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.

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An estimated blank increases the effective normal scale

Example

Assume AC, $σ_{0} > 0$ , and $τ > 0$ . Suppose the blank-corrected response is explicitly modeled by $D_{c} \sim N (c, σ_{0}^{2} + τ^{2})$ for every $c \geq 0$ , where $τ^{2} > 0$ is the declared additional variance contribution from estimating the blank. Put $σ_{e f f} = \sqrt{σ_{0}^{2} + τ^{2}}$ . The exact normal formulas use $σ_{e f f}$ in place of $σ_{0}$ : $L_{C} = z_{1 - α} σ_{e f f}, L_{D} = (z_{1 - α} + z_{1 - β}) σ_{e f f}, L_{Q} = \frac{σ_{e f f}}{k_{Q}} .$ This is a stated comparison model, not a claim that every blank-estimation procedure produces an independent normal error with this variance. The blank-correction uncertainty must be included in the response law used to calculate the limits.

Facts & Assumptions

Given: AC; $σ_{0} > 0$ , $τ > 0$ , $0 < α, β < 1 / 2$ , $k_{Q} > 0$ , and the corrected-response law in the Example.

[A1]

AC is assumed by the known-normal limit theorem used below. The Axiom of Choice Exact limits in the known-blank equal-variance normal model

[F1]

For a known-blank equal-variance normal family with scale $σ > 0$ , the three limits are $z_{1 - α} σ$ , $(z_{1 - α} + z_{1 - β}) σ$ , and $σ / k_{Q}$ . Exact limits in the known-blank equal-variance normal model

Verification

Given: AC; $σ_{0} > 0$ , $τ > 0$ , $0 < α, β < 1 / 2$ , $k_{Q} > 0$ ; and the corrected-response model in the Example.

1.1givenA1F1algebra

Set $σ_{e f f} = \sqrt{σ_{0}^{2} + τ^{2}}$ . Since $σ_{0} > 0$ and $τ > 0$ , $σ_{e f f} > σ_{0} > 0$ . Thus the corrected-response family has exactly the form required by [F1], with $σ = σ_{e f f}$ .

2.1A1F1step 1.1algebra∎

Substitution into [F1] yields the three displayed limits. Since $0 < α, β < 1 / 2$ , both $z_{1 - α}$ and $z_{1 - β}$ are positive; since $k_{Q} > 0$ , each multiplier of $σ_{e f f}$ is positive. Together with $σ_{e f f} > σ_{0}$ , each corresponding limit is strictly larger than its known-blank value.

Source notes

Eurachem §5.3.2 and Example 4 show that routine averaging and blank correction change the standard deviation used for LOD/LOQ; §5.3.5 notes that day-to-day blank variation can require intermediate-precision data. The example's exact normal law is an explicit assumption, and the limit formulas are proved locally from [F1].

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Sources

H. Cantwell (ed.), Eurachem Guide: The Fitness for Purpose of Analytical Methods, 3rd ed. (2025), §5.3.2, Example 4, and §5.3.5 (standard reference, not scraped)