LemmaStatement: AI-adaptedProof: 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.

Hamming balls below half the code distance are disjoint

Statement

Let $C \leq F^{n}$ have distance parameter $d = d (C)$ and put $t = ⌊ (d - 1) / 2 ⌋$ . For distinct codewords $c, c^{'} \in C$ , the closed Hamming balls $B_{t} (c) : = {r : d_{H} (r, c) \leq t}$ and $B_{t} (c^{'})$ are disjoint.

Facts & Assumptions

Given: A finite-field linear code $C \leq F^{n}$ , its distance parameter $d (C)$ , and $t = ⌊ (d (C) - 1) / 2 ⌋$ .

[L1]

Hamming weight and distance of a finite-field linear code: $d (C)$ is the minimum distance between distinct codewords for nonzero $C$ and is $n + 1$ for $C = {0}$ .

[L2]

Hamming distance is a metric and linear-code distance is minimum nonzero weight: Hamming distance is a metric and its minimum over distinct codeword pairs is $d (C)$ when such pairs exist.

Proof

1.1L1givenalgebra

The defining floor gives $2 t \leq d - 1 < d$ .

2.1L1L2step 1.1∎

If $r$ belonged to both balls, [L2] and the triangle inequality would give $d_{H} (c, c^{'}) \leq d_{H} (c, r) + d_{H} (r, c^{'}) \leq 2 t < d$ , contradicting the minimum-distance property [L2]. Thus the balls are disjoint. If $C = {0}$ there are no distinct codewords, so the conclusion is vacuous.

Depends on

Used by

Classical distance gives error detection, unique decoding, and erasure recovery Theorem

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Richard Cleve, Quantum Information Processing course notes, §37.1 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 7, §7.6 (standard reference, not scraped)