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The role postulates force the Fano plane (C1 Theorem 3.6)

Proved
FanoUnique.roles_force_fano

by ShapeZero · Sep 24, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricsfano-planesteiner-triple-systems

Let n≥1n \ge 1n≥1, and let SSS be a Steiner triple system on {0,…,n−1}\{0, \dots, n-1\}{0,…,n−1} that admits a role colouring: each point of each line gets a role in {0,1,2}\{0,1,2\}{0,1,2}, the three points of a line get three different roles, and every point takes every role exactly once. Then SSS is the Fano plane up to relabelling: there is a bijection eee from the points of SSS to {0,…,6}\{0, \dots, 6\}{0,…,6} with

{ e(ℓ):ℓ a line of S }={{i, i+1, i+3}:i∈Z/7}.\{\, e(\ell) : \ell \text{ a line of } S \,\} = \bigl\{\{i,\ i+1,\ i+3\} : i \in \mathbb{Z}/7\bigr\}.{e(ℓ):ℓ a line of S}={{i, i+1, i+3}:i∈Z/7}.

In particular n=7n = 7n=7. The hypothesis n≥1n \ge 1n≥1 is necessary: the empty system satisfies every other condition.

Preamble
import Mathlib
import Definitions.Def_FanoUnique_isFano
Formal statement
namespace FanoUnique

open RolesForceSeven

theorem roles_force_fano (n : ℕ) (hn : 0 < n) (S : STS n)
    (role : Fin n → Finset (Fin n) → Fin 3) (h : RoleColouring S role) :
    IsFano S := by
  sorry

end FanoUnique
Source
Shape Zero LLC, "Formal Proofs of the C1 Verification Package" (August 2026), §3.1, Theorem 3.6 ("Roles Force Fano"): https://github.com/ShapeZeroSZ/shape-zero/blob/main/01_source/proofs/ShapeZero_C1_Formal_Proofs.pdf ; corrected in "Errata — C1 Formal Proofs (Sections 3 and 6)", Section 3 (Theorem 3.6 requires a nonempty point set): https://github.com/ShapeZeroSZ/shape-zero/blob/main/01_source/proofs/ERRATUM_Theorem_6.1.md ; public references: Wikipedia, "Fano plane": https://en.wikipedia.org/wiki/Fano_plane ; Wikipedia, "Steiner system": https://en.wikipedia.org/wiki/Steiner_system
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What the Lean code literally says, in plain math · claude-opus-5-5

roles_force_fano. Let nnn be a natural number with n>0n > 0n>0, and write [n]={0,1,…,n−1}[n] = \{0, 1, \dots, n-1\}[n]={0,1,…,n−1} for the point set.

The system SSS. Let SSS be a Steiner triple system on [n][n][n] in the following sense: SSS is a finite family L\mathcal{L}L of subsets of [n][n][n] ("lines") such that

  • every line l∈Ll \in \mathcal{L}l∈L has exactly 333 elements, and
  • for every two distinct points x≠yx \neq yx=y in [n][n][n] there is exactly one line l∈Ll \in \mathcal{L}l∈L with x∈lx \in lx∈l and y∈ly \in ly∈l.

No other conditions are imposed. There is no requirement that L\mathcal{L}L be nonempty or that every point lie on a line; for instance, for n=1n = 1n=1 the empty family is such a system.

The role function. Let role\mathrm{role}role be an arbitrary function that assigns to each point x∈[n]x \in [n]x∈[n] and each subset A⊆[n]A \subseteq [n]A⊆[n] a value

role(x,A)∈{0,1,2}.\mathrm{role}(x, A) \in \{0, 1, 2\}.role(x,A)∈{0,1,2}.

It is defined on all pairs (point, subset), but only its values at pairs (x,l)(x, l)(x,l) with lll a line and x∈lx \in lx∈l appear in the hypothesis below.

