FormalConjectures/ErdosProblems/71.lean
Explanation legend Lean4 Domain Explanation Language (L4DEL)
L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.
- L4DEL
- Lean4 Domain Explanation Language
Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.
Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1
1 declarations, 1 contain sorry, 0 declared as axioms
Erdős Problem 71
References:
- erdosproblems.com/71
- [Bo77] Bollobás, Béla, Cycles modulo $k$. Bull. London Math. Soc. (1977), 97-98.
- [Er82e] Erdős, Paul, Some of my favourite problems which recently have been solved. (1982), 59--79.
- [Er95] Erdős, Paul, Some of my favourite problems in number theory, combinatorics, and geometry. Resenhas (1995), 165-186.
- [Er97b] Erdős, Paul, Some old and new problems in various branches of combinatorics. Discrete Math. (1997), 227-231.
Imports (1)
- theorem
Erdos71.erdos_71Is it true that for every infinite arithmetic progression sorry line 53 sourceSource documentation
Is it true that for every infinite arithmetic progression $P$ which contains even numbers there is some constant $c=c(P)$ such that every graph with average degree at least $c$ contains a cycle whose length is in $P$?
In [Er82e] Erdős credits this conjecture to himself and Burr. This has been proved by Bollobás [Bo77]. The best dependence of the constant $c(P)$ is unknown.
The infinite arithmetic progression is encoded as a set $P \subseteq \mathbb{N}$ satisfying
P.IsAPOfLength ⊤(which forces a positive common difference), and "contains even numbers" as the existence of an even element. The average degree of a finite simple graph isSimpleGraph.averageDegree, i.e. $(\sum_v \deg v)/|V| \in \mathbb{Q}$, and a cycle whose length is in $P$ is a cycle walkwwithw.length ∈ P.depends on sorry what this means
Statement uses:
Iff,True,Set,Nat, Set.IsAPOfLength,Top.top,ENat,Exists,And,Membership.mem,Even,Rat,Fintype,DecidableEq,SimpleGraph,DecidableRel,SimpleGraph.Adj,LE.le, SimpleGraph.averageDegree,SimpleGraph.Walkand 2 moreProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous,OfNat.ofNat11 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_71 : answer(True) ↔ ∀ P : Set ℕ, P.IsAPOfLength ⊤ → (∃ n ∈ P, Even n) → ∃ c : ℚ, ∀ (V : Type) [Fintype V] [DecidableEq V] (G : SimpleGraph V) [DecidableRel G.Adj], c ≤ G.averageDegree → ∃ (v : V) (w : G.Walk v v), w.IsCycle ∧ w.length ∈ P := by …Mentions: c