Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-25
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.

The triangle inequality

Statement

Let FF be an ordered field (Ordered field) and let x,yFx, y \in F. Then

x+yx+y.|x + y| \le |x| + |y|.

Facts & Assumptions

Given: An ordered field FF and elements x,yFx, y \in F.

[L1]

For every uFu \in F, uuu-|u| \le u \le |u|, and u|u| equals uu or u-u (Basic properties of the absolute value).

[L2]

Order compatible with addition: if aba \le b and cdc \le d, then a+cb+da + c \le b + d. Order is preserved by adding a constant and by adding inequalities states the STRICT forms and only those (a<ba+c<b+ca < b \Rightarrow a + c < b + c, and a<ba < b with c<dc < d giving a+c<b+da + c < b + d); the nonstrict form used here is those two together with the cases a=ba = b and c=dc = d, settled by trichotomy, the order being total (Ordered field). Explicitly: if a<ba < b and c<dc < d the second strict form applies; if a<ba < b and c=dc = d the first gives a+c<b+c=b+da + c < b + c = b + d; if a=ba = b and c<dc < d the first gives a+c<a+d=b+da + c < a + d = b + d; and if a=ba = b and c=dc = d the two sides are equal.

[L3]

Field and order arithmetic: (a+b)=a+(b)-(a + b) = -a + (-b), and ab    baa \le b \iff -b \le -a (Ordered field).

Proof

technique · direct
1.1

By [L1], xxx-|x| \le x \le |x| and yyy-|y| \le y \le |y|.

L1
2.1

Adding the two chains of [step 1.1] with [L2] and using (x+y)=x+(y)-(|x| + |y|) = -|x| + (-|y|) from [L3] gives (x+y)x+yx+y-(|x| + |y|) \le x + y \le |x| + |y|.

step 1.1L2L3
3.1

By [L1] the value x+y|x + y| equals x+yx + y or (x+y)-(x + y); both x+yx+yx + y \le |x| + |y| and (x+y)x+y-(x + y) \le |x| + |y| hold by [step 2.1] and [L3] (the latter from (x+y)x+y-(|x| + |y|) \le x + y), so x+yx+y|x + y| \le |x| + |y|.

step 2.1L1L3

Depends on

Used by

…and 24 more results.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 8 results over 5 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