LemmaStatement: AI-generatedProof: AI-generated

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.

Explicit eventual-stage thresholds in every fixed homotopy degree

PDF p. 4: Motivic graded spectral sequence; PDF p. 9: Eventual filtration vanishing

Statement

Assume the Axiom of Choice. Let $X$ be as above with bound $B = B (X)$ . For every fixed $q \in Z$ and every integer anchor $t$ , put $ℓ_{0} (q, t) = \max (0, B + q + 1 - t) .$ Then $π_{q} F_{m o t}^{t + ℓ} K (X) = 0$ for all integers $ℓ \geq ℓ_{0} (q, t)$ . In particular the claim holds with $t = \dim X$ and every fixed $q \leq 0$ needed in the article.

Facts & Assumptions

Given: The Axiom of Choice and fixed $q, t$ , the actual stage bound and $ℓ \geq ℓ_{0} (q, t)$ .

Goal: Exhibit a threshold for the actual stage group rather than assume that high-stage images detect it.

[F1]

The precise source stage bound and its homotopy interpretation gives vanishing if $q < r - B$ .

[F2]

The assumed The Axiom of Choice discharges the inherited Choice premise of [F1] when its bound for the selected actual filtration is invoked. Taking a maximum of two integers adds no choice requirement.

Proof

1.1F1givenalgebra

By the definition of the maximum, $t + ℓ \geq B + q + 1$ . Set $r = t + ℓ$ . Then $r - B \geq q + 1 > q$ .

2.1F1F2step 1.1∎

Under the inherited Choice premise in [F2], apply [F1] to this strict inequality. It gives the asserted zero stage group for every such $ℓ$ . The choice of $q$ was arbitrary, so the same argument covers in particular the required fixed nonpositive degrees; the threshold is allowed to depend on $q$ .

Depends on

Used by · 2

Dependency tree · 3 levels

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