ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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.

Recovering an exponential law from its tail

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

Fix $λ > 0$ and suppose a nonnegative random variable $X$ satisfies $P (X > t) = e^{- λ t} (t \geq 0) .$ Then $F_{X} (x) = {\begin{cases} 0, & x < 0, \\ 1 - e^{- λ x}, & x \geq 0, \end{cases}$ its law on $R$ is the exponential law with parameter $λ$ , and $E [X] = \frac{1}{λ} .$

Facts & Assumptions

Given: The Axiom of Countable Choice and a nonnegative random variable $X$ with $P (X > t) = e^{- λ t}$ for all $t \geq 0$ .

[L1]

The cumulative distribution function is $F_{X} (x) = P (X \leq x)$ (Cumulative distribution function of a real random variable).

[L2]

Assuming the Axiom of Countable Choice, distribution functions determine probability laws (Probability laws correspond to distribution functions).

[L3]

For nonnegative random variables, $E [X] = \int_{0}^{\infty} P (X > t) d t$ (Layer-cake formulas for random variables).

Verification

technique · direct

1.1L1given

Because $X \geq 0$ , one has $F_{X} (x) = 0$ for $x < 0$ . For $x \geq 0$ , $F_{X} (x) = 1 - P (X > x) = 1 - e^{- λ x} .$ So the displayed formula is the CDF of $X$ .

1.2L3algebra

The tail integral from [L3] gives $E [X] = \int_{0}^{\infty} e^{- λ t} d t = \frac{1}{λ} .$

2.1step 1.1L2

By [L2], that distribution function determines the law uniquely, and its interval increments are those of the exponential distribution with parameter $λ$ .

3.1step 1.1step 1.2step 2.1∎

Steps 1.1, 1.2, and 2.1 recover the law and its expectation from the tail.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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

Rick Durrett, Probability: Theory and Examples, 5th ed., Section 1.2 (standard reference, not scraped)