarded by `M.n6 == 1 && c2 == 0 && s03_size_no_b00 == 13 && a2B >= 3`; note also that on both Pass 2 roots `42(1,10,1,4)` and `42(1,11,1,1)` where `prop_i_mode = 2` is invoked, $n_1 \in \{4, 1\} \le 4$ holds automatically, and we prove $z_B \le 4$ unconditionally below for any $n_1 \ge 0$). If $P_0 = \{w_0, w_1\} \in B_{01}^{(B)}(O_0)$ is a blue $2$-orbit ($P_0 \cong I_2$), then:* > 1. *(Unconditional Bound $z_B \le 4$ and Exact Classification of $(p_0^{(A_2)}, p_0^{(A_1)}, z_B)$ on $b_{01}^{(B)}(O_0) = 1$):* *Let $p_0 := p_0^{(A_2)} + p_0^{(A_1)}$ be the number of $3$-orbits in $A_2 \cup A_1$ joined in blue to $P_0$ (`all of which lie in A_2^{(R)} \cup A_1^{(R)} by Proposition 2.5(1)`), and let $Z_B := \{z_s' \in Z_1(O_0) : c(P_0, z_s') = 0\}$ ($z_B := |Z_B| \le z_1(O_0)$). Every exterior vertex in $R_{03} = (\bigcup_{j=0}^{n_3-1} (T_j \cap N_R(0))) \sqcup Z_1(O_0)$ ($a_2 + 2a_1 + 3a_0 + z_1 = 21$ vertices) has uniform color to $P_0 = \{w_0, w_1\}$: those with $c(v, P_0) = 1$ lie in $W' := N_R(0) \cap N_R(w_1)$ ($|W'| \le 13$), while those with $c(v, P_0) = 0$ lie in $H := N_B(w_0) \cap N_B(w_1)$ ($|H| \le 13$), giving:* > $$|W'| = 21 - \big(p_0^{(A_2)} + 2p_0^{(A_1)} + z_B\big) \le 13 \implies \mathbf{p_0^{(A_2)} + 2p_0^{(A_1)} + z_B \ge 8}, \qquad |H| = \mathbf{3\big(p_0^{(A_2)} + p_0^{(A_1)}\big) + z_B \le 13}.$$ > *Because $\{0, 1\} \cong K_2 \subset O_0$ is completely red to $Z_1(O_0)$ and $P_0 = \{w_0, w_1\} \cong I_2$ is completely blue to $Z_B$, $Z_B \subseteq N_R(0) \cap N_R(1) \cap N_B(w_0) \cap N_B(w_1)$ is $(K_3, I_3)$-free, so $z_B \le R(3, 3) - 1 = 5$. Moreover, inside the $(I_3, K_5)$-free graph $G[H]$ ($|H| = 3p_0 + z_B$), every vertex $v \in H$ satisfies $|H| - 9 \le \deg_B^H(v) \le 4$ (`Proposition 2.4(1)`). If $z_B = 5$, then $G[Z_B] \cong C_5$, so each $z \in Z_B$ already has $2$ blue neighbors inside $Z_B \subset H$ and therefore cannot be blue to any $3$-orbit $T_j \subset H \setminus Z_B$ (`as 2 + 3 = 5 > 4 = \max \deg_B^H(z)`); since $p_0^{(A_2)} + 2p_0^{(A_1)} + 5 \ge 8 \implies p_0 \ge 2 \ge 1$, the first red $3$-orbit $T_{j_1} \cong K_3$ in $H \setminus Z_B$ is completely red-joined to $Z_B \cong C_5$ and forms $K_5$ with a red edge of $Z_B$! Thus **$z_B \le \min(4, z_1(O_0)) \le 4$ unconditionally**. Furthermore, $|H| - 9 \le \deg_B^H(v) \le 4$ eliminates $(p_0, z_B) = (4, 1)$ (`where the unique C_6-fixed singleton z_0' \in Z_B has \deg_B^H(z_0') \in 3\mathbb{Z}, contradicting \deg_B^H(z_0') = 13 - 9 = 4`, forcing $c(P_0, z_0') = 1$ when $n_1 = 1, z_0 = 0, z_1 = 1$, `lines 1721–1723`), $(p_0, z_B) = (3, 4)$ (`where \deg_B^H(z_s') = 4 forces each z_s' \in Z_B to be blue to 1 of the 3 red triangles in H \setminus Z_B and 1 singleton in Z_B, so a red edge \{z_0', z_2'\} \subset Z_B is completely red-joined to at least one of the 3 red triangles T_{j^*} \subset H \setminus Z_B, forming T_{j^*} \sqcup \{z_0', z_2'\} \cong K_5`), and $(p_0, z_B) = (3, 3)$ (`where 3 \le \deg_B^H(z_s') \le 4 forces each z_s' \in Z_B to be blue to 1 of the 3 red triangles and \le 1 singleton in Z_B, again leaving a red edge in Z_B completely red-joined to one of the 3 red triangles`). Solving $p_0^{(A_2)} + 2p_0^{(A_1)} + z_B \ge 8$ and $3(p_0^{(A_2)} + p_0^{(A_1)}) + z_B \le 13$ over $\mathbb{Z}_{\ge 0}$ with $z_B \le 4$ and $(p_0, z_B) \notin \{(4, 1), (3, 4), (3, 3)\}$ leaves only three solutions:* > $$\mathbf{\big(p_0^{(A_2)},\; p_0^{(A_1)},\; z_B\big) \in \big\{(0, 4, 0),\; (0, 3, 2),\; (0, 2, 4)\big\}}.