LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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.

The generated lambda-system exists and is minimal

Statement

For every set $X$ and every $E \subseteq P (X)$ , the family $λ_{X} (E)$ is a lambda-system on $X$ , contains $E$ , and is contained in every lambda-system on $X$ that contains $E$ .

Facts & Assumptions

Given: A set $X$ , a family $E \subseteq P (X)$ , and the intersection definition of $λ_{X} (E)$ in The lambda-system generated by a family of sets.

Proof

technique · direct

1.1given

A nonempty intersection of lambda-systems on $X$ contains $X$ . If $A \subseteq B$ lie in every member, then $B ∖ A$ lies in every member; and if $(A_{n})$ is increasing and lies in every member, then $⋃_{n} A_{n}$ lies in every member. Thus the intersection is a lambda-system.

1.2givenconstruct

The power set $P (X)$ is a lambda-system containing $E$ , so the family intersected in the definition of $λ_{X} (E)$ is nonempty.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 make $λ_{X} (E)$ a lambda-system. Every generator lies in every lambda-system being intersected, while an intersection is contained in each of its factors, so $E \subseteq λ_{X} (E)$ and $λ_{X} (E)$ is minimal.

Depends on

The lambda-system generated by a family of sets

Used by

Cited to discharge well-definedness by The lambda-system generated by a family of sets.

Dependency tree · two levels

2 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

A. Dembo, Probability Theory lecture notes, proof of Theorem 1.1.38 (standard reference, not scraped)