Hadwiger's conjecture
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
▸Related work1
- Hadwiger's conjecture for K6-free graphsRobertson, Seymour, Thomas · 1993
The last case resolved, by reduction to the four colour theorem.
▸For your AI
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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