LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26

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 reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) and let $x, y, z \in X$ . Then

$∣ d (x, z) - d (y, z) ∣ \leq d (x, y),$

where $∣ \cdot ∣$ is the absolute value of $R$ (Absolute value in an ordered field).

Facts & Assumptions

Given: A metric space $(X, d)$ and points $x, y, z \in X$ ; write $t : = d (x, z) - d (y, z)$ .

[A1]

The triangle inequality (M3): $d (a, c) \leq d (a, b) + d (b, c)$ for all $a, b, c \in X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[A2]

Symmetry (M2): $d (a, b) = d (b, a)$ for all $a, b \in X$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

For every real $u$ , the value $∣ u ∣$ equals $u$ or $- u$ (Basic properties of the absolute value, Absolute value in an ordered field).

[L2]

Adding a constant to an inequality: if $a \leq b$ then $a + c \leq b + c$ . Order is preserved by adding a constant and by adding inequalities states the strict form $a < b \Rightarrow a + c < b + c$ ; the nonstrict form used here is that strict form together with the case $a = b$ , in which the two sides are equal, the order being total (Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1A1

By [A1] at $(a, b, c) = (x, y, z)$ : $d (x, z) \leq d (x, y) + d (y, z)$ .

1.2A1A2

By [A1] at $(a, b, c) = (y, x, z)$ : $d (y, z) \leq d (y, x) + d (x, z)$ , and by [A2] $d (y, x) = d (x, y)$ , so $d (y, z) \leq d (x, y) + d (x, z)$ .

2.1step 1.1L2

Adding $- d (y, z)$ to both sides of step 1.1 gives $t = d (x, z) - d (y, z) \leq d (x, y)$ .

2.2step 1.2L2

Adding $- d (x, z)$ to both sides of step 1.2 gives $d (y, z) - d (x, z) \leq d (x, y)$ , that is $- t \leq d (x, y)$ .

3.1step 2.1step 2.2L1∎

By [L1] the real number $∣ t ∣$ is either $t$ or $- t$ , and both of these are at most $d (x, y)$ by steps 2.1 and 2.2, so $∣ d (x, z) - d (y, z) ∣ = ∣ t ∣ \leq d (x, y)$ .

Remarks

Read with $z$ fixed, this says the function $u \mapsto d (u, z)$ does not increase distances: its values at $x$ and at $y$ differ by at most $d (x, y)$ . That is the model for $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, which proves the same estimate with the point $z$ replaced by a nonempty set.
The inequality specialises, on $R$ with $d (u, v) = ∣ u - v ∣$ (The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded), to the familiar $∣ ∣ x - z ∣ - ∣ y - z ∣ ∣ \leq ∣ x - y ∣$ .

The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The reverse triangle inequality ∣d(x,z)−d(y,z)∣≤d(x,y)∣d(x,z)−d(y,z)∣≤d(x,y) in any metric space

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

The reverse triangle inequality $∣ d (x, z) - d (y, z) ∣ \leq d (x, y)$ in any metric space