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

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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.

Disjoint union and Cartesian product translate to addition and multiplication of ordinary generating functions

Statement

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

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

Then the disjoint union and Cartesian product of Disjoint unions and Cartesian products of combinatorial classes satisfy

$OGF (A + B) = A (x) + B (x),$

and

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

Facts & Assumptions

Given: Combinatorial classes $A$ and $B$ with counting sequences $(a_{n})$ and $(b_{n})$ and ordinary generating functions $A (x)$ and $B (x)$ .

[L1]

If $I$ is a finite set and $(A_{i})_{i \in I}$ is a family of finite sets that are pairwise disjoint, then $⋃_{i \in I} A_{i}$ is finite and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ (The sum rule: a finite disjoint union is finite with $∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣$ and $∣ ⋃_{i \in I} A_{i} ∣ = \sum_{i \in I} ∣ A_{i} ∣$ , and a sum over a finite index set splits along a partition).

[L2]

If $A$ and $B$ are finite then $A \times B$ is finite and $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ (The product rule: $∣ A \times B ∣ = ∣ A ∣ ∣ B ∣$ , and $∣ \prod_{i < m} A_{i} ∣ = \prod_{i < m} ∣ A_{i} ∣$ ).

[L3]

For formal power series, $[x^{n}] (F G) = \sum_{i = 0}^{n} [x^{i}] F [x^{n - i}] G$ (Formal power series over a commutative ring and the coefficient-extraction functional $[x^{n}]$ ).

Proof

technique · direct

1.1

For each $n \in N$ , the size- $n$ layer of $A + B$ is the disjoint union of the tagged finite sets $A_{n}$ and $B_{n}$ , so it has cardinality $a_{n} + b_{n}$ .

1.2

For each $n \in N$ , the size- $n$ layer of $A \times B$ is the disjoint union of the finite sets $A_{i} \times B_{n - i}$ for $0 \leq i \leq n$ , so its cardinality is $\sum_{i = 0}^{n} a_{i} b_{n - i}$ .

L1L2

2.1

Step 1.1 says $[x^{n}] OGF (A + B) = a_{n} + b_{n} = [x^{n}] (A (x) + B (x))$ , and step 1.2 together with [L3] says $[x^{n}] OGF (A \times B) = \sum_{i = 0}^{n} a_{i} b_{n - i} = [x^{n}] (A (x) B (x))$ . Equality of coefficients in every degree proves both displayed identities.

step 1.1step 1.2L3∎

Depends on

Used by

Binary words have generating function 1/(1-2x) Corollary
Positive-integer compositions have generating function x/(1-2x) Corollary
Rooted plane binary trees satisfy B(x)=1+xB(x)² Corollary
A product class without unique factorisation does not have generating function A(x)B(x) Counterexample
Without disjoint copies, union does not add generating functions Counterexample
If A has no size-zero objects then SEQ(A) has generating function 1/(1-A(x)) Theorem

Dependency tree · two levels

28 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)