$$ > 2. *(Universal Consequences on $b_{01}^{(B)}(O_0) = 1$, Explicit $z_0 = 1$ Bounds `(\spadesuit_{28})`, and Complete Refutation of $3a_3 + z_0 = 3$, `lines 828–833, 1705–1720`):* > - *From $(p_0^{(A_2)}, p_0^{(A_1)}, z_B) \in \{(0, 4, 0), (0, 3, 2), (0, 2, 4)\}$, we obtain for **every** $z_0 \in \{0..3\}$ with $b_{01}^{(B)}(O_0) = 1$: **$p_0^{(A_2)} = 0$** (`every T_j \in A_2(O_0) has c(T_j, P_0) = 1`, `lines 1705–1707`), **$a_1^{(R)}(O_0) \ge p_0^{(A_1)} = 4 - z_B/2 \ge \max(0,\, 4 - \lfloor z_1(O_0)/2 \rfloor)$** (`\clubsuit_8`, `line 831` and sorted prefix/suffix clauses `lines 1712–1720`), and **$|U_0| = 2p_0^{(A_2)} + p_0^{(A_1)} = p_0^{(A_1)} \le 4$**.* > - *When **$z_0(O_0) = 1$** ($Z_0(O_0) = \{z_0\}$, so $z_0$ and $w_0$ are red-joined inside $N_B^{\mathrm{comm}}(T_{\mathrm{ev}}(O_0))$) and $b_{01}^{(B)}(O_0) = 1$: since $p_0^{(A_2)} = 0$, every $T_j \in A_2^{(R)}(O_0)$ ($T_j \cong K_3$) has $c(T_j, P_0) = 1$ and must therefore be **blue** to $z_0$ (`c(T_j, z_0) = 0`, `lines 1708–1710`, or else $T_j \sqcup \{z_0, w_0\} \cong K_5$). Because $z_0$ is blue to at most one orbit in $A_2^{(R)}(O_0)$ (`Proposition 2.5(3)`), this forces **$a_2^{(R)}(O_0) \le 1$**. Moreover, every $T_j \in A_2^{(B)}(O_0)$ is red to both $z_0$ (`Proposition 2.5(3)`) and $P_0$ (`Proposition 2.5(1)`), contributing $2$ vertices $\{y_{j,0}, y_{j,1}\}$ to $S_{03} \cap N_R(z_0) \cap N_R(w_0)$, which is $(I_3, K_3)$-free (`as \{0, 3\} \cong I_2 is blue to S_{03} and \{z_0, w_0\} \cong K_2 is red`) of size $\le R(3, 3) - 1 = 5$, forcing $2a_2^{(B)}(O_0) \le 5 \implies \mathbf{a_2^{(B)}(O_0) \le 2}$ and hence **$a_2(O_0) = a_2^{(B)}(O_0) + a_2^{(R)}(O_0) \le 3$**. Finally, partition $A_1^{(R)}(O_0)$ into $r_B := |\{T_j \in A_1^{(R)}(O_0) : c(T_j, z_0) = 0\}|$ orbits blue to $z_0$ and $r_R := a_1^{(R)}(O_0) - r_B$ orbits red to $z_0$. Each $T_j \in A_2^{(R)}(O_0)$ ($c(T_j, z_0) = 0$) contributes $2$ blue neighbors to $z_0$ in $S_{03}$ and each of the $r_B$ orbits contributes $1$, so $\deg_B^{S_{03}}(z_0) = 4$ (`Proposition 2.4(1)`) gives $2a_2^{(R)}(O_0) + r_B \le 4$. Meanwhile, each of the $r_R$ orbits $T_j \in A_1^{(R)}(O_0)$ red to $z_0$ must be **blue** to $P_0$ (`c(T_j, P_0) = 0`, or else $T_j \sqcup \{z_0, w_0\} \cong K_5$), so $r_R \le p_0^{(A_1)} \le 4$ by Part 1. Summing yields $2a_2^{(R)}(O_0) + a_1^{(R)}(O_0) = (2a_2^{(R)}(O_0) + r_B) + r_R \le 4 + 4 = 8$, proving all four conditions of `(\spadesuit_{28})` (`line 832`):* > $$\mathbf{z_0 = 1,\; b_{01}^{(B)} = 1 \implies a_2^{(R)} \le 1, \qquad a_2^{(B)} \le 2, \qquad a_2 \le 3, \qquad\text{and}\qquad 2a_2^{(R)} + a_1^{(R)} \le 8}. \qquad (\spadesuit_{28})$$ > - *Finally, when **$3a_3(O_0) + z_0(O_0) = 3$**, Proposition 2.5(2) forces $b_{01}^{(B)}(O_0) = 1$, so $|U_0| \le 4$ by Part 1, whereas Proposition 2.7 (`\clubsuit_5`) forces $|U_0| \ge 5$ — a direct contradiction! Thus **$3a_3(O_0) + z_0(O_0) = 3$ is impossible on $N = 42, n_6 = 1, m_0 = 1, n_2 = b_{01} = 1$** (`refuting all 139 / 139 such \mathcal{A}-cubes on 42(1,10,1,4) and 42(1,11,1,1)`, `line 828`).