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The cell relation under row-symbol conjugation

Proved
ProofsInTheBook.Chapter33.rowSymbolConjugate_eq_some_iff

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

auxiliary-lemmabook-chapter-36combinatoricslatin-squareslean4proofs-from-the-book

Write [n]={0,…,n−1}[n]=\{0,\ldots,n-1\}[n]={0,…,n−1} for n∈Nn\in\mathbb Nn∈N, with [0]=∅[0]=\varnothing[0]=∅. A partial array of order nnn is a map P:[n]2→[n]∪{⊥}P:[n]^2\to[n]\cup\{\bot\}P:[n]2→[n]∪{⊥}, with ⊥\bot⊥ denoting an empty cell. Write F(P)={(i,j):P(i,j)≠⊥}F(P)=\{(i,j):P(i,j)\ne\bot\}F(P)={(i,j):P(i,j)=⊥} and U(P)={a∈[n]:∃i,j, P(i,j)=a}U(P)=\{a\in[n]:\exists i,j,\ P(i,j)=a\}U(P)={a∈[n]:∃i,j, P(i,j)=a}. It is partial Latin when no symbol repeats within a row or column. A completion is a map L:[n]2→[n]L:[n]^2\to[n]L:[n]2→[n] injective in each row and column, with L(i,j)=aL(i,j)=aL(i,j)=a whenever P(i,j)=a≠⊥P(i,j)=a\ne\botP(i,j)=a=⊥. For partial Latin PPP, its row-symbol conjugate P∗P^*P∗ satisfies P∗(a,j)=iP^*(a,j)=iP∗(a,j)=i if P(i,j)=aP(i,j)=aP(i,j)=a, and is empty when no such row exists; column uniqueness makes the row unique. For every n∈Nn\in\mathbb Nn∈N, partial Latin PPP of order nnn, and a,j,i∈[n]a,j,i\in[n]a,j,i∈[n],

P∗(a,j)=i⟺P(i,j)=a.P^*(a,j)=i\quad\Longleftrightarrow\quad P(i,j)=a.P∗(a,j)=i⟺P(i,j)=a.
Preamble
import Init
import Mathlib
import Definitions.Def_P2MAssembly_Chapter33
set_option autoImplicit true
open Finset
open Classical
open ProofsInTheBook.Chapter33
Formal statement
lemma ProofsInTheBook.Chapter33.rowSymbolConjugate_eq_some_iff {n : ℕ}
    {P : Fin n → Fin n → Option (Fin n)} (hP : IsPartialLatin P)
    (e c r : Fin n) :
    rowSymbolConjugate P e c = some r ↔ P r c = some e := by sorry
Source
Original formalization: https://github.com/xiangyazi24/proof_in_the_book/blob/88d88d141768cded75e782c525ef1bf04b8fe220/ProofsInTheBook/Chapter33Ryser.lean#L208. Topic: Aigner and Ziegler, Proofs from THE BOOK, 6th edition, Chapter 36, “Completing Latin squares”, pp. 253–258 (https://doi.org/10.1007/978-3-662-57265-8_36).

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