FormalConjectures/ErdosProblems/545.lean

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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

2 declarations, 1 contain sorry, 0 declared as axioms

Erdős Problem 545

References:

Imports (1)
  1. def Erdos545.knPlusTEdges The graph formed by connecting a new vertex (none) to $t$ of the vertices of $K_n$ (Fin n) line 40 source

    Source documentation

    The graph formed by connecting a new vertex (none) to $t$ of the vertices of $K_n$ (Fin n).

    standard axioms: propext what this means

    Statement uses: Nat, SimpleGraph, Option, Fin

    Proof uses: SimpleGraph.mk, Erdos545.knPlusTEdges.match_1, Ne, LT.lt, Unit, False

    4 instances, coercions and auxiliary constants left out.

    Tactic trace: 5 steps

    Open to load the checked before-and-after goal states.

    def knPlusTEdges (n t : ℕ) : SimpleGraph (Option (Fin n)) where
    …
  2. theorem Erdos545.erdos_545 Let $m$ be sufficiently large and let $G$ be a graph with $m$ edges and no isolated vertices sorry line 65 source

    Source documentation

    Let $m$ be sufficiently large and let $G$ be a graph with $m$ edges and no isolated vertices. Is the Ramsey number $R(G)$ maximised when $G$ is 'as complete as possible'? That is, if $m=\binom{n}{2}+t$ edges with $0\leq t < n$ then is $$R(G)\leq R(H),$$ where $H$ is the graph formed by connecting a new vertex to $t$ of the vertices of $K_n$?

    A question of Erdős and Graham. The restriction to sufficiently large $m$ excludes the small counterexamples recorded on the source page.

    This problem is #10 in Ramsey Theory in the graphs problem collection.

    depends on sorry what this means

    Statement uses: Iff, True, Filter.Eventually, Nat, LT.lt, Eq, HAdd.hAdd, Nat.choose, OfNat.ofNat, Fintype, SimpleGraph, DecidableRel, SimpleGraph.Adj, SimpleGraph.degree, Membership.mem, Set, SimpleGraph.neighborSet, Set.ncard, Sym2, SimpleGraph.edgeSet and 6 more

    Proof uses: sorryAx, Lean.Name, Bool.false, Lean.Name.num, Lean.Name.str, Lean.Name.anonymous

    22 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

    Open to load the checked before-and-after goal states.

    theorem erdos_545 : answer(sorry) ↔
        ∀ᶠ m : ℕ in atTop, ∀ (n t : ℕ), t < n → m = n.choose 2 + t →
          ∀ (V : Type) [Fintype V] (G : SimpleGraph V) [DecidableRel G.Adj],
            (∀ v, 0 < G.degree v) →
            G.edgeSet.ncard = m →
            SimpleGraph.diagonalGraphRamsey G ≤
              SimpleGraph.diagonalGraphRamsey (knPlusTEdges n t) := by
    …

    Mentions: SimpleGraph.diagonalGraphRamsey, Erdos545.knPlusTEdges