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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Over a commutative $Q$ -algebra, $PSET (A)$ has generating function $\exp (\sum_{k \geq 1} (- 1)^{k - 1} A (x^{k}) / k)$

Statement

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

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

Over a commutative $Q$ -algebra,

$OGF (PSET (A)) = \exp (\sum_{k \geq 1} \frac{(- 1)^{k - 1} A (x^{k})}{k}) .$

Facts & Assumptions

Given: The hypotheses and notation of the statement above.

[L1]

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

[L2]

Formal $\exp$ and $\log$ are inverse homomorphisms, and $\log ((1 + u) (1 + v)) = \log (1 + u) + \log (1 + v)$ (Formal $\exp$ and $\log$ are inverse homomorphisms and formal binomial powers obey the expected addition laws).

[L3]

The formal logarithm is $\log (1 + u) = \sum_{j \geq 1} (- 1)^{j - 1} u^{j} / j$ (Formal exponential, logarithm, and binomial powers over a commutative $Q$ -algebra).

Proof

technique · direct

1.1

Let $P (x)$ denote the powerset generating function. By [L1], $P (x) = \prod_{n \geq 1} (1 + x^{n})^{a_{n}}$ , so applying $\log$ and using [L2] gives $\log P (x) = \sum_{n \geq 1} a_{n} \log (1 + x^{n})$ .

L1L2

2.1

By [L3], $\log (1 + x^{n}) = \sum_{k \geq 1} (- 1)^{k - 1} x^{n k} / k$ , so $\log P (x) = \sum_{n \geq 1} \sum_{k \geq 1} (- 1)^{k - 1} a_{n} x^{n k} / k = \sum_{k \geq 1} ((- 1)^{k - 1} / k) \sum_{n \geq 1} a_{n} (x^{k})^{n} = \sum_{k \geq 1} (- 1)^{k - 1} A (x^{k}) / k$ . Again the rearrangement is coefficientwise finite in every degree.

step 1.1L3algebra

3.1

Exponentiating step 2.1 and using the inverse relation of [L2] gives the stated formula for $P (x)$ .

step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

10 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)

Over a commutative QQ-algebra, PSET⁡(A)PSET(A) has generating function exp⁡(∑k≥1(−1)k−1A(xk)/k)exp(∑k≥1​(−1)k−1A(xk)/k)