ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge 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 small left-preconditioned residual can hide a large physical residual

Example

For any ordered field $F$ and $0 < ϵ < 1$ , take the scalar system $A = [1]$ , $M = [ϵ^{- 1}]$ , $b = [1]$ , and trial value $x = 0$ . The physical residual has absolute value $1$ , while the left-preconditioned residual has absolute value $ϵ$ .

Facts & Assumptions

Given: An ordered field $F$ , a scalar $0 < ϵ < 1$ , and the scalar linear system $A x = b$ with $A = 1$ , preconditioner $M = ϵ^{- 1}$ , right-hand side $b = 1$ , and trial value $x = 0$ .

[F1]

Invertible preconditioners give equivalent linear systems, with the transformed residuals and errors written explicitly: for invertible $M$ , the left-preconditioned residual is $M^{- 1} r$ , where $r = b - A x$ .

[F2]

Field: every nonzero field element has a multiplicative inverse.

[F3]

Ordered field: the hypothesis $0 < ϵ < 1$ gives $0 < ϵ$ and $0 < 1 - ϵ$ , so $ϵ$ and $1 - ϵ$ lie in the positive cone; closure under addition gives $1 = ϵ + (1 - ϵ) > 0$ as well.

[F4]

Absolute value in an ordered field: if $z \geq 0$ then $∣ z ∣ = z$ .

Verification

1.1givenF2F3algebra

Since $ϵ > 0$ , it is nonzero; [F2] gives its inverse $ϵ^{- 1}$ . With $M = ϵ^{- 1}$ , commutativity gives $M ϵ = ϵ M = 1$ , so $M^{- 1} = ϵ$ . The physical residual is $r = b - A x = 1 - 1 \cdot 0 = 1$ .

2.1F1F3F4step 1.1algebra

By [F1], the left-preconditioned residual is $M^{- 1} r = ϵ$ . Both $1$ and $ϵ$ are positive by [F3], so [F4] gives absolute values $1$ and $ϵ$ , respectively.

3.1givenstep 1.1step 2.1algebra∎

Since $0 < ϵ < 1$ , the left-preconditioned residual has strictly smaller absolute value than the physical residual. The ratio can be made as small as the chosen $ϵ$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Yousef Saad, Iterative Methods for Sparse Linear Systems, 2nd ed., §9.3.4, printed pp. 285–286 (standard reference, not scraped)
Richard Barrett et al., Templates for the Solution of Linear Systems: Building Blocks for Iterative Methods, 2nd ed., §3.1.2, printed pp. 35–36 (standard reference, not scraped)