TheoremStatement: AI-adaptedProof: AI-adaptedSession-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.

$\log$ is the unique $f$ with $f (x y) = f (x) + f (y)$ that is differentiable at $1$ with $f^{'} (1) = 1$

Statement

The natural logarithm is the unique function $f : (0, \infty) \to R$ satisfying

$f (x y) = f (x) + f (y) (x, y > 0)$

that is differentiable at $1$ with $f^{'} (1) = 1$ .

Facts & Assumptions

Given: A function $f$ satisfying the displayed equation, differentiable at $1$ , with $f^{'} (1) = 1$ .

[F1]

The derivative is the limit of the difference quotient (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L1]

$L^{'} (x) = 1 / x$ and $L (1) = 0$ (The integral logarithm satisfies $L^{'} (x) = 1 / x$ for $x > 0$ and $L (1) = 0$ ).

[L2]

$L (x y) = L (x) + L (y)$ (The integral logarithm satisfies $L (x y) = L (x) + L (y)$ for all positive $x$ and $y$ ).

[L3]

A differentiable function is continuous (A function differentiable at $c$ is continuous at $c$ ).

[L4]

A continuous function on an interval with zero derivative is constant (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

[L5]

$L = \log$ on $(0, \infty)$ (The integral logarithm $L$ is the published natural logarithm).

[L6]

Sums, differences and scalar multiples of functions differentiable at a point are differentiable there, with the corresponding derivatives (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

Proof

technique · direct

1.1

Setting $x = y = 1$ in the functional equation gives $f (1) = 2 f (1)$ , hence $f (1) = 0$ .

givenalgebra

1.2

Conversely, [L2] and [L5] give the product equation for $\log$ , while [L1] and [L5] give differentiability at $1$ with derivative $1$ .

L2L1L5

2.1

Fix $x > 0$ . For $h$ sufficiently close to $0$ , $x + h > 0$ , and the functional equation gives $f (x + h) - f (x) = f (1 + h / x) - f (1) .$

step 1.1givenalgebra

3.1

For $h \neq 0$ , divide step 2.1 by $h$ : $\frac{f (x + h) - f (x)}{h} = \frac{1}{x} \frac{f (1 + h / x) - f (1)}{h / x} .$ As $h \to 0$ , [F1] and $f^{'} (1) = 1$ show that $f^{'} (x) = 1 / x$ .

step 2.1F1givenalgebra

4.1

Both $f$ and $L$ are differentiable on $(0, \infty)$ , so [L6] makes $f - L$ differentiable, and step 3.1 with [L1] gives $(f - L)^{'} = 0$ . By [L3], $f - L$ is continuous, so [L4] makes it constant.

step 3.1L1L3L4L6algebra

5.1

At $1$ , step 1.1 and [L1] give $(f - L) (1) = 0$ , so step 4.1 yields $f = L = \log$ by [L5].

step 1.1step 4.1L1L5

6.1

Steps 5.1 and 1.2 prove existence and uniqueness.

step 5.1step 1.2∎

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: 77 results over 23 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.

Sources

John M. Lee, The Logarithm (standard reference, not scraped)

log⁡log is the unique ff with f(xy)=f(x)+f(y)f(xy)=f(x)+f(y) that is differentiable at 11 with f′(1)=1f′(1)=1