LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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.

Triangle inequality for finite sums

Statement

Let $a : \mathbb{N} \to \mathbb{R}$ be a sequence of reals and $n \in \mathbb{N}$ , with finite sums as in Finite sums and finite products, by recursion and the absolute value as in Absolute value in an ordered field. Then

$\Big|\sum_{k<n} a_k\Big| \le \sum_{k<n} |a_k|.$

Facts & Assumptions

Given: A sequence $a : \mathbb{N} \to \mathbb{R}$ and a natural $n$ .

[L1]

Recursion clauses (Finite sums and finite products, by recursion): $\sum_{k<0} a_k = 0$ and $\sum_{k<n+1} a_k = \sum_{k<n} a_k + a_n$ .

[L2]

Two-term triangle inequality: $|x + y| \le |x| + |y|$ (The triangle inequality).

[L3]

Basic absolute value facts: $|x| \ge 0$ and $|0| = 0$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L4]

Adding a constant preserves the order: Order is preserved by adding a constant and by adding inequalities gives $x < y \Rightarrow x + c < y + c$ and states no nonstrict form, so the form used below, $x \le y \Rightarrow x + c \le y + c$ , is that statement together with the case $x = y$ , which trichotomy settles; and $\le$ is transitive (Ordered field).

[L5]

Induction principle (The principle of mathematical induction).

Proof

technique · induction

1.1

Base case $n = 0$ : both sides are $0$ , since $\big|\sum_{k<0} a_k\big| = |0| = 0$ and $\sum_{k<0}|a_k| = 0$ , so the inequality holds.

baseL1L3

1.2

Inductive hypothesis: fix $n \in \mathbb{N}$ and assume $\big|\sum_{k<n} a_k\big| \le \sum_{k<n} |a_k|$ .

2.1

Splitting off the last term and applying the two-term triangle inequality: $\big|\sum_{k<n+1} a_k\big| = \big|\sum_{k<n} a_k + a_n\big| \le \big|\sum_{k<n} a_k\big| + |a_n|$ .

step 1.2L1L2

3.1

Adding $|a_n|$ to both sides of the inductive hypothesis gives $\big|\sum_{k<n} a_k\big| + |a_n| \le \sum_{k<n}|a_k| + |a_n| = \sum_{k<n+1}|a_k|$ , so transitivity yields $\big|\sum_{k<n+1} a_k\big| \le \sum_{k<n+1}|a_k|$ .

step 1.2step 2.1L1L4

4.1

By the induction principle the inequality holds for every $n \in \mathbb{N}$ .

step 1.1step 3.1L5discharge-induction∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 31 results over 10 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. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
Young, Hölder, and Minkowski inequalities (Oregon State University) (standard reference, not scraped)
Triangle inequality (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §7.1 (standard reference, not scraped)