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Even alternating-deletion count for a noninjective label sequence

Proved
ProofsInTheBook.Chapter39.labelSeq_deletionParity_of_not_injective_of_noOpposite

by xiangyazi24 · Sep 12, 2026 · Mathlib c5ea003 (Lean v4.30.0)

auxiliary-lemmabook-chapter-43combinatoricsgraph-theorylean4proofs-from-the-book

Write [a]={0,…,a−1}[a]=\{0,\ldots,a-1\}[a]={0,…,a−1} for a∈Na\in\mathbb Na∈N (empty when a=0a=0a=0). A signed label on [m][m][m] is a pair (ε,j)(\varepsilon,j)(ε,j) with ε∈{+,−}\varepsilon\in\{+,-\}ε∈{+,−} and j∈[m]j\in[m]j∈[m]; negation reverses its sign. For a sequence M:[ℓ]→{+,−}×[m]M:[\ell]\to\{+,-\}\times[m]M:[ℓ]→{+,−}×[m], write Alt⁡+(M)\operatorname{Alt}_+(M)Alt+​(M) when there exists a strictly increasing u:[ℓ]→[m]u:[\ell]\to[m]u:[ℓ]→[m] such that im⁡M={((−1)a,u(a)):a∈[ℓ]}\operatorname{im}M=\{((-1)^a,u(a)):a\in[\ell]\}imM={((−1)a,u(a)):a∈[ℓ]}, and define Alt⁡−(M)\operatorname{Alt}_-(M)Alt−​(M) by reversing all these signs. Here (−1)a(-1)^a(−1)a denotes the positive sign for even aaa. These predicates concern the label set ordered by index, not the input order. Let k,m∈Nk,m\in\mathbb Nk,m∈N and L:[k+1]→{+,−}×[m]L:[k+1]\to\{+,-\}\times[m]L:[k+1]→{+,−}×[m]. Assume LLL is not injective and L(i)≠−L(j)L(i)\ne-L(j)L(i)=−L(j) for all i,j∈[k+1]i,j\in[k+1]i,j∈[k+1]. Let Li^L^{\widehat i}Li be the sequence obtained by deleting position iii and retaining the order of the other entries. Then

2∣∣{i∈[k+1]:Alt⁡+(Li^)}∣,¬Alt⁡+(L),¬Alt⁡−(L).2\mid|\{i\in[k+1]:\operatorname{Alt}_+(L^{\widehat i})\}|,\qquad\neg\operatorname{Alt}_+(L),\qquad\neg\operatorname{Alt}_-(L).2∣∣{i∈[k+1]:Alt+​(Li)}∣,¬Alt+​(L),¬Alt−​(L).
Preamble
import Init
import Mathlib
import Mathlib.Data.Fin.Tuple.Sort
import Definitions.Def_P2MAssembly_Chapter39
set_option autoImplicit true
open ProofsInTheBook.Chapter39
open SignedPermutation
Formal statement
theorem ProofsInTheBook.Chapter39.labelSeq_deletionParity_of_not_injective_of_noOpposite {k m : ℕ}
    {L : Fin (k + 1) → SignedLabel m} (hnot : ¬ Function.Injective L)
    (_hno : NoOppositeLabelSeq L) :
    Even (labelSeqAltPosDeletionSet L).card ∧
      ¬ IsAltPosLabelSeq L ∧ ¬ IsAltNegLabelSeq L := by sorry
Source
Original formalization: https://github.com/xiangyazi24/proof_in_the_book/blob/88d88d141768cded75e782c525ef1bf04b8fe220/ProofsInTheBook/Chapter39Tucker.lean#L2024. Topic: Aigner and Ziegler, Proofs from THE BOOK, 6th edition, Chapter 43, “The chromatic number of Kneser graphs”, pp. 301–305 (https://doi.org/10.1007/978-3-662-57265-8_43).

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