CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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A product class without unique factorisation does not have generating function $A (x) B (x)$

Counterexample

Let $A = {a}$ with $∣ a ∣ = 1$ , let $B = {b_{0}, b_{1}}$ with $∣ b_{0} ∣ = ∣ b_{1} ∣ = 1$ , and let $C = {c}$ with $∣ c ∣ = 2$ . Suppose a would-be product construction sends both pairs $(a, b_{0})$ and $(a, b_{1})$ to the same object $c$ . Then

$OGF (A) = x, OGF (B) = 2 x, OGF (C) = x^{2} \neq 2 x^{2} .$

Facts & Assumptions

Given: The symbolic product rule applies to the Cartesian product, where the ordered pair itself records both components (Disjoint union and Cartesian product translate to addition and multiplication of ordinary generating functions, Disjoint unions and Cartesian products of combinatorial classes).

Verification

technique · direct

1.1

The class $A \times B$ has two size- $2$ objects, namely $(a, b_{0})$ and $(a, b_{1})$ , so its generating function is $2 x^{2}$ .

given

2.1

The collapsed construction $C$ has only one size- $2$ object, so its generating function is $x^{2}$ . The missing factor of $2$ is exactly the loss of unique factorisation.

step 1.1given∎

Depends on

Used by

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Dependency tree · two levels

6 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.

A product class without unique factorisation does not have generating function A(x)B(x)A(x)B(x)