The graceful tree conjecture
Can every tree be labelled so its edge differences are all distinct?
▸Motivation
Ringel 1963 and Kotzig; the term 'graceful' is Golomb's. Verified for every tree on at most 35 vertices.
It implies Ringel's conjecture on decomposing complete graphs into copies of a tree — which was itself proved in 2020 for large trees by Montgomery, Pokrovskiy and Sudakov, without settling gracefulness.
Ringel 1963; Kotzig. Verified to 35 vertices.
▸Lean API
import Mathlib.Combinatorics.SimpleGraph.Acyclic
namespace Conjectura.GT007
/-- Ringel and Kotzig, 1963: can the vertices of every tree on `n` vertices be
labelled `0` to `n−1` so that the `n−1` edge differences are exactly `1` through
`n−1`, each once? Verified for all trees on at most 35 vertices. -/
def goal : Prop :=
∀ (n : ℕ) (T : SimpleGraph (Fin n)) [DecidableRel T.Adj], T.IsTree →
∃ f : Fin n → Fin n, Function.Bijective f ∧
Function.Bijective (fun e : T.edgeSet =>
Sym2.lift ⟨fun u v => max (f u : ℕ) (f v) - min (f u : ℕ) (f v),
by intro u v; simp [max_comm, min_comm]⟩ (e : Sym2 (Fin n)))
end Conjectura.GT007▸Definition6
▸Related work0
▸For your AI
I am proving a theorem in Lean 4 and submitting it to Conjectura.
## Problem GT007 — The graceful tree conjecture
Can every tree be labelled so its edge differences are all distinct?
## 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 : ∀ (n : ℕ) (T : SimpleGraph (Fin n)) [DecidableRel T.Adj], T.IsTree →
∃ f : Fin n → Fin n, Function.Bijective f ∧ Function.Bijective (...) := by
sorry
```
## The file I submit
```lean
import Conjectura.Problems.GT007.Statement
namespace Submission
theorem solution : Conjectura.GT007.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.
### Adj
structure
```lean
T.Adj
```
Defined in Mathlib.
### IsTree
structure
```lean
T.IsTree
```
Defined in Mathlib.
### Bijective
def
```lean
Function.Bijective
```
Defined in Mathlib.
### edgeSet
abbrev
```lean
T.edgeSet
```
Defined in Mathlib.
### lift
def
```lean
Sym2.lift
```
Defined in Mathlib.
### Sym2
abbrev
```lean
Sym2
```
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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