LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If $F$ does not shatter $T$ then $x_{T}$ agrees on ${v_{F} : F \in F}$ with a combination of the $x_{S}$ for $S ⊊ T$

Statement

Let $K$ be a field, let $F \subseteq P ([n])$ , and let $T \subseteq [n]$ . If $F$ does not shatter $T$ , then on the set of incidence vectors ${v_{F} : F \in F}$ the monomial

$x_{T} : = \prod_{i \in T} x_{i}$

agrees with a $K$ -linear combination of the monomials $x_{S}$ with $S ⊊ T$ .

Facts & Assumptions

Given: a field $K$ , a family $F \subseteq P ([n])$ , and a set $T \subseteq [n]$ that is not shattered by $F$ ; incidence vectors and polynomials are taken over $K$ .

[F1]

Since $T$ is not shattered, there is some subset $A \subseteq T$ that is not realised as $F \cap T$ by any member $F$ of $F$ (Shattering and the Vapnik–Chervonenkis dimension of a set family).

[F2]

The incidence vector $v_{F}$ has coordinate $1$ exactly on the elements of $F$ (The incidence vector $v_{A} \in F^{n}$ of a subset $A \subseteq [n]$ over a stated field).

Proof

technique · direct

1.1

Choose $A \subseteq T$ as in [F1], and define $g (x) : = \prod_{i \in A} x_{i} \prod_{i \in T ∖ A} (1 - x_{i}) .$

F1construct

2.1

For any $F \in F$ , the value $g (v_{F})$ is $1$ exactly when $F \cap T = A$ , and it is $0$ otherwise. Since no member of $F$ realises the trace $A$ , step 1.1 gives $g (v_{F}) = 0$ for every $F \in F$ .

F2step 1.1

3.1

Expanding the product in step 1.1 gives $g = (- 1)^{∣ T ∖ A ∣} x_{T} + \sum_{A \subseteq S ⊊ T} (\pm 1) x_{S} .$ Since step 2.1 says $g$ is the zero function on the incidence vectors of $F$ , this rearranges there to an expression of $x_{T}$ as a linear combination of the $x_{S}$ with $S ⊊ T$ .

step 1.1step 2.1∎

Depends on

Used by

A second proof of Sauer–Shelah, from the multilinear polynomial space Theorem

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

L. Babai and P. Frankl, Linear Algebra Methods in Combinatorics, §7.4 (standard reference, not scraped)

If FF does not shatter TT then xTxT​ agrees on {vF:F∈F}{vF​:F∈F} with a combination of the xSxS​ for S⊊TS⊊T