Plectis Atlas
Conditional winning templates from finite partial positions
Family 187: Snaky in 21 Maker moves
Candidate pairing
Proposition
The 35-move certificate has 616 cards, including its six base cards, and 1,837 placement terms. Every card satisfies A_i^{35}\subsetneq T_i^{35}, o\in T_i^{35}\setminus A_i^{35}, and h_i^{35}\le34. Four of its conditional interfaces are
Each row applies at any Maker turn with its required point owned and its entire envelope free of Breaker, regardless of all other ownership. For row 615 there is also a Breaker-turn conclusion: if Maker owns \{o,(1,0)\} and Breaker avoids T_{615}^{35}, then after any legal next reply Maker wins within 33 further claims, unless already victorious.
Snaky in 21 Maker moves
TeX label r35:conditional, build/route35.tex, line 150
The Lean statement says
theorem snaky_winning_strategy_with_legal_states : ∃ σ : Policy Cell, (∀ r M B, σ r M B ∉ M ∧ σ r M B ∉ B) ∧ ∀ β : ℕ → Cell, LegalRepliesBeforeFinal σ 35 β ∅ ∅ → HasSnaky (playState σ 35 β ∅ ∅ 35).1 ∧ (playState σ 35 β ∅ ∅ 35).1.card = 35 ∧ Disjoint (playState σ 35 β ∅ ∅ 35).1 (playState σ 35 β ∅ ∅ 34).2 ∧ ∀ k, k < 35 → Disjoint (playState σ 35 β ∅ ∅ k).1 (playState σ 35 β ∅ ∅ k).2 ∧ (playState σ 35 β ∅ ∅ k).2.card = k := by sorryA Lean statement states the theorem and leaves its proof as sorry on purpose. OpenAI’s proof is in a separate solution module, and a program called Comparator checks that proof against this statement.