Conjectura
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← problemsGT004openMathematics/ Combinatorics/ Graph theory

Hadwiger's conjecture

Maintainer — open

If a graph has no K_t minor, is it (t−1)-colourable?

Motivation

Hugo Hadwiger, 1943. Bollobás called it one of the deepest unsolved problems in graph theory.

The case t = 5 is equivalent to the four colour theorem, so any proof subsumes it. Robertson, Seymour and Thomas proved t = 6 in 1993 by reducing it to the four colour theorem. Nothing is known for t ≥ 7.

Hugo Hadwiger, 1943. t = 6 by Robertson, Seymour and Thomas, 1993.

Lean API
import Mathlib.Combinatorics.SimpleGraph.Coloring.Vertex
import Mathlib.Combinatorics.SimpleGraph.Finite

namespace Conjectura.GT004

/-- If a graph has no `Kₜ` minor, is it `(t−1)`-colourable? The case `t = 5` is
equivalent to the four colour theorem; `t = 6` was proved by Robertson, Seymour and
Thomas; everything beyond is open. -/
def goal : Prop :=
  ∀ (V : Type) [Fintype V] (G : SimpleGraph V) (t : ℕ),
    (¬ ∃ f : Fin t → Finset V,
        (∀ i, (f i).Nonempty) ∧
        (∀ i j, i ≠ j → Disjoint (f i) (f j)) ∧
        (∀ i j, i ≠ j → ∃ u ∈ f i, ∃ v ∈ f j, G.Adj u v)) →
      G.Colorable (t - 1)

end Conjectura.GT004
Definition6
FintypeFintype

class

FinsetFinset

structure

NonemptyNonempty

abbrev

DisjointDisjoint

def

AdjG.Adj

structure

ColorableG.Colorable

def

Related work1
For your AI
Download as .md
I am proving a theorem in Lean 4 and submitting it to Conjectura.

## Problem GT004 — Hadwiger's conjecture

If a graph has no K_t minor, is it (t−1)-colourable?

## Environment (fixed — do not assume anything newer)

- Lean toolchain: `leanprover/lean4:v4.33.0-rc1`
- Mathlib: `v4.33.0-rc1`

If a lemma you want does not exist in that Mathlib, prove it inline instead of
importing something newer.

## The exact statement I must prove

```lean
theorem solution : ∀ (V : Type) [Fintype V] (G : SimpleGraph V) (t : ℕ), ... → G.Colorable (t - 1) := by
  sorry
```

## The file I submit

```lean
import Conjectura.Problems.GT004.Statement

namespace Submission

theorem solution : Conjectura.GT004.goal := by
  sorry

end Submission

```

## The Lean definitions of every term in this problem

These are the actual definitions your proof will be checked against. Do not
substitute your own version of any of them.

### Fintype

class

```lean
Fintype
```
Defined in Mathlib.

### Finset

structure

```lean
Finset
```
Defined in Mathlib.

### Nonempty

abbrev

```lean
Nonempty
```
Defined in Mathlib.

### Disjoint

def

```lean
Disjoint
```
Defined in Mathlib.

### Adj

structure

```lean
G.Adj
```
Defined in Mathlib.

### Colorable

def

```lean
G.Colorable
```
Defined in Mathlib.

## Rules — submissions violating these are rejected automatically

1. **Do not change the name or type of `solution`.** It must satisfy the
   statement above exactly.
2. **Do not redefine or shadow anything from the problem's Statement module.**
   Declaring your own `goal`, or redefining a name it depends on, produces a
   proof of a *different* statement and is rejected. This is the single most
   common rejection.
3. **No `sorry`** anywhere, including in helper lemmas. It surfaces as the
   axiom `sorryAx` and is detected transitively through imports.
4. **No `native_decide`** — it surfaces as `Lean.ofReduceBool` and is not
   accepted, because it trusts compiled code rather than the kernel.
5. Only these axioms are permitted: `propext`, `Classical.choice`,
   `Quot.sound`.
6. Follow Mathlib style: hypotheses left of the colon, explicit types,
   `snake_case` theorem names, `UpperCamelCase` types.

## What I want from you

Here is my argument in informal mathematics:

> [PASTE YOUR PROOF SKETCH HERE]

Turn it into Lean 4 that compiles under the environment above and satisfies the
statement exactly. Where you are unsure a lemma exists in this Mathlib version,
say so explicitly rather than guessing a name.

Submissions are not open yet

Conjectura is in beta. You can read every statement, every definition and the Lean behind them, and download the exact files the checker uses — but proofs are not being accepted yet.

The reason is a deliberate order of operations. Accepting a proof means running a stranger’s code and standing behind a verdict, and no statement here yet carries a researcher’s name. A machine-checked answer to a question nobody has vouched for is worth very little, so the vouching comes first.

The English write-ups are also switched off during the beta. Nothing on this page is generated by a model.

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