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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.

Noise, Fault Tolerance, Concatenation, and Thresholds — Examples

1 · Prerequisites

2 · Summary

Objective

Work through one finite geometric-series instance of the iid-Z, perfect-syndrome planar memory bound, with all decoder and noise hypotheses stated. The example is a logical leaf and does not infer a faulty-gate threshold.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

Surface-Code Memory Bound Under IID Z Noise

Statement

Assume the distance-five relative square planar patch, which has n=2(5)2=50 data qubits, iid Z-error rate ϵ=1/144, one perfect syndrome round, and minimum-weight recovery. The finite series in the planar memory theorem gives the conditional upper bound Plogical≤6∑ℓ=550(1/2)ℓ=38(1−2−46)<38. This calculation concerns an ideal-syndrome memory round only.

Example

Take the stated finite parameters. Since 36ϵ=1/4, each summand (36ϵ)ℓ/2 equals 2−ℓ. The theorem's prefactor is d+1=6.

Facts & Assumptions

Given: d=5, n=50, ϵ=1/144, iid Z-noise, perfect syndrome extraction, and minimum-weight recovery.

[F1]

Planar Surface-Code Memory Bound Under IID Z Noise: Under those hypotheses, Plogical≤(d+1)∑ℓ=dn(36ϵ)ℓ/2.

Verification

1.1givenalgebra

Compute 36ϵ=36/144=1/4, so (36ϵ)ℓ/2=((1/4)1/2)ℓ=2−ℓ.

2.1givenstep 1.1algebra

The finite geometric sum is ∑ℓ=5502−ℓ=2−5(1−2−46)/(1−1/2)=116(1−2−46). Multiplying by d+1=6 gives 38(1−2−46)<38.

3.1F1step 2.1∎

Applying [F1] yields the stated conditional failure-probability bound. The hypotheses include perfect syndrome extraction; this example makes no faulty-gate threshold claim.

Sources