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

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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.

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

$⟨ v_{A}, v_{B} ⟩$ is the image of $∣ A \cap B ∣$ in $F$ ; over $F_{2}$ it is $0$ or $1$ according to the parity of $∣ A \cap B ∣$

Statement

Let $F$ be a field, let $A, B \subseteq [n]$ , and let $v_{A}, v_{B} \in F^{n}$ be their incidence vectors. Then

$⟨ v_{A}, v_{B} ⟩ = ∣ A \cap B ∣ \cdot 1_{F} .$

In particular:

over $R$ one has $⟨ v_{A}, v_{B} ⟩ = ∣ A \cap B ∣$ ;
over $F_{2}$ one has $⟨ v_{A}, v_{B} ⟩ = 1$ exactly when $∣ A \cap B ∣$ is odd, and it is $0$ exactly when $∣ A \cap B ∣$ is even;
taking $B = A$ gives $⟨ v_{A}, v_{A} ⟩ = ∣ A ∣ \cdot 1_{F}$ .

Facts & Assumptions

Given: a field $F$ , a natural number $n$ , and subsets $A, B \subseteq [n]$ .

[F1]

The incidence vector satisfies $(v_{A})_{i} = 1_{F}$ when $i \in A$ and $(v_{A})_{i} = 0_{F}$ otherwise (The incidence vector $v_{A} \in F^{n}$ of a subset $A \subseteq [n]$ over a stated field).

[F2]

The standard form is $⟨ x, y ⟩ = \sum_{i < n} x_{i} y_{i}$ (The standard bilinear form $⟨ x, y ⟩ = \sum_{i < n} x_{i} y_{i}$ on $F^{n}$ ).

Proof

technique · direct

1.1

For each index $i < n$ , the product $(v_{A})_{i} (v_{B})_{i}$ equals $1_{F}$ when $i \in A \cap B$ and equals $0_{F}$ otherwise.

2.1

Therefore the sum in [F2] contains exactly $∣ A \cap B ∣$ copies of $1_{F}$ and all remaining terms are $0_{F}$ , so $⟨ v_{A}, v_{B} ⟩ = ∣ A \cap B ∣ \cdot 1_{F}$ .

F2step 1.1

3.1

The three stated consequences follow immediately: over $R$ the scalar $∣ A \cap B ∣ \cdot 1_{R}$ is the integer itself, over $F_{2}$ it is $1$ or $0$ according to the parity of $∣ A \cap B ∣$ , and setting $B = A$ gives the final clause.

step 2.1∎

Remarks

This is the page's basic dictionary item. Every parity or intersection-size hypothesis below is rewritten through this lemma before any linear algebra is applied.

Depends on

Used by

Dependency tree · two levels

29 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, §§1.1, 2.3.1 (standard reference, not scraped)
J. Matousek, Thirty-three Miniatures, Miniature 3 (standard reference, not scraped)

⟨vA,vB⟩⟨vA​,vB​⟩ is the image of ∣A∩B∣∣A∩B∣ in FF; over F2F2​ it is 00 or 11 according to the parity of ∣A∩B∣∣A∩B∣