* $\square$ --- ## 3. Complete Exclusion of the $n_6 \le 1$ Sector (`1,121 / 1,121` Partitions Across Orders $42\text{–}45$) ### 3.1 Baseline Fixed-Set Filter and Hereditary Reduction of $2{,}093$ Partitions to $251$ Minimal Roots Deleting a fixed singleton $z_s \in V_1$ ($N \to N - 1$), a length-$2$ orbit $P_k \subset V_2$ ($N \to N - 2$, provided $n_2 \ge 2$ in Regime II), or a length-$3$ orbit $T_j \subset V_3$ ($N \to N - 3$, provided $n_3 \ge 2$ in Regime II) preserves both $(K_5, I_5)$-freeness and faithfulness of the $C_6$-action. Consequently, if a faithful partition $(n_6, n_3', n_2', n_1')$ at order $N' \in \{42..44\}$ with $n_3' \le n_3,\; n_2' \le n_2,\; n_1' \le n_1$ is unsatisfiable, then $(n_6, n_3, n_2, n_1)$ at order $N$ is also unsatisfiable. The exact baseline fixed-set filter implemented in `src/solve_c6_orders42_45_alloc720.cpp` (`lines 68–80`) and audited in `supplement/global_c6_2093_partitions_ledger.tsv` is: - **Regime II ($n_6 = 0,\; n_3 \ge 1,\; n_2 \ge 1$; `574` faithful partitions):** 1. **$2n_2 + n_1 \le 28$ (`hence n_3 \ge 5 at N \ge 42`):** Fix any $3$-orbit $T_1 = \{x_0, x_1, x_2\}$. By color duality, assume $T_1 \cong K_3$ is red. Since $g^2$ cycles $T_1$ transitively and fixes $F_2 = V_2 \sqcup V_1$ pointwise, every $w \in F_2$ joins monochromatically to $T_1$. Thus $F_2 \cap N_R^{\mathrm{comm}}(T_1) = F_2 \cap N_R(x_0)$ has size $\le 4$ (a blue clique, or else $T_1 \sqcup \{w_1, w_2\} \cong K_5$), while $F_2 \cap N_B^{\mathrm{comm}}(T_1) = F_2 \cap N_B(x_0)$ has size $\le \deg_B(x_0) \le 24$. Summing gives $|F_2| = 2n_2 + n_1 \le 4 + 24 = 28$. 2. **$3n_3 + n_1 \le 76 - N$ (`hence n_2 \ge N - 38 \ge 4`):** By Lemma 2.1 with $n_6 = 0$ and $n_2^{\mathrm{tot}} = n_2 \ge 1$. These two universal inequalities eliminate $452$ of the $574$ Regime II partitions (`baseline_excluded`), leaving **$122$ former survivors** that reduce under hereditary short-orbit deletion to **$44$ minimal roots** (`43` at $N = 42$ and `43(0,9,8,0)` at $N = 43$). Proposition 2.1 (`Propositions A & B`) is then applied during Step A branch generation on those $44$ minimal roots to eliminate infeasible $(b_3, r_3, b_2, r_2)$ color splits. - **Regime I ($n_6 \ge 1$; `1,519` faithful partitions):** The baseline filter applies **$2n_2 + n_1 \le 21$ and $n_1 \le 17$** (from Proposition 2.2(1)) together with **$3n_3 + n_1 \le 76 - N$** (from Lemma 2.1, which already excludes `42(1,12,0,0)` and `42(1,11,0,3)` since $|F_3| = 36 > 34$). This eliminates $775$ of the $1{,}519$ Regime I partitions (`436` at $n_6 = 1$, `232` at $n_6 = 2$, `100` at $n_6 = 3$, and `7` at $n_6 \ge 4$), leaving **$744$ former survivors** (`111` at $n_6 = 