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The role postulates force exactly seven points

Proved
RolesForceSeven.roles_force_seven

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

combinatoricsfano-planesteiner-triple-systems

Let n≥1n \ge 1n≥1. Let SSS be a Steiner triple system on {0,…,n−1}\{0, \dots, n-1\}{0,…,n−1} — every line has exactly 333 points and every two distinct points lie on exactly one line — and suppose SSS has a role colouring: the three points of each line get three different roles in {0,1,2}\{0,1,2\}{0,1,2}, every point takes every role at least once, and every point takes every role at most once. Then

n=7.n = 7 .n=7.

The conclusion is the point count only; it does not assert that SSS is the Fano plane. The hypothesis n≥1n \ge 1n≥1 is necessary: the empty system satisfies every other condition and has 000 points.

Preamble
import Mathlib
import Definitions.Def_RolesForceSeven_sts
Formal statement
namespace RolesForceSeven
theorem roles_force_seven (n : ℕ) (hn : 0 < n) (S : STS n)
    (role : Fin n → Finset (Fin n) → Fin 3) (h : RoleColouring S role) :
    n = 7 := by sorry
end RolesForceSeven
Source
Shape Zero LLC, "Formal Proofs of the C1 Verification Package" (August 2026), §3.1, Theorem 3.6: 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, "Steiner system" (Steiner triple systems, replication number): https://en.wikipedia.org/wiki/Steiner_system ; Wikipedia, "Fano plane": https://en.wikipedia.org/wiki/Fano_plane ; Wikipedia, "Octonion" (Fano plane mnemonic for the multiplication of imaginary units): https://en.wikipedia.org/wiki/Octonion
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What the Lean code literally says, in plain math · claude-opus-5-5

Theorem roles_force_seven. 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 structure SSS. In the code, SSS is a "Steiner triple system on [n][n][n]". It is given by a finite family L\mathcal{L}L of subsets of [n][n][n], called lines, such that:

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

Nothing else is required. There is no divisibility or admissibility condition on nnn, and no stated requirement that every point lies on a line. (For n≥2n \ge 2n≥2 that follows from the pair condition. For n=1n = 1n=1 there are no pairs, and since [1][1][1] has no 333-element subsets, L=∅\mathcal{L} = \emptysetL=∅. For n=2n = 2n=2 no such structure exists.)

The role function. The theorem also takes an arbitrary function

role:[n]×P([n])→{0,1,2},\mathrm{role} : [n] \times \mathcal{P}([n]) \to \{0, 1, 2\},role:[n]×P([n])→{0,1,2},

written role(x,ℓ)\mathrm{role}(x, \ell)role(x,ℓ). It is defined on every pair of a point and a subset of [n][n][n], but the hypothesis below only uses its values at pairs (x,ℓ)(x, \ell)(x,ℓ) with ℓ∈L\ell \in \mathcal{L}ℓ∈L and x∈ℓx \in \ellx∈ℓ. The hypothesis hhh says that role\mathrm{role}role is a "role colouring" of SSS, meaning all three of the following conditions hold:

  1. Distinct roles within a line: for every line ℓ∈L\ell \in \mathcal{L}ℓ∈L and all x,y∈ℓx, y \in \ellx,y∈ℓ,
role(x,ℓ)=role(y,ℓ)  ⟹  x=y.\mathrm{role}(x, \ell) = \mathrm{role}(y, \ell) \implies x = y.role(x,ℓ)=role(y,ℓ)⟹x=y.

Since lines have 333 points, the three points of each line receive the three roles 0,1,20, 1, 20,1,2 in some order. 2. Every role occurs at every point: for every point x∈[n]x \in [n]x∈[n] and every role ρ∈{0,1,2}\rho \in \{0, 1, 2\}ρ∈{0,1,2}, there is a line ℓ∈L\ell \in \mathcal{L}ℓ∈L with x∈ℓx \in \ellx∈ℓ and role(x,ℓ)=ρ\mathrm{role}(x, \ell) = \rhorole(x,ℓ)=ρ. 3. Each role at most once per point: for every point x∈[n]x \in [n]x∈[n] and all lines ℓ,ℓ′∈L\ell, \ell' \in \mathcal{L}ℓ,ℓ′∈L with x∈ℓx \in \ellx∈ℓ and x∈ℓ′x \in \ell'x∈ℓ′,

role(x,ℓ)=role(x,ℓ′)  ⟹  ℓ=ℓ′.\mathrm{role}(x, \ell) = \mathrm{role}(x, \ell') \implies \ell = \ell'.role(x,ℓ)=role(x,ℓ′)⟹ℓ=ℓ′.

Together, conditions 2 and 3 say that for each point xxx, the map ℓ↦role(x,ℓ)\ell \mapsto \mathrm{role}(x, \ell)ℓ↦role(x,ℓ) is a bijection from the lines through xxx onto {0,1,2}\{0, 1, 2\}{0,1,2}. So every point lies on exactly three lines.

Conclusion. For every such n>0n > 0n>0, every such SSS and every such role\mathrm{role}role satisfying conditions 1–3,

n=7.n = 7.n=7.

The conclusion is only about nnn. It says nothing more about SSS or role\mathrm{role}role, and it does not claim that a structure satisfying the hypotheses exists for n=7n = 7n=7.

Degenerate cases. The case n=0n = 0n=0 is excluded by the hypothesis n>0n > 0n>0. For n=1n = 1n=1 the hypotheses cannot all hold: L=∅\mathcal{L} = \emptysetL=∅, so condition 2 fails. For n=2n = 2n=2 no SSS exists at all. In both of these cases the statement is vacuously true.

Human review
  • Endorsed by Shuze Chen · Sep 24, 2026

    Confirmed by the moderator at approval.

  • Endorsed by ShapeZero · Sep 24, 2026

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

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