Alphabeta Math
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→R be a sequence of reals and n∈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

∣∑k<nak∣≤∑k<n∣ak∣.

Facts & Assumptions

Given: A sequence a:N→R and a natural n.

[L1]

Recursion clauses (Finite sums and finite products, by recursion): ∑k<0ak=0 and ∑k<n+1ak=∑k<nak+an.

[L2]

Two-term triangle inequality: ∣x+y∣≤∣x∣+∣y∣ (The triangle inequality).

[L3]

Basic absolute value facts: ∣x∣≥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⇒x+c<y+c and states no nonstrict form, so the form used below, x≤y⇒x+c≤y+c, is that statement together with the case x=y, which trichotomy settles; and ≤ is transitive (Ordered field).

[L5]

Proof

technique · induction
1.1

Base case n=0: both sides are 0, since ∣∑k<0ak∣=∣0∣=0 and ∑k<0∣ak∣=0, so the inequality holds.

baseL1L3
1.2

Inductive hypothesis: fix n∈N and assume ∣∑k<nak∣≤∑k<n∣ak∣.

ih
2.1

Splitting off the last term and applying the two-term triangle inequality: ∣∑k<n+1ak∣=∣∑k<nak+an∣≤∣∑k<nak∣+∣an∣.

step 1.2L1L2
3.1

Adding ∣an∣ to both sides of the inductive hypothesis gives ∣∑k<nak∣+∣an∣≤∑k<n∣ak∣+∣an∣=∑k<n+1∣ak∣, so transitivity yields ∣∑k<n+1ak∣≤∑k<n+1∣ak∣.

step 1.2step 2.1L1L4
4.1

By the induction principle the inequality holds for every n∈N.

step 1.1step 3.1L5discharge-induction∎

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources