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.

Equal classical syndromes imply a codeword difference

Statement

Let $e, f \in F^{n}$ and let $H$ be a parity-check matrix of $C$ . If $H e = H f$ , then $e - f \in C$ . If also $wt (e), wt (f) \leq t$ and $2 t < d (C)$ , then $e = f$ .

Facts & Assumptions

Given: A code $C \leq F^{n}$ with distance parameter $d (C)$ , a parity-check matrix $H$ , errors $e, f \in F^{n}$ , and a nonnegative integer $t$ with $2 t < d (C)$ .

[L1]

Syndrome of a received word over a finite field: The syndrome of a vector $u$ is $H u$ .

[L2]

Syndrome of a received word over a finite field: Its hypothesis makes $H$ a parity-check matrix of $C$ ; the matrix convention in that definition's prerequisites is $\ker H = C$ .

[L3]

Hamming weight and distance of a finite-field linear code: The weight counts nonzero coordinates, and every nonzero codeword has weight at least $d (C)$ ; the zero code has $d (C) = n + 1$ .

Proof

1.1L1L2givenalgebra

If $H e = H f$ , linearity gives $H (e - f) = 0$ . By [L1] and [L2], $e - f \in \ker H = C$ . This uses subtraction in $F^{n}$ ; in characteristic two it equals addition.

2.1L3step 1.1givenalgebra∎

The support of $e - f$ is contained in the union of the supports of $e$ and $f$ , so $wt (e - f) \leq wt (e) + wt (f) \leq 2 t < d (C)$ . If $e - f \neq 0$ , [L3] contradicts that it is a codeword; for the zero code, its only codeword is already zero. Thus $e - f = 0$ and $e = f$ .

Depends on

Used by

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

Dependency tree · two levels

12 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, §37.1.3 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 7, §7.6 (standard reference, not scraped)