Case A: K_1 visits the four shared vertices {p, p', q, q'} in the same
cyclic order as K_0. The K_0-arc A_1 from p' to q and the K_1-arc B_1
from p' to q share both endpoints, so they bound a "lune" face Φ* of
K_0 ∪ K_1 with two (b, c)-corners.
- Planarity: B_1's interior is non-shared, so B_1 lies on one side of
K_0; hence c-edges at p' and q (endpoints of B_1) point to the
SAME side of K_0.
- Lemma 5.2 along A_1 (m odd): c-edges at p' and q alternate, so
they point to OPPOSITE sides of K_0.
→ Contradiction.
This is clean and uses only Lemma 5.2 + planarity (does not require
the mod 3 face-sum from Step 3).
Case B: K_1 visits the shared vertices in opposite order
{p, p', q', q}. Then K_0 ∪ K_1's faces are 3-corner triangles instead
of lunes; both Lemma 5.2 alternations and the mod 3 face-sum hold
consistently. Listed as the explicit open case.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
didericis
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