ExampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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.

Dropping $f (e) = 1$ leaves the whole family $c \log$ , including logarithms to other bases and the zero function

Example

Without the normalisation $f (e) = 1$ , the continuous product-to-sum functions are exactly

$f_{c} (x) = c \log x (c \in R) .$

They include the zero function and, for suitable nonzero $c$ , logarithms to other bases.

Facts & Assumptions

Given: $c \in R$ .

[L1]

Every continuous product-to-sum function is uniquely $c \log x$ , including $c = 0$ (Every continuous $f$ with $f (x y) = f (x) + f (y)$ is $f (x) = c \log x$ for a unique $c$ , including $c = 0$ ).

[L2]

The natural logarithm is continuous and satisfies $\log (x y) = \log x + \log y$ (Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm).

[L3]

The natural logarithm satisfies $\log (e) = 1$ ( $\log$ is the unique continuous $f : (0, \infty) \to R$ with $f (x y) = f (x) + f (y)$ and $f (e) = 1$ ).

[F1]

For $b > 0$ , $b \neq 1$ , $\log_{b} x = \log x / \log b$ (The logarithm to a positive base other than one).

Verification

technique · direct

1.1

By [L2], $f_{c} (x y) = c \log (x y) = c \log x + c \log y = f_{c} (x) + f_{c} (y)$ , and $f_{c}$ is continuous.

L2algebra

1.2

Also $f_{c} (e) = c \log (e) = c$ , so only $c = 1$ meets the normalisation $f (e) = 1$ .

L3algebra

1.3

If $b > 0$ , $b \neq 1$ , then [F1] identifies $\log_{b}$ with the member $c = 1 / \log b$ . The case $c = 0$ is the zero function and cannot equal $\log_{b}$ , because $\log_{b} b = 1$ .

F1algebra

2.1

The classification theorem [L1] shows that steps 1.1 through 1.3 exhaust all continuous product-to-sum functions, not merely a subfamily.

step 1.1step 1.2step 1.3L1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Dropping f(e)=1f(e)=1 leaves the whole family clog⁡clog, including logarithms to other bases and the zero function