ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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.

Derivative of the heaviside function is dirac delta

Example

Assume Countable Choice for Lebesgue integration. On $R$ , let $H = 1_{(0, \infty)}$ . Then $D u_{H} = δ_{0}$ . Any assigned value of $H (0)$ gives the same regular distribution.

Facts & Assumptions

[F1]

Locally integrable functions have regular functionals $u_{f} (φ) = \int f φ$ (Regular distribution from a locally integrable function), and under Countable Choice these functionals are distributions (Locally integrable functions embed in distributions).

[F2]

$(D u) (φ) = - u (φ^{'})$ , and $δ_{0} (φ) = φ (0)$ (Distributional derivative, Dirac delta and its derivatives).

[F3]

Under Countable Choice, the complex FTC gives $\int_{a}^{b} φ^{'} = φ (b) - φ (a)$ (Complex integration by parts on intervals and decaying lines, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F4]

A singleton is Lebesgue null (A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in $R^{n}$ ).

Proof

Given: the bounded measurable function $H$ and Countable Choice.

1.1

Since $∣ H ∣ \leq 1$ , its absolute integral on every compact interval is finite, so F1 makes $u_{H}$ a distribution. Altering its value only at zero changes no integral by F4. For a test $φ$ , choose $R > 0$ beyond its compact support. Then F2 and F3 give $D u_{H} (φ) = - \int_{0}^{R} φ^{'} (x) d x = φ (0) - φ (R) = φ (0) = δ_{0} (φ)$ . This calculation applies to complex tests componentwise.

givenF1F2F3F4

2.1

The equality on every test proves the distribution identity. The sign is positive because the negative transpose sign cancels the lower-endpoint sign. A test supported away from zero gives zero; the zero test gives zero; only the finite interval $[0, R]$ enters, so no endpoint at infinity is evaluated. $□$

step 1.1F2

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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

Semyon Dyatlov, Lecture notes for 18.155 (2022) (standard reference, not scraped)