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 change of basis gives different factors with the same adapter update

Example

Take $d = k = 2$ , $r = 1$ , $c = 1$ , $W_{0} = 0$ , $B = (1, 2)^{T}$ , and $A = (3, 4)$ . Then

$B A = (\begin{array}{cc} 3 & 4 \\ 6 & 8 \end{array}) .$

The invertible one-dimensional change of basis $M = (3)$ gives $B^{'} = B M = (3, 6)^{T}$ and $A^{'} = M^{- 1} A = (1, 4 / 3)$ . These different factors produce the same adapter update.

Facts & Assumptions

Given: The specified matrices and the invertible scalar $M = 3$ .

[F1]

Rank capacity and factorization ambiguity of a low-rank adapter: for every invertible $M$ , the transformed factors $B M$ and $M^{- 1} A$ represent the same update.

Verification

1.1givenalgebra

Direct multiplication gives $B A = (\begin{array}{cc} 3 & 4 \\ 6 & 8 \end{array})$ .

2.1givenF1step 1.1algebra∎

Since $M = 3$ is invertible, $B^{'} = (3, 6)^{T}$ and $A^{'} = (1, 4 / 3)$ , and direct multiplication gives $B^{'} A^{'} = (\begin{array}{cc} 3 & 4 \\ 6 & 8 \end{array}) = B A$ , as also guaranteed by [F1].

Depends on

Rank capacity and factorization ambiguity of a low-rank adapter

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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