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 line constraint removes one kernel direction

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 $C_{0} = {(t, 0) : t \in R}$ . Every scalar datum $y \in R$ has exactly one admissible preimage, and $(C_{0} - C_{0}) \cap \ker A = {0}$ ; this constraint removes the vertical null direction.

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 $C_{0} = {(t, 0) : t \in R}$ .

[F1]

Admissible constraint classes and restricted identifiability: restricted identifiability means injectivity of $A ∣_{C}$ , and $C - C = {x - x^{'} : x, x^{'} \in C}$ .

[F2]

A constraint removes kernel ambiguity exactly when its differences avoid the kernel: $A ∣_{C}$ is injective if and only if $(C - C) \cap \ker A = {0}$ .

Verification

1.1givenF1algebra

From $A (s, t) = s$ , $\ker A = {(0, t) : t \in R}$ . Also $C_{0} - C_{0} = {(t, 0) : t \in R}$ , whose intersection with this vertical kernel is ${(0, 0)}$ .

2.1givenF1F2step 1.1constructalgebra∎

By [F2] and step 1.1, $A ∣_{C_{0}}$ is injective. For any $y \in R$ , $(y, 0) \in C_{0}$ and $A (y, 0) = y$ , so it is the unique admissible preimage of $y$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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