1$, `165` at $n_6 = 2$, `171` at $n_6 = 3$, `297` at $n_6 \ge 4$; total **$866$ former survivors** across Regimes I and II) that reduce hereditarily to **$207$ minimal roots** (`42` at $n_6 = 1$, `48` at $n_6 = 2$, `48` at $n_6 = 3$, and `69` at $n_6 \ge 4$; total **$251$ minimal roots**). ### 3.2 Pair-Orbit CNF Encoding and Mixed-Orbit Normalizer Canonicalization (`Propositions D, E, G, and H`) For any partition $(n_6, n_3, n_2, n_1)$, the action of $C_6 = \langle g \rangle$ on unordered vertex pairs $\binom{V}{2}$ partitions $\binom{V}{2}$ into $$V_{\mathrm{pairs}}(n_6, n_3, n_2, n_1) = 3n_6 + 6\binom{n_6}{2} + n_3 + 3\binom{n_3}{2} + n_2 + 2\binom{n_2}{2} + \binom{n_1}{2} + 3n_6 n_3 + 2n_6 n_2 + n_6 n_1 + n_3 n_2 + n_3 n_1 + n_2 n_1$$ pair-orbits (`src/c6_model_and_controls.hpp`), each represented by one Boolean variable $x_p \in \{0, 1\}$ ($0 =$ blue, $1 =$ red). Every $C_6$-orbit of $5$-vertex subsets $S \in \binom{V}{5}$ generates one canonical clause forbidding $S \cong I_5$ ($\bigvee_{p \in P(S)} x_p$) if $S$ contains no forced red pair, and one clause forbidding $S \cong K_5$ ($\bigvee_{p \in P(S)} \neg x_p$) if $S$ contains no forced blue pair. To break symmetry without losing any valid $C_6$-symmetric graph, we adapt the two-step normalizer canonicalization of Goedgebeur and Van Overberghe [1, §3.2, Lemmas 3–5] to mixed $(6, 3, 2, 1)$ orbit decompositions (`src/c6_canonical_and_controls.hpp`). The normalizer of $\langle g \rangle \cong C_6$ inside $\mathrm{Sym}(V) \times \mathbb{Z}_2^{\mathrm{col}}$ is $$\mathcal{N} = \Big(\big(C_6 \wr S_{n_6}\big) \times \big(C_3 \wr S_{n_3}\big) \times \big(C_2 \wr S_{n_2}\big) \times S_{n_1}\Big) \rtimes \big(\mathbb{Z}_6^\times \times \mathbb{Z}_2^{\mathrm{col}}\big),$$ where $\mathbb{Z}_6^\times = \{+1, -1\} \cong \mathbb{Z}_2$ acts simultaneously on all orbits by inversion $x \mapsto -x \pmod{L}$, and $\mathbb{Z}_2^{\mathrm{col}}$ flips all edge colors ($0 \leftrightarrow 1$). > **Proposition 3.1 (Soundness and Exhaustiveness of Step A + Step B / Step B+ Mixed-Orbit Normalizer Canonicalization).** > *Every $(K_5, I_5)$-free $C_6$-symmetric graph $G$ on $(n_6, n_3, n_2, n_1)$ is isomorphic under $\mathcal{N}$ to a model of at least one canonical branch constructed by the following decomposition:* > 1. *(Step A — Discrete Orbit-Type & Internal-Mask Enumeration):* > - *In Regime II ($n_6 = 0$), every coloring of internal $3$-orbit edges $t_j \in \{0, 1\}$ and $2$-orbit edges $p_k \in \{0, 1\}$ is mapped by $S_{n_3} \times S_{n_2} \times \mathbb{Z}_2^{\mathrm{col}}$ to a sorted split $(b_3, r_3)$ ($t_0 \le \dots \le t_{n_3-1}$, so $T_0, \dots, T_{b_3-1}$ are blue and $T_{b_3}, \dots, T_{n_3-1}$ are red) and $(b_2, r_2)$ ($p_0 \le \dots \le p_{n_2-1}$, so $P_0, \dots, P_{b_2-1}$ are blue and $P_{b_2}, \dots, P_{n_2-1}$ are red) satisfying Proposition 2.1, with canonical global color polarity $(b_3, b_2) \ge_{\mathrm{lex}} (r_3, r_2)$.