ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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 half-space constraint leaves a continuous kernel ambiguity

Example

Use $X = R^{2}$ with its Euclidean metric, $Y = R$ with its absolute-value metric, and $U = X$ . Let $A : X \to Y$ be $A (s, t) = s$ , and restrict the truth class to the half-space $C_{+} = {(s, t) \in R^{2} : t \geq 0}$ . For every datum $y \in R$ , the constrained fiber is ${(y, t) : t \geq 0}$ , a continuous half-line of admissible truths; moreover $(C_{+} - C_{+}) \cap \ker A = {(0, t) : t \in R} \neq {0}$ .

Facts & Assumptions

Given: $X = R^{2}$ and $Y = R$ with the stated Euclidean and absolute-value metrics, $U = X$ , $A (s, t) = s$ and the nonempty constraint class $C_{+} = {(s, t) \in R^{2} : t \geq 0}$ .

[F1]

Admissible constraint classes and restricted identifiability: $C - C = {x - x^{'} : x, x^{'} \in C}$ , $\ker A = {x : A x = 0}$ , and restricted identifiability is injectivity on $C$ .

[F2]

A constraint removes kernel ambiguity exactly when its differences avoid the kernel: injectivity on $C$ implies $(C - C) \cap \ker A = {0}$ .

Verification

1.1givenF1constructalgebra

We have $\ker A = {(0, t) : t \in R}$ . If $(a, b) \in R^{2}$ with $b \geq 0$ , then $(a, b) = (a, b) - (0, 0)$ with both points in $C_{+}$ . If $b < 0$ , then $(a, b) = (a, 0) - (0, - b)$ with both points in $C_{+}$ . Thus $C_{+} - C_{+} = R^{2}$ , so its intersection with $\ker A$ is the nonzero vertical axis.

2.1givenF1F2step 1.1algebra∎

Since this intersection is nontrivial, the forward implication in [F2] rules out injectivity of $A ∣_{C_{+}}$ . Directly, $A (s, t) = y$ and $t \geq 0$ hold exactly when $(s, t) = (y, t)$ with $t \geq 0$ ; distinct such $t$ give distinct admissible truths with the same datum.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

5 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.