LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-24

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.

Absolute value and the triangle inequality

Statement

For $x \in \mathbb{Q}$ set $|x| = x$ if $x \ge 0$ and $|x| = -x$ otherwise. Then $|xy| = |x|\,|y|$ , $|x + y| \le |x| + |y|$ , and $\bigl||x| - |y|\bigr| \le |x - y|$ .

The proof uses only the axioms of a totally ordered field, so it applies verbatim to any such field (in particular to $\mathbb{R}$ once constructed).

Facts & Assumptions

Given: Elements $x, y$ of the totally ordered field $\mathbb{Q}$ .

[L1]

Ordered-field arithmetic: trichotomy, compatibility of the order with addition, products of positives are positive, $(-u)v = -(uv)$ (The rationals form a totally ordered field).

Proof

technique · cases

1.1

For every $u$ : $u \le |u|$ , $-u \le |u|$ , $|u| \ge 0$ , and $|-u| = |u|$ , directly from the definition and trichotomy.

1.2

Case $x$ and $y$ have the same sign (both $\ge 0$ or both $< 0$ ): $xy \ge 0$ and $xy = |x||y|$ (for both negative, $xy = (-x)(-y)$ ).

assume-case sameL1

1.3

Case $x, y$ have mixed signs, say $x \ge 0 > y$ (the other order is identical): $xy \le 0$ , so $|xy| = -(xy) = x(-y) = |x||y|$ .

assume-case mixedL1

2.1

In all cases $|xy| = |x|\,|y|$ .

step 1.2step 1.3cases-exhaustive

2.2

Adding $x \le |x|$ and $y \le |y|$ gives $x + y \le |x| + |y|$ ; adding $-x \le |x|$ and $-y \le |y|$ gives $-(x+y) \le |x| + |y|$ ; since $|x+y|$ is one of $\pm(x+y)$ , $|x+y| \le |x| + |y|$ .

step 1.1L1

3.1

From $x = (x - y) + y$ : $|x| \le |x-y| + |y|$ , so $|x| - |y| \le |x-y|$ ; symmetrically $|y| - |x| \le |y-x| = |x-y|$ ; hence $\bigl||x|-|y|\bigr| \le |x-y|$ .

step 2.2L1∎

Depends on

The rationals form a totally ordered field

Used by

Cauchy sequence of rationals Definition
Null sequence Definition
FALSE: the rationals are complete False statement
A non-null Cauchy sequence is eventually bounded away from zero, with constant sign Lemma
Every Cauchy sequence of rationals is bounded Lemma
Null sequences are Cauchy Lemma
Null sequences form an ideal Lemma
The null ideal is maximal Lemma
Cauchy sequences form a commutative ring Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 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

T. Tao, Analysis I, 3rd ed., §4.3 (standard reference, not scraped)
Absolute value — ordered rings (Wikipedia) (standard reference, not scraped)