PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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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.

The generating-series identity $E (- t) H (t) = 1$

Statement

In the formal power-series ring $R [x_{1}, \dots, x_{n}] ⟦ t ⟧$ , put

$E (- t) : = \sum_{i = 0}^{n} (- 1)^{i} e_{i} t^{i} = \prod_{j = 1}^{n} (1 - x_{j} t), H (t) : = \sum_{k \geq 0} h_{k} t^{k} .$

Then

$E (- t) H (t) = 1.$

Equivalently, for every $k \geq 1$ ,

$\sum_{i = 0}^{\min (k, n)} (- 1)^{i} e_{i} h_{k - i} = 0.$

Facts & Assumptions

Given: A commutative ring $R$ and variables $x_{1}, \dots, x_{n}$ .

[L1]

For $0 \leq i \leq n$ , the polynomial $e_{i}$ is the sum $\sum_{1 \leq j_{1} < \dots < j_{i} \leq n} x_{j_{1}} \dots x_{j_{i}}$ over the $i$ -element index sets, and $e_{0} = 1$ (The elementary symmetric polynomials $e_{0}, e_{1}, \dots, e_{n}$ ).

[L2]

The polynomial $h_{k}$ is the sum of all monomials of total degree $k$ , and $h_{0} = 1$ (Power sums $p_{k}$ and complete homogeneous symmetric polynomials $h_{k}$ ).

Proof

technique · direct

1.1givenL1algebra

Expand $\prod_{j = 1}^{n} (1 - x_{j} t)$ by choosing either $1$ or $- x_{j} t$ from each factor. The choices taking $- x_{j} t$ at exactly the indices of an $i$ -element set $S$ contribute $(- 1)^{i} (\prod_{j \in S} x_{j}) t^{i}$ , so summing over $∣ S ∣ = i$ and then over $i$ gives $\prod_{j = 1}^{n} (1 - x_{j} t) = \sum_{i = 0}^{n} (- 1)^{i} e_{i} t^{i}$ by [L1], and both displayed descriptions of $E (- t)$ therefore agree.

1.2givenalgebra

For one variable, $(1 - x_{j} t) (1 + x_{j} t + x_{j}^{2} t^{2} + \dots) = 1$ coefficientwise as a formal power series.

2.1step 1.2L2

Multiplying the one-variable geometric series over $j = 1, \dots, n$ gives $\prod_{j} (1 - x_{j} t)^{- 1}$ , whose coefficient of $t^{k}$ is the sum of $x_{1}^{a_{1}} \dots x_{n}^{a_{n}}$ over $a_{1} + \dots + a_{n} = k$ , namely $h_{k}$ .

3.1step 1.1step 2.1L2algebra∎

Thus $H (t) = \prod_{j} (1 - x_{j} t)^{- 1}$ , which is $E (- t)^{- 1}$ by step 1.1, so $E (- t) H (t) = 1$ . Comparing the coefficient of $t^{k}$ gives the displayed recurrence, including $k = 0$ as $e_{0} h_{0} = 1$ .

Depends on

Used by

Dependency tree · two levels

5 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

D. Grinberg, An Introduction to Algebraic Combinatorics, Chapter 7, Section 7.1 (standard reference, not scraped)

The generating-series identity E(−t)H(t)=1E(−t)H(t)=1