False statementConstruction: AI-adaptedVerification: AI-generatedSession-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.
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FALSE: $SEQ (A)$ is a combinatorial class even when $A$ has a size-zero object

Statement

False claim: $SEQ (A)$ is always a combinatorial class, even when $A$ has an object of size $0$ .

The theorem If $A$ has no size-zero objects then $SEQ (A)$ has generating function $1 / (1 - A (x))$ excludes exactly this case, and the exclusion is necessary.

Facts & Assumptions

Given: The sequence construction (The sequence construction $SEQ (A)$ ) and its generating function theorem, which assumes that $A$ has no size-zero objects (If $A$ has no size-zero objects then $SEQ (A)$ has generating function $1 / (1 - A (x))$ ).

Refutation

technique · direct

1.1

Let $A = {e}$ with $∣ e ∣ = 0$ . Then for every $r \geq 0$ , the length- $r$ sequence $(e, \dots, e)$ lies in $SEQ (A)$ and has total size $0$ .

given

2.1

These sequences are all distinct because their lengths differ, so the size- $0$ level of $SEQ (A)$ is infinite. Hence $SEQ (A)$ is not a combinatorial class.

step 1.1given

3.1

The claim is therefore false, and the no-size-zero hypothesis in the sequence theorem is load bearing.

step 2.1given∎

Depends on

Used by

Nothing in the library uses this result yet.

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.

Sources

Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics (standard reference, not scraped)

FALSE: SEQ⁡(A)SEQ(A) is a combinatorial class even when AA has a size-zero object