FormalConjectures/ErdosProblems/722.lean

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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 722

References:

Imports (1)
  1. theorem Erdos722.erdos_722 Let $k>r$ and $n$ be sufficiently large in terms of $k$ and $r$ sorry line 49 source

    Source documentation

    Let $k>r$ and $n$ be sufficiently large in terms of $k$ and $r$. Does there always exist a block $r-(n,k,1)$ design (or Steiner system with parameters $(n,k,r)$), provided the trivial necessary divisibility conditions $\binom{k-i}{r-i}\mid \binom{n-i}{r-i}$ are satisfied for every $0\leq i<r$? That is, can one find a family of $\binom{n}{r}\binom{k}{r}^{-1}$ many subsets of $\{1,\ldots,n\}$, all of size $k$, such that any $A\subseteq \{1,\ldots,n\}$ of size $r$ is contained in exactly one set in the family?

    This was proved by Keevash [Ke14] for all $(r,k)$.

    depends on sorry what this means

    Statement uses: Iff, True, Nat, LE.le, OfNat.ofNat, LT.lt, Filter.Eventually, Dvd.dvd, Nat.choose, HSub.hSub, Exists, Finset, Fin, Finset.IsBlockDesign, Filter.atTop

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

    14 instances, coercions and auxiliary constants left out.

    Tactic trace: 1 steps

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

    theorem erdos_722 :
        answer(True) ↔ ∀ r k : ℕ, 1 ≤ r → r < k → ∀ᶠ n : ℕ in atTop,
        (∀ i < r, (k - i).choose (r - i) ∣ (n - i).choose (r - i)) →
          ∃ H : Finset (Finset (Fin n)), H.IsBlockDesign r k := by
    …