LemmaStatement: AI-adaptedProof: AI-adaptedprecheck 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 $x \in Q$ set $∣ x ∣ = x$ if $x \geq 0$ and $∣ x ∣ = - x$ otherwise. Then $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ , $∣ x + y ∣ \leq ∣ x ∣ + ∣ y ∣$ , and $∣ ∣ x ∣ - ∣ y ∣ ∣ \leq ∣ x - y ∣$ .

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

Facts & Assumptions

Given: Elements $x, y$ of the totally ordered field $Q$ .

[L1]

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

Proof

technique · cases

1.1L1

For every $u$ : $u \leq ∣ u ∣$ , $- u \leq ∣ u ∣$ , $∣ u ∣ \geq 0$ , and $∣ - u ∣ = ∣ u ∣$ , directly from the definition and trichotomy.

1.2assume-case sameL1

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

1.3assume-case mixedL1

Case $x, y$ have mixed signs, say $x \geq 0 > y$ (the other order is identical): $x y \leq 0$ , so $∣ x y ∣ = - (x y) = x (- y) = ∣ x ∣ ∣ y ∣$ .

2.1step 1.2step 1.3cases-exhaustive

In all cases $∣ x y ∣ = ∣ x ∣ ∣ y ∣$ .

2.2step 1.1L1

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

3.1step 2.2L1∎

From $x = (x - y) + y$ : $∣ x ∣ \leq ∣ x - y ∣ + ∣ y ∣$ , so $∣ x ∣ - ∣ y ∣ \leq ∣ x - y ∣$ ; symmetrically $∣ y ∣ - ∣ x ∣ \leq ∣ y - x ∣ = ∣ x - y ∣$ ; hence $∣ ∣ x ∣ - ∣ y ∣ ∣ \leq ∣ x - y ∣$ .

Depends on

The rationals form a totally ordered field

Absolute value and the triangle inequality

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources