Alphabeta Math
PropositionStatement: 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.

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⊆P(X). Then D is a lambda-system on X if and only if

  1. X∈D;
  2. X∖A∈D whenever A∈D;
  3. ⋃n∈NAn∈D whenever A0,A1,⋯∈D are pairwise disjoint.

Facts & Assumptions

Given: A set X and a family D⊆P(X).

[L1]

A lambda-system on X is a family D⊆P(X) such that X∈D; if A,B∈D and A⊆B, then B∖A∈D; and if A0⊆A1⊆⋯ with every An∈D, then ⋃n∈NAn∈D (Lambda-systems, or Dynkin systems).

Proof

technique · direct
1.1L1given

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

1.2given

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∈D, which is the first lambda-system clause of [L1], and ∅=X∖X∈D by clauses 1 and 2.

2.1L1step 1.1algebra

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 A0,A1,⋯∈D be pairwise disjoint and put Bn:=⋃k≤nAk. We show Bn∈D by induction on n. For n=0, B0=A0∈D. Suppose Bn∈D. Disjointness gives An+1⊆X∖Bn, and X∖Bn∈D by step 1.1, so (X∖Bn)∖An+1∈D by [L1]. Its complement in X is X∖((X∖Bn)∖An+1)=Bn∪An+1=Bn+1, which lies in D by step 1.1.

2.2L1givenstep 1.2algebra

Still under the clauses 1, 2 and 3 assumed in step 1.2, let A,B∈D with A⊆B. Then X∖B∈D by clause 2, and A∩(X∖B)=∅ because A⊆B. The sequence A, X∖B, ∅, ∅,… is therefore a pairwise disjoint sequence in D by step 1.2, so clause 3 gives A∪(X∖B)∈D, and clause 2 then gives X∖(A∪(X∖B))=B∖A∈D, which is the relative-difference clause of [L1].

3.1L1step 1.1step 2.1

Still in the forward direction, the sets Bn of step 2.1 increase and satisfy ⋃nBn=⋃nAn, so ⋃nAn∈D by the increasing-union clause of [L1]. This is clause 3, and with step 1.1 it proves the forward direction.

4.1L1givenstep 1.2step 2.2algebra∎

Still under the clauses assumed in step 1.2, let A0⊆A1⊆⋯ lie in D. Put C0:=A0 and Cn+1:=An+1∖An; each Cn+1 lies in D by step 2.2, the Cn are pairwise disjoint, and ⋃nCn=⋃nAn. Clause 3 then gives ⋃nAn∈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.

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