PropositionStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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.

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Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution

Statement

Let $R$ be a commutative ring, $f, g \in R ⟦ x ⟧$ , $r \in R$ , and $n, k \in N$ . Then

$[x^{n}] (f + g) = [x^{n}] f + [x^{n}] g, [x^{n}] (r f) = r [x^{n}] f,$

and $f = g$ if and only if $[x^{n}] f = [x^{n}] g$ for every $n$ . Moreover,

$[x^{n}] (x^{k} f) = {\begin{cases} [x^{n - k}] f, & k \leq n, \\ 0, & k > n, \end{cases}$

and

$[x^{n}] (f g) = \sum_{i = 0}^{n} [x^{i}] f [x^{n - i}] g .$

These formulas include $n = 0$ , $k = 0$ , and $k > n$ .

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[F1]

For $f = \sum a_{n} x^{n}$ , coefficient extraction is evaluation: $[x^{n}] f = a_{n}$ (Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ ).

[F2]

The series $x$ is supported at degree $1$ , and $x^{n}$ is supported at degree $n$ (Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ ).

[F3]

Cauchy multiplication is the finite convolution $[x^{n}] (f g) = \sum_{i = 0}^{n} [x^{i}] f [x^{n - i}] g$ (Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ ).

Proof

technique · unfold the definitions

1.1

The two linearity identities are the pointwise definitions of addition and scalar multiplication. Equality of all extracted coefficients is equality of the underlying functions $N \to R$ , proving both directions of extensionality.

givenF1

1.2

In the convolution for $x^{k} f$ , the first factor has one nonzero coefficient, at $k$ . It contributes $[x^{n - k}] f$ when $k \leq n$ and there is no contributing index when $k > n$ ; for $k = 0$ this says $1 f = f$ .

givenF2

1.3

The last display is the defining finite convolution, whose $n = 0$ instance is $[x^{0}] (f g) = [x^{0}] f [x^{0}] g$ .

givenF3

2.1

Steps 1.1-1.3 establish every asserted clause and every listed boundary case.

step 1.1step 1.2step 1.3∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Herbert S. Wilf, generatingfunctionology (standard reference, not scraped)

Coefficient extraction is RR-linear, separates formal series, shifts under multiplication by xkxk, and converts products to finite convolution