Replaces the earlier proof sketch with a clearer attempt that:
1. Corrects the sharp threshold: m_1 >= m - 1 (not m_1 >= m).
Empirically verified for m ∈ {4, 6} across m_1 ≥ m - 1.
2. Proves the ⊆ direction cleanly: each O-face dual vertex has
degree m/2 ≤ 3 in T'_{f'}, so its incident spokes must be
pairwise distinct, putting σ in the "perms-per-half" set P_m.
3. Reduces the ⊇ direction to a cyclic 2-SAT solvability claim
(Conjecture 2sat): for each σ ∈ P_m, find an "orientation"
o ∈ {0,1}^m at the m D-positions such that each length-1 gap
has R_j = L_{j+1} and each length-2 gap has R_j ≠ L_{j+1}.
4. Acknowledges the gap: a naive "all-zero orientation" fails
(e.g. rainbow at m_1 = 6 has the all-zero attempt fail at
gap (4,5)). A satisfying assignment exists in every tested
case (6-18 per σ at m=6, m_1=6) but a clean general proof
awaits. Two routes outlined: S_3-equivariant case analysis,
or global implication-graph analysis.
5. Confirms sharpness with explicit forcing-propagation
counterexample at m=6, m_1=4 (rainbow not in π_D).
6. States provisional corollary: π_D = P_m at m_1 ≥ m - 1
(conditional on Conjecture 2sat); chain pigeonhole reduces
to π_U meeting P_m.
Honest about what's proven (the ⊆ half, the 2-SAT reduction,
the sharpness counterexample) and what's left (the 2-SAT
solvability proof).
Note: rainbow_proof.tex (4 pages).
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
didericis
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