PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

A family is a lambda-system exactly when it contains $X$ and is closed under complements and countable disjoint unions

Statement

Let $X$ be a set and let $D \subseteq P (X)$ . Then $D$ is a lambda-system on $X$ if and only if

$X \in D$ ;
$X ∖ A \in D$ whenever $A \in D$ ;
$⋃_{n \in N} A_{n} \in D$ whenever $A_{0}, A_{1}, \dots \in D$ are pairwise disjoint.

Facts & Assumptions

Given: A set $X$ and a family $D \subseteq P (X)$ .

[L1]

A lambda-system on $X$ is a family $D \subseteq P (X)$ such that $X \in D$ ; if $A, B \in D$ and $A \subseteq B$ , then $B ∖ A \in D$ ; and if $A_{0} \subseteq A_{1} \subseteq \dots$ with every $A_{n} \in D$ , then $⋃_{n \in N} A_{n} \in D$ (Lambda-systems, or Dynkin systems).

Proof

technique · direct

1.1

Forward direction: assume, in this step and in every later step that cites it, that $D$ is a lambda-system. Then $X \in D$ by [L1], which is clause 1. For $A \in D$ we have $A \subseteq X$ and $X \in D$ , so $X ∖ A \in D$ by the relative-difference clause of [L1]; this is clause 2.

L1given

1.2

Reverse direction, whose hypothesis is independent of the forward branch: assume, in this step and in every later step that cites it, clauses 1, 2 and 3. Then $X \in D$ , which is the first lambda-system clause of [L1], and $\emptyset = X ∖ X \in D$ by clauses 1 and 2.

given

2.1

Return to the forward direction, so that the hypothesis in force is again the one of step 1.1, namely that $D$ is a lambda-system. Let $A_{0}, A_{1}, \dots \in D$ be pairwise disjoint and put $B_{n} : = ⋃_{k \leq n} A_{k}$ . We show $B_{n} \in D$ by induction on $n$ . For $n = 0$ , $B_{0} = A_{0} \in D$ . Suppose $B_{n} \in D$ . Disjointness gives $A_{n + 1} \subseteq X ∖ B_{n}$ , and $X ∖ B_{n} \in D$ by step 1.1, so $(X ∖ B_{n}) ∖ A_{n + 1} \in D$ by [L1]. Its complement in $X$ is $X ∖ ((X ∖ B_{n}) ∖ A_{n + 1}) = B_{n} \cup A_{n + 1} = B_{n + 1}$ , which lies in $D$ by step 1.1.

L1step 1.1algebra

2.2

Still under the clauses 1, 2 and 3 assumed in step 1.2, let $A, B \in D$ with $A \subseteq B$ . Then $X ∖ B \in D$ by clause 2, and $A \cap (X ∖ B) = \emptyset$ because $A \subseteq B$ . The sequence $A, X ∖ B, \emptyset, \emptyset, \dots$ is therefore a pairwise disjoint sequence in $D$ by step 1.2, so clause 3 gives $A \cup (X ∖ B) \in D$ , and clause 2 then gives $X ∖ (A \cup (X ∖ B)) = B ∖ A \in D$ , which is the relative-difference clause of [L1].

L1givenstep 1.2algebra

3.1

Still in the forward direction, the sets $B_{n}$ of step 2.1 increase and satisfy $⋃_{n} B_{n} = ⋃_{n} A_{n}$ , so $⋃_{n} A_{n} \in D$ by the increasing-union clause of [L1]. This is clause 3, and with step 1.1 it proves the forward direction.

L1step 1.1step 2.1

4.1

Still under the clauses assumed in step 1.2, let $A_{0} \subseteq A_{1} \subseteq \dots$ lie in $D$ . Put $C_{0} : = A_{0}$ and $C_{n + 1} : = A_{n + 1} ∖ A_{n}$ ; each $C_{n + 1}$ lies in $D$ by step 2.2, the $C_{n}$ are pairwise disjoint, and $⋃_{n} C_{n} = ⋃_{n} A_{n}$ . Clause 3 then gives $⋃_{n} A_{n} \in D$ , which is the increasing-union clause of [L1]. With steps 1.2 and 2.2 this proves the reverse direction. This proves the stated claim.

L1givenstep 1.2step 2.2algebra∎

Depends on

Lambda-systems, or Dynkin systems

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 1 result over 1 level. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. Dembo, Probability Theory lecture notes, Definition 1.1.36 and the Remark following it (standard reference, not scraped)

A family is a lambda-system exactly when it contains XX and is closed under complements and countable disjoint unions