TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If $A$ has no size-zero objects then $PSET (A)$ has generating function $\prod_{n \geq 1} (1 + x^{n})^{a_{n}}$

Statement

Let $A$ be a combinatorial class with no size-zero objects, and write

$A (x) = \sum_{n \geq 1} a_{n} x^{n} .$

Then $PSET (A)$ is a combinatorial class and

$OGF (PSET (A)) = \prod_{n \geq 1} (1 + x^{n})^{a_{n}} .$

Facts & Assumptions

Given: A combinatorial class $A$ with no size-zero objects and counting sequence $(a_{n})_{n \geq 1}$ .

[L1]

Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products (Summable formal families may be regrouped and rearranged, distribute over multiplication, and have well-defined locally finite products).

Proof

technique · direct

1.1construct

For one fixed object $u \in A$ of size $n \geq 1$ , a powerset object either omits $u$ or includes it once, so the contribution of $u$ is the two-term series $1 + x^{n}$ .

2.1step 1.1L1

A powerset object is a simultaneous yes-or-no choice for every object of $A$ . Since every object has positive size, only finitely many such choices can affect a fixed degree, so the per-object factors of step 1.1 form a locally finite product that can be regrouped by [L1]. This also shows that each size layer of $PSET (A)$ is finite.

3.1step 2.1L1∎

For each $n \geq 1$ there are exactly $a_{n}$ objects of size $n$ , and each contributes one factor $1 + x^{n}$ . Regrouping the locally finite product of step 2.1 therefore gives $OGF (PSET (A)) = \prod_{n \geq 1} (1 + x^{n})^{a_{n}}$ .

Depends on

Used by

Dependency tree · two levels

12 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 AA has no size-zero objects then PSET⁡(A)PSET(A) has generating function ∏n≥1(1+xn)an∏n≥1​(1+xn)an​