TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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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.
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If $B (0) = 0$ then $A \circ B$ has generating function $A (B (x))$

Statement

Let $A$ and $B$ be combinatorial classes with ordinary generating functions

$A (x) = \sum_{r \geq 0} a_{r} x^{r}, B (x) = \sum_{n \geq 0} b_{n} x^{n},$

and suppose $B (0) = 0$ . Then $A \circ B$ is a combinatorial class and

$OGF (A \circ B) = A (B (x)) .$

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[L1]

Formal composition is $f \circ g = \sum_{r \geq 0} [x^{r}] f g^{r}$ , and it is admissible when $g (0) = 0$ (Composition $f \circ g$ of formal series when the outer series is a polynomial or the inner series has zero constant term).

[L2]

Substitution by a zero-constant series is a ring homomorphism (Substitution by a zero-constant series is a ring homomorphism, and composition is associative when both inner series have zero constant coefficient).

Proof

technique · direct

1.1

Fix $r \geq 0$ . An object of size $r$ in $A$ contributes one ordered list of $r$ slots, and filling those slots with $B$ -objects is counted by $B (x)^{r}$ . Since there are $a_{r}$ choices for the outer object, the total contribution of all outer objects of size $r$ is $a_{r} B (x)^{r}$ .

construct

2.1

Because $B (0) = 0$ , every $B$ -object has positive size. Therefore an object of total size $n$ in $A \circ B$ can only come from outer size $r \leq n$ , so each size layer is finite and the total generating function is $\sum_{r \geq 0} a_{r} B (x)^{r}$ .

step 1.1given

3.1

The series of step 2.1 is exactly the admissible formal composition $A (B (x))$ by [L1], and [L2] records that substitution by a zero-constant series is the corresponding ring operation on formal series. Hence $OGF (A \circ B) = A (B (x))$ .

step 2.1L1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics (standard reference, not scraped)
Stephen Melczer, An Invitation to Enumeration, Chapter 5: Combinatorial Constructions (standard reference, not scraped)

If B(0)=0B(0)=0 then A∘BA∘B has generating function A(B(x))A(B(x))