PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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.

$\deg (ω_{n}) = n$ for every integer $n$

Statement

$\deg (ω_{n}) = n$ for every integer $n$ .

Facts & Assumptions

Given: An integer $n$ and the standard loop $ω_{n}$ .

[L1]

For every integer $n$ , define ${\tilde{ω}}_{n} (t) = n t$ and $ω_{n} = p \circ {\tilde{ω}}_{n}$ (The standard circle loops $ω_{n} (t) = [n t]$ for $n \in Z$ ).

[L2]

For a based circle loop $α$ with its lift $\tilde{α}$ beginning at zero, define $\deg (α) = \tilde{α} (1)$ (The degree of a based circle loop).

[L3]

Given a covering $p : E \to B$ , a path $α : I \to B$ , and $e_{0} \in E$ above $α (0)$ , there is a unique path $\tilde{α} : I \to E$ with $\tilde{α} (0) = e_{0}$ and $p \circ \tilde{α} = α$ (Existence and uniqueness of path lifts through a covering map).

Proof

technique · direct

1.1L1L3

The path $t \mapsto n t$ starts at zero and projects to $ω_{n}$ by [L1]. The uniqueness clause of [L3] therefore identifies it with the lift used to define the degree of $ω_{n}$ .

2.1step 1.1L2algebra∎

Its terminal value is $n \cdot 1 = n$ , so [L2] gives $\deg (ω_{n}) = n$ . This calculation is uniform for $n = 0$ , positive $n$ , and negative $n$ .

$\deg (ω_{n}) = n$ for every integer $n$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

deg⁡(ωn)=ndeg(ωn​)=n for every integer nn

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$\deg (ω_{n}) = n$ for every integer $n$