ExampleConstruction: AI-generatedVerification: 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.

A field as an associative algebra and an A∞-algebra

Example

Let $k$ be a field, concentrated in degree zero. Set $A^{0} = k$ and $A^{j} = 0$ for $j \neq 0$ . Define $m_{1} = 0$ , let $m_{2}$ be field multiplication, and set $m_{n} = 0$ for $n \geq 3$ . These maps make $A$ an uncurved cohomological A∞-algebra.

Facts & Assumptions

Given: A field $k$ and the A∞ identities from Uncurved cohomological A∞-algebra in the unsuspended convention.

[F1]

Uncurved cohomological A∞-algebra in the unsuspended convention: $m_{n}$ has degree $2 - n$ and the Stasheff sum vanishes in every arity $n \geq 1$ .

[F2]

Field: field multiplication is associative and has an identity.

Verification

1.1givenF1F2algebra

The zero map $m_{1}$ is homogeneous of degree $+ 1$ , field multiplication $m_{2}$ has degree zero on $A^{0}$ , and every zero map $m_{n}$ for $n \geq 3$ is homogeneous of degree $2 - n$ . Thus the degrees match [F1].

1.2F1F2algebra

For $n = 3$ , the only possibly nonzero terms are $m_{2} (m_{2} \otimes 1) - m_{2} (1 \otimes m_{2})$ . They agree by associativity in [F2], so their difference is zero.

2.1F1step 1.1algebra

For $n = 1$ every Stasheff summand contains $m_{1}$ , so the sum is zero. For $n = 2$ , the only potentially nonzero outer operation is $m_{1} m_{2}$ , but $m_{1} = 0$ ; all summands containing an inner $m_{1}$ also vanish.

2.2F1step 1.1algebra

For $n \geq 4$ , a nonzero summand would require both inner and outer operations to have arity two, which forces $n = s + (r + 1 + t) - 1 \leq 3$ . Therefore all terms vanish in these arities as well.

3.1F1step 2.1step 1.2step 2.2∎

The defining identity holds in every arity by steps 2.1, 1.2 and 2.2, so the stated operations form an A∞-algebra.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

3 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

Bernhard Keller, Introduction to A-infinity algebras and modules, §3.1 (standard reference, not scraped)
Jean-Louis Loday and Bruno Vallette, Algebraic Operads, §9.2.5 (standard reference, not scraped)