ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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 boundary score cone differs from its tangent space

Example

Assume the general DQM density-version convention uses AC and the mean-zero Hilbert-space convention uses Countable Choice; the two-point densities below are explicit, so this proof itself makes no choice. Let $X = {- 1, + 1}$ with all subsets measurable and let $P_{θ} {x} = (1 + θ x) / 2$ for $θ \in [0, 1]$ . At the boundary law $P_{0}$ , consider allowed one-sided paths $t \mapsto P_{θ (t)}$ with $θ (0) = 0$ , $θ (t) \in [0, 1]$ , and an existing right derivative at zero. Their DQM score set is the cone ${a X : a \geq 0}$ , while its closed linear tangent space is all of $L_{0}^{2} (P_{0}) = {b X : b \in R}$ . Thus the score set and its linear tangent space are distinct.

Facts & Assumptions

Given: AC and Countable Choice, the restricted model ${P_{θ} : 0 \leq θ \leq 1}$ on the two-point sample space, based at $P_{0}$ .

[A1]

AC is the axiom that every family of nonempty sets has a choice function (The Axiom of Choice).

[A2]

Countable Choice gives a choice function for every countable family of nonempty sets (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

A one-sided DQM score satisfies $\sqrt{p_{t}} = \sqrt{p} + (t / 2) s \sqrt{p} + o_{L^{2}} (t)$ (A dominated differentiable-in-quadratic-mean submodel).

[F2]

Such a score has mean zero and is unique $P_{0}$ -almost surely (A DQM score has mean zero and is unique).

[F3]

The tangent set records the scores of allowed paths; the tangent space is the closure of their linear span (Tangent set of a statistical model at a law, Closed tangent space).

[F4]

The full power set is a sigma-algebra, and a measurable space is a set with a specified sigma-algebra (Sigma-algebras, Measurable spaces and measurable sets).

[F5]

Counting measure assigns a finite set its cardinality; on this two-point sample space its total mass is two (Counting measure on an arbitrary set).

[F6]

Counting measure is a measure, and a finite measure is sigma-finite (Counting measure is a measure, Finite, sigma-finite, and semifinite measures).

[F7]

For every $x > 0$ , the square-root function is differentiable with derivative $1 / (2 \sqrt{x})$ (Continuity and derivatives of positive-base real powers).

[F8]

Under Countable Choice, $L_{0}^{2} (P_{0})$ is closed in $L^{2} (P_{0})$ (The mean-zero L² space is a Hilbert space).

Verification

technique · Identify every one-sided score and compare its cone with its real linear span

1.1givenF4F5F6algebra

Every subset of ${- 1, + 1}$ and its complement lie in the full power set, and every countable union of such subsets is still a subset; thus the power set is a sigma-algebra by [F4], and with this sigma-algebra the sample set is a measurable space by [F4]. Counting measure on the two-point set is a measure of total mass two by [F5, F6], hence sigma-finite by the constant sequence of covers $E_{n} = {- 1, + 1}$ ; its only null set is empty. For every $θ \in [0, 1]$ , the masses $(1 + θ) / 2$ and $(1 - θ) / 2$ are nonnegative and sum to one, so $P_{θ}$ is a probability measure dominated by counting measure. At $θ = 0$ both masses are $1 / 2$ , so $\sqrt{p_{0} (x)} = 1 / \sqrt{2}$ .

2.1F1step 1.1algebra

If a one-sided DQM path has score $s$ , [F1] on each atom gives $\sqrt{P_{θ (t)} {x}} = 1 / \sqrt{2} + t s (x) / (2 \sqrt{2}) + o (t)$ . Squaring each expansion yields $P_{θ (t)} {x} = 1 / 2 + t s (x) / 2 + o (t)$ .

2.2givenA1F1F7step 1.1algebra

Conversely, for each $a \geq 0$ , take $θ (t) = a t$ for sufficiently small $t \geq 0$ . Then $P_{θ (t)} {x} = (1 + a t x) / 2$ , and [F7] at $1 / 2$ gives $\sqrt{P_{θ (t)} {x}} = 1 / \sqrt{2} + t a x / (2 \sqrt{2}) + o (t)$ on each atom. The two-coordinate counting-measure $L^{2}$ remainder is $o (t)$ , so this path has score $a X$ . Its densities are explicit, so [A1] requires no selection in this construction. Thus every vector on the nonnegative ray is an allowed score.

3.1step 2.1F2algebra

Subtract the two mass expansions from step 2.1. Since $θ (t) = P_{θ (t)} {+ 1} - P_{θ (t)} {- 1}$ , we get $θ (t) / t \to (s (+ 1) - s (- 1)) / 2$ . By [F2], $P_{0} s = (s (+ 1) + s (- 1)) / 2 = 0$ , so this limit equals $s (+ 1)$ . The constraints $θ (t) \geq 0$ and $t > 0$ force the limit $a : = s (+ 1)$ to be nonnegative, and $s (- 1) = - a$ ; hence every score is $s (x) = a x$ with $a \geq 0$ .

4.1A2F3F8step 2.2algebra∎

The real span of ${a X : a \geq 0}$ is ${b X : b \in R}$ , and every mean-zero function on the equiprobable two-point space has values $(b, - b)$ , so this span is $L_{0}^{2} (P_{0})$ . This space is closed in $L^{2} (P_{0})$ by [F8], hence its closure—and therefore the tangent space by [F3]—is $L_{0}^{2} (P_{0})$ . It strictly contains the non-linear score cone because $X$ is in the cone but $- X$ is not.

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