LemmaStatement: AI-adaptedProof: AI-adaptedprecheck 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 $a : N \to R$ be a sequence of reals and $n \in N$ , 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

$∣ \sum_{k < n} a_{k} ∣ \leq \sum_{k < n} ∣ a_{k} ∣ .$

Statement references

Finite sums and finite products, by recursion

Absolute value in an ordered field

Facts & Assumptions

Given: A sequence $a : N \to R$ 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: $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ (The triangle inequality).

[L3]

Basic absolute value facts: $∣ x ∣ \geq 0$ 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, $x \leq y \Rightarrow x + c \leq y + c$ , is that statement together with the case $x = y$ , which trichotomy settles; and $\leq$ is transitive (Ordered field).

[L5]

Induction principle (The principle of mathematical induction).

Proof

technique · induction

1.1baseL1L3

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

1.2ih

Inductive hypothesis: fix $n \in N$ and assume $∣ \sum_{k < n} a_{k} ∣ \leq \sum_{k < n} ∣ a_{k} ∣$ .

2.1step 1.2L1L2

Splitting off the last term and applying the two-term triangle inequality: $∣ \sum_{k < n + 1} a_{k} ∣ = ∣ \sum_{k < n} a_{k} + a_{n} ∣ \leq ∣ \sum_{k < n} a_{k} ∣ + ∣ a_{n} ∣$ .

3.1step 1.2step 2.1L1L4

Adding $∣ a_{n} ∣$ to both sides of the inductive hypothesis gives $∣ \sum_{k < n} a_{k} ∣ + ∣ a_{n} ∣ \leq \sum_{k < n} ∣ a_{k} ∣ + ∣ a_{n} ∣ = \sum_{k < n + 1} ∣ a_{k} ∣$ , so transitivity yields $∣ \sum_{k < n + 1} a_{k} ∣ \leq \sum_{k < n + 1} ∣ a_{k} ∣$ .

4.1step 1.1step 3.1L5discharge-induction∎

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

Triangle inequality for finite sums

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources