ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 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.

A residue of p over q at a simple zero is p(a)/q'(a)

Example

For

$p (z) = z + 1, q (z) = z^{2} - 1,$

the function $p / q$ has a simple pole at $a = 1$ , and

$Res ⁣ (\frac{z + 1}{z^{2} - 1}, 1) = 1.$

Facts & Assumptions

Given: The quotient $f (z) = (z + 1) / (z^{2} - 1)$ .

[L1]

If $q (a) = 0$ and $q^{'} (a) \neq 0$ , then $Res (p / q, a) = p (a) / q^{'} (a)$ (Residues of p over q at a simple zero of q).

Verification

technique · direct

1.1givenalgebra

At $a = 1$ one has $q (1) = 0$ , $q^{'} (z) = 2 z$ , and therefore $q^{'} (1) = 2 \neq 0$ .

2.1step 1.1L1algebra∎

Applying [L1] gives $Res ⁣ (\frac{z + 1}{z^{2} - 1}, 1) = \frac{p (1)}{q^{'} (1)} = \frac{2}{2} = 1.$

Depends on

Residues of p over q at a simple zero of q

Used by

Nothing in the library uses this result yet.

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.