TheoremStatement: 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 distance is a metric and linear-code distance is minimum nonzero weight

Statement

For every finite field $F$ and $n \in N$ , $d_{H} (x, y) = wt (x - y)$ is a metric on $F^{n}$ . If $n \geq 1$ and $C \leq F^{n}$ is the codeword subspace of a classical linear code, then for nonzero $C$ , $\min_{x \neq y \in C} d_{H} (x, y) = \min_{0 \neq c \in C} wt (c) = d (C)$ . For $C = {0}$ there are no distinct codewords, and the distance parameter is the stated convention $n + 1$ .

Facts & Assumptions

Given: A finite field $F$ and $n \in N$ for the metric claim. For the code-distance claim, assume $n \geq 1$ , let $C \leq F^{n}$ be a classical linear code, and use the weight and distance functions of Hamming weight and distance of a finite-field linear code.

[L1]

Classical linear code over a finite field: A code $C$ is a linear subspace of $F^{n}$ , so it contains $0$ and is closed under subtraction.

[L2]

Hamming weight and distance of a finite-field linear code: $wt (x)$ counts nonzero coordinates, $d_{H} (x, y) = wt (x - y)$ , and $d (C)$ is the minimum nonzero weight or $n + 1$ for the zero code.

Proof

1.1L2algebra

Weight is nonnegative and vanishes exactly on the zero vector. Also $wt (- z) = wt (z)$ because $z_{j} = 0$ exactly when $- z_{j} = 0$ . Hence $d_{H} (x, y) \geq 0$ , $d_{H} (x, y) = 0$ exactly when $x = y$ , and $d_{H} (x, y) = d_{H} (y, x)$ .

1.2L2algebra

If $(x - z)_{j} \neq 0$ , then at least one of $(x - y)_{j}$ or $(y - z)_{j}$ is nonzero, since their sum is $(x - z)_{j}$ . Therefore the support of $x - z$ is contained in the union of the supports of $x - y$ and $y - z$ , so $d_{H} (x, z) \leq d_{H} (x, y) + d_{H} (y, z)$ .

2.1L1L2∎

If $C \neq {0}$ and $x, y \in C$ are distinct, then [L1] gives $x - y \in C ∖ {0}$ , so $d_{H} (x, y) \geq d (C)$ . Conversely every $c \in C ∖ {0}$ occurs as the difference of the pair $(c, 0)$ , and $d_{H} (c, 0) = wt (c)$ . Thus the two nonempty sets of distances have the same minimum. When $C = {0}$ no distinct pair exists, and [L2] supplies the separate convention.

Remarks

This proof is independent of the field characteristic and uses no choice principle.

Depends on

Used by

Hamming balls below half the code distance are disjoint Lemma

Dependency tree · two levels

11 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, §§37.1–37.1.3 (standard reference, not scraped)
John Preskill, Quantum Computation, Chapter 7, §7.6 (standard reference, not scraped)