TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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.

Probability is additive on every finite pairwise-disjoint family of events

Statement

Let $(A_{i})_{i \in I}$ be a finite pairwise-disjoint family of events in a finite probability space. Then $P ⁣ (⋃_{i \in I} A_{i}) = \sum_{i \in I} P (A_{i}) .$ This includes the empty and one-member families.

Facts & Assumptions

Given: A finite probability space $(Ω, w)$ and a finite pairwise-disjoint family $(A_{i})_{i \in I}$ .

[L1]

Event probability is the finite sum of outcome weights (Finite probability spaces, outcome weights, events, and event probabilities).

[L2]

A finite sum may be reindexed by a bijection, split over a disjoint union, and evaluated in either order over a finite product (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

Proof

technique · direct

1.1

Pairwise disjointness makes $(i, ω) \mapsto ω$ a bijection from ${(i, ω) : i \in I, ω \in A_{i}}$ onto $⋃_{i \in I} A_{i}$ .

given

2.1

Reindexing by this bijection and summing first over each fibre gives $P (⋃_{i} A_{i}) = \sum_{i \in I} \sum_{ω \in A_{i}} w (ω) = \sum_{i \in I} P (A_{i})$ .

step 1.1L1L2

3.1

If $I = \emptyset$ , both sides are the empty sum $0$ ; if $I$ has one member, step 2.1 is the identity $P (A) = P (A)$ .

step 2.1∎

Depends on

Used by

Dependency tree · two levels

11 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

C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd ed., Theorem 1.2 (standard reference, not scraped)
H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 1.3.2 (standard reference, not scraped)