CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A symmetric chain decomposition gives a second proof of Sperner's bound

Statement

If $A$ has $n$ elements, every antichain in $B (A)$ has cardinality at most $(\binom{n}{⌊ n / 2 ⌋})$ .

Facts & Assumptions

Given: An $n$ -element set $A$ and an antichain $F$ in $B (A)$ .

[L1]

The Boolean lattice has a partition into symmetric chains (Every finite Boolean lattice has a symmetric chain decomposition).

[F1]

An antichain contains at most one element from any chain (Antichains, chain covers, and antichain covers of a poset).

[F2]

Rank $⌊ n / 2 ⌋$ consists of the $⌊ n / 2 ⌋$ -subsets of $A$ and has cardinality $(\binom{n}{⌊ n / 2 ⌋})$ (The Boolean lattice of subsets of a finite set and its rank levels, The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

Proof

technique · direct

1.1givenL1

Fix the symmetric saturated-chain decomposition supplied by [L1]. Every chain in it meets rank $⌊ n / 2 ⌋$ exactly once, because its consecutive ranks run from some $r \leq ⌊ n / 2 ⌋$ through $n - r \geq ⌊ n / 2 ⌋$ .

2.1step 1.1F2

Consequently the number of chains in the decomposition equals the cardinality of rank $⌊ n / 2 ⌋$ , hence equals $(\binom{n}{⌊ n / 2 ⌋})$ .

3.1step 2.1F1∎

By [F1], the antichain $F$ contains at most one member from each chain. Its cardinality is therefore at most the number in step 2.1, which is Sperner's bound.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

M. Keller and W. T. Trotter, Applied Combinatorics, §6.2 (standard reference, not scraped)