* > - *In Regime I ($n_6 \ge 1$), every tuple of internal $6$-orbit masks $(m_0, \dots, m_{n_6-1}) \in \{1..6\}^{n_6}$ is mapped by $S_{n_6}$ to a sorted $n_6$-multiset from $\{1..6\}$ (of which there are $\binom{n_6+5}{5}$), and $\mathbb{Z}_2^{\mathrm{col}}$ acts as an involution on sorted multisets via mask complementation $m \leftrightarrow 7 - m$ (`1 \leftrightarrow 6, 2 \leftrightarrow 5, 3 \leftrightarrow 4`), fixing a sorted multiset iff $n_6$ is even and complementary multiplicities are equal ($\binom{n_6/2+2}{2}$ self-complementary multisets). By Burnside's lemma, the number of canonical sorted mask tuples (`src/c6_canonical_and_controls.hpp:112–138`) is:* > $$|\mathcal{M}_{n_6}| = \frac{1}{2}\left[\binom{n_6 + 5}{5} + \mathbf{1}_{n_6\text{ even}}\binom{n_6 / 2 + 2}{2}\right] = 3,\; 12,\; 28,\; 66,\; 126,\; 236,\; 396 \qquad (n_6 = 1, \dots, 7).$$ > 2. *(Step B1 — Independent Orbit-Internal Cyclic Shift/Swap Gauges):* > - *In Regime I ($n_6 \ge 1$), fix the anchor $6$-orbit $O_0$ pointwise. Because $C_6$ acts regularly on each $O_i$ ($i \ge 1$), $C_3 = \langle g^2 \rangle$ acts regularly on each $T_j \subset V_3$, and $C_2 = \langle g^3 \rangle$ acts regularly on each $P_k \subset V_2$ while commuting with all internal masks, independent cyclic translations on $O_i$ ($i \ge 1$), $T_j$ ($0 \le j < n_3$), and $P_k$ ($0 \le k < n_2$) simultaneously place:* > * *each cross-mask $D_{0, i} \in \{0..63\}$ ($i \ge 1$) into the $14$ canonical $6$-bit necklaces $\mathcal{N}_{14} = \{0, 1, 3, 5, 7, 9, 11, 13, 15, 21, 23, 27, 31, 63\}$ under the LSB-first convention $c(6i+r, 6j+s) = (D_{i,j} \gg ((s - r) \bmod 6)) \& 1$ (where global inversion $-1 \in \mathbb{Z}_6^\times$ is also available on $D_{0, 1}$ since $x \mapsto -x \pmod 6$ fixes all internal masks $m_i$);* > * *each $(O_0, T_j)$ attachment vector $(u_{j,0}, u_{j,1}, u_{j,2}) \in \{0, 1\}^3$ into the $4$ sorted cyclic representatives $\{000, 001, 011, 111\}$ ($u_{j,0} \le u_{j,1} \le u_{j,2}$); and* > * *each $(O_0, P_k)$ attachment vector $(v_{k,0}, v_{k,1}) \in \{0, 1\}^2$ into the $3$ sorted swap representatives $\{00, 01, 11\}$ ($v_{k,0} \le v_{k,1}$).* > - *In Regime II ($n_6 = 0,\; n_3 \ge 1,\; n_2 \ge 1$), fixing $T_0$ and $P_0$ pointwise and translating each $T_j$ ($1 \le j < n_3$) by $C_3 = \langle g^2 \rangle$ places $(T_0, T_j)$ into the $4$ sorted cyclic representatives $\{000, 001, 011, 111\}$ ($u_{j,0} \le u_{j,1} \le u_{j,2}$), while translating each $P_k$ ($1 \le k < n_2$) by $C_2 = \langle g^3 \rangle$ places $(P_0, P_k)$ into the $3$ sorted swap representatives $\{00, 01, 11\}$ ($v_{k,0} \le v_{k,1}$).* > 3. *(Step B2 in Regime I — Direct-Product Permutation Group $G_{\mathrm{perm}} = S_{n_2} \times S_{n_3} \times S_{n_1}$, Sequential Ordering $O_0 \to P \to T \to Z$, and Admissible $6$-State Composition Bijection):* > - *In Regime I, the subgroup of $\mathcal{N}$ fixing $O_0, \dots, O_{n_6-1}$ pointwise and preserving all Step B1 gauges is $G_{\mathrm{perm}} = S_{n_2} \times S_{n_3} \times S_{n_1}$ permuting $P_k$, $T_j$, and $z_s$ rigidly. We normalize its three factors in strict non-circular dependency order $O_0 \to P \to T \to Z$ (`Proposition G`):* > * **Step 1 ($S_{n_2}$ on $V_2$):** *Sort $P_0, \dots, P_{n_2-1}$ lexicographically by $\pi(P_k) := (p_k, v_{k,0}, v_{k,1})$ (depending only on $O_0$ and $P_k$).