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.

Zero-zero initialization stalls one step, while nonzero A can move

Example

Take a scalar layer with $d = k = r = 1$ , $c = 1$ , $W_{0} = 0$ , one data point $x_{1} = y_{1} = 1$ , loss $ℓ_{1} (z) = \frac{1}{2} (z - y_{1})^{2}$ , one update $T = 1$ , minibatch $I_{0} = (1)$ , and finite step size $η_{0} > 0$ . With $A_{0} = B_{0} = 0$ , the finite-budget procedure returns $A_{1} = B_{1} = 0$ . With the same data and $A_{0} = 1$ , $B_{0} = 0$ , it instead returns $A_{1} = 1$ and $B_{1} = η_{0}$ .

Facts & Assumptions

Given: The scalar data, loss, scale, initial factors, one-step horizon, and positive finite step size in the statement.

[F1]

Exact merging and gradients for a low-rank adapter: $\nabla_{B} L = c G A^{T}$ and $\nabla_{A} L = c B^{T} G$ .

[F2]

Finite-budget low-rank adaptation by gradient descent: the procedure initializes $B_{0} = 0$ and updates each factor by subtracting the step size times its reverse-mode gradient.

Verification

1.1givenalgebra

At either initialization, $W_{0} + c B_{0} A_{0} = 0$ , so the scalar loss derivative with respect to the weight is $G = ℓ_{1}^{'} (0) = 0 - 1 = - 1$ . The loss is a polynomial defined and differentiable on all of $R$ , so the one-step reverse-mode call is admissible.

2.1F1F2step 1.1algebra

If $A_{0} = B_{0} = 0$ , [F1] gives $\nabla_{B} L = (- 1) 0 = 0$ and $\nabla_{A} L = 0 (- 1) = 0$ . By [F2], subtracting these zero gradients leaves $A_{1} = B_{1} = 0$ .

3.1F1F2step 1.1algebra∎

If $A_{0} = 1$ and $B_{0} = 0$ , [F1] gives $\nabla_{A} L = 0$ and $\nabla_{B} L = (- 1) (1) = - 1$ . The update in [F2] therefore gives $A_{1} = 1 - η_{0} \cdot 0 = 1$ and $B_{1} = 0 - η_{0} (- 1) = η_{0}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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