ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge 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.

Polynomial differential forms on the line

Example

Let $S = Z [t, ϵ] = (Z [t]) [ϵ]$ and $A = S / (ϵ^{2})$ , where $(ϵ^{2})$ is the ideal generated by $ϵ^{2}$ . Write $ϵ$ also for its class in $A$ . Each class has a unique representative $f (t) + g (t) ϵ$ , so $A = Z [t] \oplus Z [t] ϵ$ , with $∣ t ∣ = 0$ , $∣ ϵ ∣ = 1$ , $A^{0} = Z [t]$ , $A^{1} = Z [t] ϵ$ , and $A^{n} = 0$ for $n \notin {0, 1}$ . Multiplication is induced by polynomial multiplication; equivalently, $ϵ^{2} = 0$ and $t ϵ = ϵ t$ . The unit is the class of $1$ in degree zero. Define the map $d : A \to A$ by $d (f (t) + g (t) ϵ) = f^{'} (t) ϵ,$ where the formal derivative of $f (t) = \sum_{i = 0}^{n} a_{i} t^{i}$ is $f^{'} (t) = \sum_{i = 1}^{n} i a_{i} t^{i - 1}$ .

Facts & Assumptions

Given: The graded algebra and map above. Verify that they form a cohomological DGA.

[F1]

The integers form a commutative ring: The standard addition and multiplication on $Z$ make it a commutative ring with multiplicative identity.

[F2]

Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides and Commutative ring: A commutative ring has an abelian additive group, an associative unital multiplication, both distributive laws, and commutative multiplication.

[F3]

Polynomial convolution makes $R [x]$ a commutative ring containing $R$ as its constant subring: If $R$ is a commutative ring, then the coefficientwise and convolution operations make $R [x]$ a commutative ring.

[F4]

In a commutative ring, $(S)$ consists of finite sums $\sum r_{i} s_{i}$ , and $(a) = R a$ : In a commutative ring, $(a)$ consists of the multiples of $a$ .

[F5]

For a two-sided ideal $I$ , the additive cosets form a ring $R / I$ with identity $1 + I$ : For a ring $R$ and a two-sided ideal $I$ , $R / I$ is a ring with identity.

[F6]

Cohomological differential graded algebra: For homogeneous $a \in A^{p}$ and $b \in A^{q}$ , require the signed Leibniz rule $d^{p + q} (a b) = d^{p} (a) b + (- 1)^{p} a d^{q} (b) .$

[F7]

Cohomological differential graded algebra: The defining cochain differential satisfies $d^{n + 1} d^{n} = 0$ .

Verification

1.1F1F2

By [F1] and [F2], $Z$ with its standard operations is a commutative ring with unit, so it is an admissible base for [F6].

2.1F3F4F5step 1.1

Applying [F3] first to $Z$ and then to $Z [t]$ shows that $S = (Z [t]) [ϵ]$ is a commutative ring. By [F4], $(ϵ^{2}) = S ϵ^{2}$ is a two-sided ideal; [F5] therefore gives the quotient ring $A$ with its associative multiplication and unit. For cosets, $[s] [t] = [s t] = [t s] = [t] [s]$ , so $A$ is commutative. The constant classes define a unital ring homomorphism $Z \to A$ ; its induced scalar action is central, and associativity and distributivity make multiplication $Z$ -bilinear.

3.1F3F4step 2.1algebra

Every polynomial in $ϵ$ with coefficients in $Z [t]$ has a unique form $f_{0} (t) + f_{1} (t) ϵ + ϵ^{2} h (t, ϵ)$ . By [F4], the last term is in $(ϵ^{2})$ , and a polynomial of degree at most one in $ϵ$ lies in $(ϵ^{2})$ only when both its coefficients vanish. Thus every quotient class has a unique representative $f + g ϵ$ , giving the stated direct sum of $Z$ -modules. The quotient multiplication is $(f + g ϵ) (h + k ϵ) = f h + (f k + g h) ϵ$ ; this preserves the grading and gives $ϵ^{2} = 0$ and $t ϵ = ϵ t$ .

4.1givenstep 3.1constructalgebra

The unique expression in step 3.1 makes the formula for $d$ well defined. Coefficientwise differentiation is additive and commutes with multiplication by each integer, so $d$ is $Z$ -linear; it sends $A^{0}$ to $A^{1}$ and $A^{1}$ to $A^{2} = 0$ , hence has degree $+ 1$ .

5.1F7step 4.1algebra

For every $f + g ϵ$ , $d^{2} (f + g ϵ) = d (f^{'} ϵ) = 0$ , because $d$ vanishes on $A^{1}$ . Thus [F7] holds.

5.2givenF6algebrastep 4.1

If $f = \sum_{i} a_{i} t^{i}$ and $h = \sum_{j} b_{j} t^{j}$ have degree zero, the finite coefficient sums give $(f h)^{'} = \sum_{i + j \geq 1} (i + j) a_{i} b_{j} t^{i + j - 1} = f^{'} h + f h^{'}$ . Hence $d (f h) = d (f) h + f d (h)$ .

5.3F6step 3.1step 4.1algebra

If $a = f$ and $b = g ϵ$ have degrees zero and one, both $d (a b)$ and $d (a) b + a d (b)$ vanish because $d (g ϵ) = 0$ and $(f^{'} ϵ) (g ϵ) = 0$ . If $a = g ϵ$ and $b = f$ , then both sides of $d (a b) = d (a) b - a d (b)$ vanish because $d (g ϵ) = 0$ and $(g ϵ) (f^{'} ϵ) = 0$ . If both have degree one, their product and both terms on the right vanish. These are all remaining homogeneous degree pairs, so [F6] holds.

6.1F6step 1.1step 2.1step 3.1step 4.1step 5.1step 5.2step 5.3∎

step 1.1, step 2.1, step 3.1 establish the graded unital associative $Z$ -algebra structure, and step 4.1, step 5.1, step 5.2, step 5.3 establish the degree-one, square-zero differential and signed Leibniz rule. Thus all conditions of [F6] hold, so $A$ is a cohomological DGA.

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