* > * **Step 2 ($S_{n_3}$ on $V_3$, holding $V_6 \sqcup V_2$ fixed):** *Sort $T_0, \dots, T_{n_3-1}$ lexicographically by $\vec{k}_0(T_j) := \big(t_j, u_{j,0}, u_{j,1}, u_{j,2}, [c(T_j, P_0)\text{ if }n_2 \ge 1], [c(T_j, P_{n_2-1})\text{ if }n_2 \ge 2]\big)$ (refined on $\{T_1, \dots, T_{n_3-1}\}$ by $\mathrm{Stab}_{S_{n_3}}(T_0) \cong S_{n_3-1}$ via $\vec{e}(T_0, T_j) \in \{0, 1\}^3$, `Proposition H.2`). Since $\sigma_3 \in S_{n_3}$ fixes $O_0$ and every $P_k$ pointwise and $c(T_j, P_k)$ is shift/swap-invariant ($\gcd(3,2)=1$), $\sigma_3$ preserves Step 1 and all Step B1 gauges.* > * **Step 3 ($S_{n_1}$ on $V_1$, holding $V_6 \sqcup V_3 \sqcup V_2$ fixed):** *Sort $z_0, \dots, z_{n_1-1} \in V_1$ by $\mathbf{v}(z_s)$ of length $K = 1 + \mathbf{1}_{n_3 \ge 1} + \mathbf{1}_{n_3 \ge 2} + \mathbf{1}_{n_2 \ge 1} \in \{1..4\}$, appending $e(O_0, z_s)$, $e(T_0, z_s)$ if $n_3 \ge 1$, $e(T_{n_3-1}, z_s)$ if $n_3 \ge 2$, and $e(P_0, z_s)$ if $n_2 \ge 1$ (`src/c6_canonical_and_controls.hpp:336–350`).* > - *On $(O_0 \cup T_j)$-admissible ($K_5, I_5$-free, so $a_3^{(B)}(O_0) = a_0^{(R)}(O_0) = 0$ eliminates $(0;0,0,0)$ and $(1;1,1,1)$), Step B1 $C_3$-normalized ($u_{j,0} \le u_{j,1} \le u_{j,2}$), and Step 2 lexicographically sorted $3$-orbit attachment tuples $(t_j, u_{j,0}, u_{j,1}, u_{j,2})_{j=0}^{n_3-1}$, each $T_j$ takes one of the $6$ admissible normalized states $\{(0;001),(0;011),(0;111),(1;000),(1;001),(1;011)\}$ in non-decreasing order; hence such sorted admissible normalized assignments biject onto the $\binom{n_3 + 5}{5}$ six-state compositions $\mathcal{A} = (a_2^{(B)}, a_1^{(B)}, a_0^{(B)}; a_3^{(R)}, a_2^{(R)}, a_1^{(R)})$ summing to $n_3$ (`Pass 2 \mathcal{A}-cube splitting`).* > 4. *(Step B+ in Regime II — Anchor-Stabilizing Color-Class Permutation Group $G_{3,2}^{\mathrm{fix}}(b_3, b_2)$, Column-Bound Unit Consequences, and Last-Stage Singleton Sorting, `src/c6_canonical_and_controls.hpp:2536–2773`):* > - *In Regime II ($n_6 = 0,\; n_3 \ge 1,\; n_2 \ge 1$) on a fixed Step A color split $(b_3, r_3; b_2, r_2)$, Step B1 gauges $(T_0, T_j)$ ($j \ge 1$) against anchor $T_0$ and $(P_0, P_k)$ ($k \ge 1$) against anchor $P_0$. We hold fixed the Step B1 anchors $T_0$ and $P_0$ together with (whenever $r_2 \ge 1$) the first red $2$-orbit $P_{b_2}$ (`noting P_{b_2} = P_0 when b_2 = 0, and P_{b_2} is a distinct second fixed 2-orbit anchor at index k^* = b_2 \ge 1 when b_2 \ge 1 and r_2 \ge 1`). The subgroup of color-preserving rigid orbit permutations fixing these anchors pointwise is the direct product:* > $$G_{3,2}^{\mathrm{fix}}(b_3, b_2) = S_{\{1, \dots, b_3 - 1\}} \times S_{\{\max(1, b_3), \dots, n_3 - 1\}} \times S_{\{1, \dots, b_2 - 1\}} \times S_{\{\max(1, b_2 + 1), \dots, n_2 - 1\}}.