Hypothesis ("role colouring"). Assume all three of the following:

  1. Distinct roles within a line: for every line l∈Ll \in \mathcal{L}l∈L and all x,y∈lx, y \in lx,y∈l,
role(x,l)=role(y,l)  ⟹  x=y.\mathrm{role}(x, l) = \mathrm{role}(y, l) \implies x = y.role(x,l)=role(y,l)⟹x=y.
  1. Every role occurs at every point: for every point x∈[n]x \in [n]x∈[n] and every ρ∈{0,1,2}\rho \in \{0,1,2\}ρ∈{0,1,2} there is a line l∈Ll \in \mathcal{L}l∈L with x∈lx \in lx∈l and role(x,l)=ρ\mathrm{role}(x, l) = \rhorole(x,l)=ρ.
  2. Distinct roles at a point: for every point x∈[n]x \in [n]x∈[n] and all lines l,l′∈Ll, l' \in \mathcal{L}l,l′∈L with x∈lx \in lx∈l and x∈l′x \in l'x∈l′,
role(x,l)=role(x,l′)  ⟹  l=l′.\mathrm{role}(x, l) = \mathrm{role}(x, l') \implies l = l'.role(x,l)=role(x,l′)⟹l=l′.

Taken together, (2) and (3) say that for each point xxx, the map l↦role(x,l)l \mapsto \mathrm{role}(x, l)l↦role(x,l) is a bijection from the lines through xxx onto {0,1,2}\{0,1,2\}{0,1,2}. Condition (1) says that on each line, x↦role(x,l)x \mapsto \mathrm{role}(x, l)x↦role(x,l) is injective. In the degenerate case n=1n = 1n=1 the only admissible SSS has no lines, and then (2) cannot hold, so the hypothesis is unsatisfiable for n=1n = 1n=1. The case n=0n = 0n=0 is excluded by n>0n > 0n>0.

Conclusion ("SSS is the Fano plane"). There exists a bijection e:[n]→{0,1,…,6}e : [n] \to \{0, 1, \dots, 6\}e:[n]→{0,1,…,6} such that the family of images of the lines under eee is exactly the Fano line set:

{ e(l):l∈L }={ Fi:i∈Z/7 },Fi={ i, i+1, i+3 } (arithmetic mod 7).\{\, e(l) : l \in \mathcal{L} \,\} = \{\, F_i : i \in \mathbb{Z}/7 \,\}, \qquad F_i = \{\, i,\ i+1,\ i+3 \,\} \ (\text{arithmetic mod } 7).{e(l):l∈L}={Fi​:i∈Z/7},Fi​={i, i+1, i+3} (arithmetic mod 7).

Explicitly, the seven Fano lines are

{0,1,3}, {1,2,4}, {2,3,5}, {3,4,6}, {4,5,0}, {5,6,1}, {6,0,2}.\{0,1,3\},\ \{1,2,4\},\ \{2,3,5\},\ \{3,4,6\},\ \{4,5,0\},\ \{5,6,1\},\ \{6,0,2\}.{0,1,3}, {1,2,4}, {2,3,5}, {3,4,6}, {4,5,0}, {5,6,1}, {6,0,2}.

This is an equality of families of sets. Because a bijection [n]→{0,…,6}[n] \to \{0,\dots,6\}[n]→{0,…,6} exists, the conclusion in particular forces n=7n = 7n=7.

In summary, the statement asserts: for every n>0n > 0n>0, every Steiner triple system SSS on [n][n][n] as above, and every role function satisfying (1)–(3), the number of points is n=7n = 7n=7 and SSS is isomorphic, via a relabelling of the points, to the Fano plane with lines {i,i+1,i+3} mod 7\{i, i+1, i+3\} \bmod 7{i,i+1,i+3}mod7.

Human review
  • Endorsed by Shuze Chen · Sep 25, 2026

    Confirmed by the moderator at approval.

  • Endorsed by ShapeZero · Sep 25, 2026

    Confirmed by the mission captain (proposal self-audit).

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