ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31

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.

Computing a Dirichlet inverse recursively

Example

Let $1$ be the constant-one function. The inverse recursion from An arithmetic function has a Dirichlet inverse exactly when its value at $1$ is nonzero gives

g (1) = 1, g (2) = - 1, g (3) = - 1, g (4) = 0, g (5) = - 1, g (6) = 1.

Facts & Assumptions

Given: The constant-one function $1$ and the recursion for its Dirichlet inverse $g$ .

Verification

technique · direct

1.1

Since $1 (1) = 1$ , the inverse criterion gives $g (1) = 1$ . For $n > 1$ the recursion is $g (n) = - \sum_{d ∣ n, d < n} g (d)$ . Therefore $g (2) = - g (1) = - 1$ , $g (3) = - g (1) = - 1$ , $g (4) = - (g (1) + g (2)) = 0$ , $g (5) = - g (1) = - 1$ , and $g (6) = - (g (1) + g (2) + g (3)) = 1$ .

givenalgebra

2.1

Using the displayed values, $(1 * g) (1) = 1$ , $(1 * g) (2) = 1 + (- 1) = 0$ , $(1 * g) (3) = 1 + (- 1) = 0$ , $(1 * g) (4) = 1 + (- 1) + 0 = 0$ , $(1 * g) (5) = 1 + (- 1) = 0$ , and $(1 * g) (6) = 1 + (- 1) + (- 1) + 1 = 0$ . So $(1 * g) (n) = ε (n)$ for $1 \leq n \leq 6$ , where $ε$ is the identity of The Dirichlet-convolution identity and the constant-one function.

step 1.1algebra

3.1

Dirichlet convolution is commutative, so the same finite check also gives $(g * 1) (n) = ε (n)$ on this displayed range.

step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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

Victor Shoup, A Computational Introduction to Number Theory and Algebra, Exercise 2.54 (standard reference, not scraped)