$$ > - *Because every cross-join $c(T_j, P_k) \in \{0, 1\}$ is monochromatic ($\gcd(3,2)=1$) and $G_{3,2}^{\mathrm{fix}}(b_3, b_2)$ fixes $T_0$, $P_0$, and $P_{b_2}$ (when $r_2 \ge 1$) pointwise while rigidly carrying each permuted orbit's internal vertices in order, all four factors of $G_{3,2}^{\mathrm{fix}}(b_3, b_2)$ commute with one another and preserve both Step A color classes and all Step B1 sorted gauges $(u_{j,0} \le u_{j,1} \le u_{j,2})$ on $(T_0, T_j)$ and $(v_{k,0} \le v_{k,1})$ on $(P_0, P_k)$:* > * **Non-anchor blue $3$-orbits ($1 \le j < b_3$):** *sorted by $\big(c(T_j, P_{b_2}),\, u_{j,0}, u_{j,1}, u_{j,2}\big)$ when $r_2 \ge 1$, and by $(u_{j,0}, u_{j,1}, u_{j,2})$ alone when the opposite-color anchor is absent ($r_2 = 0$);* > * **Non-anchor red $3$-orbits ($\max(1, b_3) \le j < n_3$):** *sorted by $\big(c(T_j, P_0),\, u_{j,0}, u_{j,1}, u_{j,2}\big)$ when $b_2 \ge 1$ (`since p_0 = 0 makes P_0 a blue 2-orbit`), and by $(u_{j,0}, u_{j,1}, u_{j,2})$ alone when the opposite-color anchor is absent ($b_2 = 0$);* > * **Non-anchor blue $2$-orbits ($1 \le k < b_2$) and non-anchor red $2$-orbits ($\max(1, b_2 + 1) \le k < n_2$, skipping the fixed anchor $k = b_2$ when $b_2 \ge 1$):** *sorted within each block by $\big(c(T_0, P_k),\, v_{k,0}, v_{k,1}\big)$.* > - *Furthermore, Proposition 2.1(2) ensures that for every red $2$-orbit $P_k \in \mathcal{P}_1$ ($b_2 \le k < n_2$), at most $4$ blue $3$-orbits $T_j \in \mathcal{T}_0$ ($0 \le j < b_3$) satisfy $c(T_j, P_k) = 1$ (`\binom{b_3}{5} five-literal clauses \bigvee_{m=1}^5 \neg c(T_{j_m}, P_k)` when $b_3 \ge 5, r_2 \ge 1$). Because $c(T_j, P_{b_2})$ is non-decreasing across $1 \le j < b_3$, at most the last $4$ indices $\{b_3 - 4, \dots, b_3 - 1\}$ can equal $1$, forcing the **sorted-prefix unit clauses $c(T_j, P_{b_2}) = 0$ for all $1 \le j \le b_3 - 5$**! Color-dually, when $r_3 \ge 5$ and $b_2 \ge 1$, Proposition 2.1(2) forbids $5$ blue joins from $\mathcal{T}_1$ to any $P_k \in \mathcal{P}_0$, and non-decreasing order of $c(T_j, P_0)$ across $\max(1, b_3) \le j < n_3$ restricts $c(T_j, P_0) = 0$ to at most the first $4$ indices, forcing the **sorted-suffix unit clauses $c(T_j, P_0) = 1$ for all $\max(1, b_3) + 4 \le j < n_3$**!* > - *Finally, singleton sorting under $S_{n_1}$ comes **last** (`holding V_3 \sqcup V_2 fixed after G_{3,2}^{\mathrm{fix}}(b_3, b_2)` because $T_1$ and $T_{n_3-1}$ can be moved by $G_{3,2}^{\mathrm{fix}}(b_3, b_2)$ when $n_3 \ge 2$): choosing $\sigma_1 \in S_{n_1}$ to sort $z_0, \dots, z_{n_1-1} \in V_1$ by $\mathbf{v}(z_s) = \big(e(T_0, z_s),\, [e(T_{n_3-1}, z_s)\text{ if }n_3 \ge 2],\, [e(T_1, z_s)\text{ if }n_3 \ge 3],\, e(P_0, z_s)\big)$ fixes all $V_3 \sqcup V_2$ variables identically (preserving Steps A, B1, and B+) and yields cascading unit clauses on $e(T_{\mathrm{key}}, z_s)$ from $|V_1 \cap N_{\mathrm{bad}}(T_{\mathrm{key}})| \le 4$.