Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generated
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.

The computed KKT residual certificate bounds the exact linear-system residual

Statement

Suppose Finite-precision reduced-QR KKT solve with certified residual status successfully returns z^ and a certified η for a nonsingular square matrix K and vector b. Then ∥Kz^−b∥2≤η. If z∗=K−1b, then also ∥z^−z∗∥2≤∥K−1∥2η. The certificate may be large; the theorem makes no accuracy claim without an additional bound on ∥K−1∥2 and a small certified η.

Facts & Assumptions

Given: The operationwise arithmetic hypotheses, successful reduced QR and residual computation, and the returned certificate η.

[F1]

Finite-precision reduced-QR KKT solve with certified residual status returns the minimum of the rounded-residual and QR backward-error bounds.

[F2]

Rounded matrix products give certified linear and normal-equation residual allowances bounds the difference between the exact residual and the computed rounded residual.

[F3]

Computed reduced QR least squares is backward stable for full column rank supplies the QR backward representation and its perturbation bounds.

[F5]

The SQP KKT matrix is nonsingular under positive reduced curvature provides nonsingularity for a full-row-rank active KKT matrix with positive reduced curvature.

[A1]

The KKT matrix has full-row-rank constraints and positive curvature on their null space; all successful-operation hypotheses hold, and the stated certificate is computed without overflow.

Proof

1.1A1F1F2F3F4F5algebra

By [F1], a successful return includes successful exact-storage admission, so its stored K,b equal the intended data. By [F5], K is nonsingular. Put r=Kz^−b for those same intended data. By [F2], ∥r∥2≤∥r^∥2+W. By [F3], r=δb−ΔKz^, so ∥r∥2≤βb∥b∥2+βK∥K∥F∥z^∥2. These are precisely the two bounds in [F1].

2.1F1step 1.1algebra∎

Since [F1] returns their minimum, ∥Kz^−b∥2≤η. The exact solution satisfies Kz∗=b, hence K(z^−z∗)=r; multiplying by K−1 gives ∥z^−z∗∥2≤∥K−1∥2∥r∥2≤∥K−1∥2η.

Depends on

Used by

Dependency tree · two levels

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Sources