LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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.

Small coordinate-by-coordinate increments stay inside a Euclidean ball and telescope the total increment

Statement

Let $B(a,r)\subseteq U\subseteq\mathbb R^m$ and let $\|h\|_2<r$ . Define $p_0=a$ and $p_j=a+\sum_{i<j}h_ie_i$ for $1\le j\le m$ . Then every $p_j$ lies in $B(a,r)$ , $p_m=a+h$ , and for every map $f$ on $U$ ,

$f(a+h)-f(a)=\sum_{j<m}\bigl(f(p_{j+1})-f(p_j)\bigr).$

Facts & Assumptions

Given: The displayed ball, vector $h$ , and coordinate-prefix points $p_j$ .

[L1]

The coordinate list of $x\in F^{n}$ with respect to the ordered basis $e$ is its ordinary coordinate list (The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$ ; hence $\dim_F F^{n} = n$ , and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$ ).

[L2]

The Euclidean norm of $x$ is $\|x\|_2=\sqrt{\sum_{i<n}x_i^2}$ (The Euclidean inner product $\langle x,y\rangle = \sum_{k<n} x_k y_k$ on $\mathbb{R}^n$ ).

Proof

technique · direct

1.1

The standard-basis coordinate formula makes $p_m=a+h$ , while $p_j-a$ has coordinates $h_i$ for $i<j$ and $0$ otherwise.

L1L2

2.1

Hence $\|p_j-a\|_2^2=\sum_{i<j}h_i^2\le\sum_{i<m}h_i^2=\|h\|_2^2<r^2$ , so every prefix point is in $B(a,r)$ .

step 1.1L2algebra

3.1

Summing $f(p_{j+1})-f(p_j)$ cancels all intermediate values and leaves $f(p_m)-f(p_0)=f(a+h)-f(a)$ .

step 1.1step 2.1∎

Depends on

Used by

If all partial derivatives exist on a neighbourhood and are continuous at a point, then the map is totally differentiable there with Jacobian derivative Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 95 results over 21 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

J. Lebl, Basic Analysis I, §8.4 (standard reference, not scraped)