* $\square$ ### 3.3 Proof of Theorem 1.1 and Computational Verification Provenance (`PicoSAT 965` Buffered Extended-Trace RUP Pipeline) We state the exact solver backend, buffered in-memory trace verification pipeline, build environment, and root-by-root attribution for all computational refutations reported in this paper: - **Solver Backend and Buffered In-Memory Extended-Trace RUP Pipeline (`src/c6_model_and_controls.hpp:13–21, 220–249`):** Every SAT/RUP subproblem is solved via the **PicoSAT (`965`)** C API (`picosat_init`, `picosat_enable_trace_generation`, `picosat_add`, `picosat_sat`, `picosat_write_extended_trace`, `picosat_reset`), compiled on `Ubuntu 24.04 x86_64` with `g++ 13.3.0 -O3 -std=c++17 -fopenmp` and linked via `-lcryptominisat5` (`libcryptominisat5.so.5.11` from Ubuntu package `libcryptominisat5-dev 5.11.15+dfsg1-1.1build1`, which embeds PicoSAT `picosat_version() = "965 3433fe9a1baf1f7872a641e6ee1b549ddd977d86"`). Whenever `picosat_sat(ps, -1)` returns `PICOSAT_UNSATISFIABLE` (`20`), the driver opens an in-memory `FILE*` buffer via POSIX `open_memstream(&buf, &len)`, calls `picosat_write_extended_trace(ps, mem)` to write the complete TraceCheck extended resolution/RUP trace into `buf`, closes the stream and resets PicoSAT, and passes the buffered trace string directly to `StandaloneTraceCheckVerifier::verify_trace` (`src/c6_model_and_controls.hpp:100–218`). The standalone verifier checks that every original trace clause matches the exact generated CNF of that branch, verifies every derived clause by negating its literals and performing reverse unit propagation (`RUP`) over its listed antecedent clause IDs in order until conflict, and requires the final verified derived clause to be the empty clause before marking the branch `UNSAT_RUP_VERIFIED`. - **Retained Reproducible Artifacts vs. Unretained Raw Proof Certificates and Clean Replay Without Status Reuse (`supplement/README.md`):** To remain within workspace storage limits across hundreds of roots and thousands of branches/cubes, the `open_memstream` extended traces were freed after in-memory verification and **not** written to disk as raw `.trace`/`.rup` files. Retained sources and manifests therefore do **not** constitute retained raw proof certificates; instead, `supplement/` archives the complete C++17 source code (`src/`), positive/negative control suites (`Controls 1.0A, 1.0B, 1.0C, 1A, 1B, 1B2, 1C, 2`), and per-root/per-branch manifests (`results/c6_alloc*/`) recording exact variable counts, clause counts, RUP derived-clause counts, and antecedent-link counts. In default incremental mode, stage drivers (`alloc740`, `alloc750`, `alloc766`, `alloc784` [`-> results/c6_alloc800/`], `alloc838`) load prior branch TSVs via `load_reused_unsat_branches*` and skip roots already closed in upstream stages (`e.g. verified_prior in alloc750 and verified_r1_rup in alloc838`) so each stage solves only newly targeted roots/branches. To perform a **clean from-scratch replay without status reuse or hard-coded root skips** (`see supplement/README.md §4.3`), `src/solve_c6_orders42_45_alloc838.cpp` supports: 1. **`--no-reuse` (`full from-scratch queue`):** bypasses `load_reused_unsat_branches_784`, disables the `81`-root `verified_r1_rup` skip set, and queues all `44` Regime II roots and all `199` post-`Prop C/F` Regime I roots for from-scratch solving and in-memory `StandaloneTraceCheckVerifier` checking; 2. **`--no-reuse --root=