Alphabeta Math
TheoremStatement: AI-adaptedProof: 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.

Pathwise error decomposition along a reference chain

Statement

Let θ=r0,r1,…,rm,r~ be the finite scalar reference chain and returned value of Successive statistical and computational error targets, with m≥0. Then, pointwise, r~−θ=∑j=1m(rj−rj−1)+(r~−rm), and consequently ∣r~−θ∣≤∑j=1m∣rj−rj−1∣+∣r~−rm∣. The sum is empty when m=0. No independence, unbiasedness, or distributional assumption is needed.

Facts & Assumptions

Given: A finite chain θ=r0,r1,…,rm,r~ of real values, with m≥0. Prove the stated identity and bound.

[F1]

The successive signed errors are ej=rj−rj−1 for 1≤j≤m, the last numerical error is efp=r~−rm, and r0=θ. (Successive statistical and computational error targets)

Proof

1.1givenF1algebra

By [F1], the finite sum telescopes: ∑j=1m(rj−rj−1)=rm−r0; for m=0 both sides are 0.

2.1step 1.1F1algebra

Adding r~−rm to the equality in step 1.1 and using r0=θ gives r~−θ=∑j=1m(rj−rj−1)+(r~−rm).

3.1step 2.1algebra∎

Apply the triangle inequality to the finite sum in step 2.1 to obtain the absolute-error bound.

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Sources