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-initialized descent on an underdetermined system

Example

Take $X = [1 1] \in R^{1 \times 2}$ , $y = 1$ , and $F (w) = \frac{1}{2} (w_{1} + w_{2} - 1)^{2}$ . Starting from $w_{0} = (0, 0)$ , use fixed step $η = \frac{1}{4}$ . The iterates are $w_{k} = \frac{1}{2} (1 - 2^{- k}) (1, 1)$ and converge to $(\frac{1}{2}, \frac{1}{2})$ , the unique minimum-Euclidean-norm least-squares solution among the nonunique minimizers $w_{1} + w_{2} = 1$ .

Facts & Assumptions

Given: The stated matrix, response, objective, initial vector, and fixed step.

[F1]

Zero-initialized gradient descent selects minimum-norm least squares: when $X \neq 0$ and $0 < η \leq 1 / λ_{\max} (X^{T} X)$ , zero-initialized gradient descent converges to the unique minimum-norm least-squares minimizer.

Verification

1.1givenF1algebra

Here $X^{T} X = (\begin{array}{cc} 1 & 1 \\ 1 & 1 \end{array})$ has eigenvalues $2$ and $0$ , so $λ_{\max} = 2$ and $0 < η = \frac{1}{4} \leq \frac{1}{2}$ . Thus the general selector theorem [F1] applies.

2.1givenstep 1.1algebrainduction

If $w_{k} = (a_{k}, a_{k})$ , then $\nabla F (w_{k}) = (2 a_{k} - 1) (1, 1)$ and the update gives $a_{k + 1} = a_{k} - \frac{1}{4} (2 a_{k} - 1) = \frac{1}{2} a_{k} + \frac{1}{4}$ . Starting with $a_{0} = 0$ , induction yields $a_{k} = \frac{1}{2} (1 - 2^{- k})$ , so $w_{k} \to (\frac{1}{2}, \frac{1}{2})$ .

3.1givenstep 2.1algebra∎

The least-squares minimum is zero, attained exactly when $w_{1} + w_{2} = 1$ . For every such vector, $w_{1}^{2} + w_{2}^{2} = \frac{1}{2} + \frac{1}{2} (w_{1} - w_{2})^{2} \geq \frac{1}{2}$ , with equality exactly when $w_{1} = w_{2} = \frac{1}{2}$ . Hence $(\frac{1}{2}, \frac{1}{2})$ is the unique minimum-norm solution reached in step 2.1.

Depends on

Zero-initialized gradient descent selects minimum-norm least squares

Used by

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