TheoremStatement: AI-adaptedProof: AI-adaptedprecheck 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 · two levels

2 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., equation (4.2) (standard reference, not scraped)
H. Pishro-Nik, Introduction to Probability, Statistics, and Random Processes, Section 1.4 (standard reference, not scraped)