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 lambda-system generated by a pi-system is closed under finite intersections

Statement

If $P$ is a pi-system on $X$ , then its generated lambda-system $λ_{X} (P)$ is closed under binary intersections.

Facts & Assumptions

Given: A pi-system $P$ on $X$ and $D : = λ_{X} (P)$ .

[L1]

A pi-system is nonempty and closed under binary intersections (Pi-systems).

[L2]

The family $D$ is the smallest lambda-system on $X$ containing $P$ (The generated lambda-system exists and is minimal).

[L3]

For $A$ in a lambda-system $D$ , the family $D_{A} = {B \in D : A \cap B \in D}$ is a lambda-system (For a member A of a lambda-system D, the sets B with A intersection B in D form a lambda-system).

Proof

technique · direct

1.1L1L2L3

Fix $A \in P$ . By [L3], $D_{A}$ is a lambda-system. If $B \in P$ , then $A \cap B \in P \subseteq D$ by [L1] and [L2], so $P \subseteq D_{A}$ . Minimality in [L2] yields $D \subseteq D_{A}$ .

2.1step 1.1L2L3algebra

Now fix $B \in D$ . Symmetry of intersection and step 1.1 show $A \cap B \in D$ for every $A \in P$ , so $P \subseteq D_{B}$ . By [L3] and [L2], $D_{B}$ is a lambda-system containing $P$ and therefore contains $D$ .

3.1step 2.1∎

Thus for arbitrary $A, B \in D$ one has $A \in D_{B}$ , which means $A \cap B \in D$ .

Depends on

Used by

Dynkin's pi-lambda theorem Theorem

Dependency tree · two levels

5 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)