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

The multiplication rule and finite chain rule for conditional probability

Statement

If $P (B) > 0$ , then $P (A \cap B) = P (A ∣ B) P (B) .$ More generally, let $A_{0}, \dots, A_{n - 1}$ be events such that $P (A_{0} \cap \dots \cap A_{k - 1}) > 0$ for every $1 \leq k < n$ . Then $P ⁣ (⋂_{i < n} A_{i}) = P (A_{0}) \prod_{k = 1}^{n - 1} P ⁣ (A_{k} ∣ ⋂_{i < k} A_{i}) .$ For $n = 0$ the intersection and product both have probability or value $1$ ; for $n = 1$ the formula is $P (A_{0}) = P (A_{0})$ .

Facts & Assumptions

Given: Events satisfying the positivity hypotheses in the Statement.

[L1]

For $P (B) > 0$ , $P (A ∣ B) = P (A \cap B) / P (B)$ (Conditional probability $P (A ∣ B)$ for $P (B) > 0$ ).

Proof

technique · induction

1.1

Multiplying the identity in [L1] by the positive denominator gives $P (A \cap B) = P (A ∣ B) P (B)$ .

L1algebra

1.2

For $n = 0$ the empty intersection is $Ω$ and the empty product is $1$ , while for $n = 1$ the asserted identity is immediate.

basealgebra

1.3

Assume the chain formula holds for $n = k \geq 1$ , and suppose the stated prefix probabilities are positive for $A_{0}, \dots, A_{k}$ .

2.1

Apply step 1.1 to $A = A_{k}$ and $B = ⋂_{i < k} A_{i}$ , then substitute the induction hypothesis for $P (B)$ ; this gives the chain formula for $k + 1$ .

step 1.1step 1.3algebra

3.1

Induction proves the formula for every finite $n$ , and every displayed conditional probability has a positive conditioning event by hypothesis.

step 1.2step 2.1discharge-induction∎

Depends on

Conditional probability $\mathbb P(A\mid B)$ for $\mathbb P(B)>0$

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 6 results over 3 levels. 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

C. M. Grinstead and J. L. Snell, Introduction to Probability, 2nd ed., equation (4.2) (standard reference, not scraped)
H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 1.4 (standard reference, not scraped)