Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-26
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Nonnegativity of a metric is a consequence of the other axioms, not an axiom

Statement

Let XX be a set and let p:X×XRp : X \times X \to \mathbb{R} satisfy the reflexivity axiom (M1') p(x,x)=0p(x,x) = 0 and the symmetry axiom (M2) p(x,y)=p(y,x)p(x,y) = p(y,x) of Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric. Then:

  1. If pp satisfies the triangle inequality (M3), then p(x,y)0p(x,y) \ge 0 for all x,yXx, y \in X.
  2. If pp satisfies the strong triangle inequality (M3'), then p(x,y)0p(x,y) \ge 0 for all x,yXx, y \in X.

In particular every metric, every pseudometric and every ultrametric (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric) takes only nonnegative values. Nonnegativity is therefore a theorem about the axiom list this library uses, not a fourth axiom, and no statement on this page needs to assume it separately.

Facts & Assumptions

Given: A set XX, points x,yXx, y \in X, and a function p:X×XRp : X \times X \to \mathbb{R} satisfying (M1') p(a,a)=0p(a,a) = 0 for every aXa \in X and (M2) p(a,b)=p(b,a)p(a,b) = p(b,a) for all a,bXa, b \in X (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric).

[A1]

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

[A2]

(M3') The strong triangle inequality p(a,c)max{p(a,b),p(b,c)}p(a,c) \le \max\{p(a,b), p(b,c)\} holds for all a,b,cXa, b, c \in X (Metric space: d(x,y)=0d(x,y) = 0 iff x=yx = y, symmetry, and the triangle inequality; pseudometric and ultrametric).

[L1]

Trichotomy of the order of R\mathbb{R}: for reals a,ba, b exactly one of a<ba < b, a=ba = b, b<ab < a holds, so a0a \ge 0 fails exactly when a<0a < 0 (Order on the reals, Complete ordered field (least-upper-bound property), Ordered field).

[L2]

Adding two strict inequalities: if a<ba < b and c<dc < d then a+c<b+da + c < b + d (Order is preserved by adding a constant and by adding inequalities).

[L3]

A two-element subset {a,b}\{a,b\} of R\mathbb{R} has a maximum, and that maximum is aa or bb; if a=ba = b it is aa (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

Proof

technique · direct
1.1

Instantiate [A1] at a=xa = x, b=yb = y, c=xc = x: p(x,x)p(x,y)+p(y,x)p(x,x) \le p(x,y) + p(y,x).

A1
1.2

Instantiate [A2] at a=xa = x, b=yb = y, c=xc = x: p(x,x)max{p(x,y),p(y,x)}p(x,x) \le \max\{p(x,y), p(y,x)\}.

A2
1.3

Suppose, towards ruling it out, that p(x,y)<0p(x,y) < 0.

assume-hyp
2.1

By (M1') the left side of step 1.1 is 00 and by (M2) the right side is p(x,y)+p(x,y)p(x,y) + p(x,y), so 0p(x,y)+p(x,y)0 \le p(x,y) + p(x,y).

step 1.1given
2.2

By (M2) the two entries of the maximum in step 1.2 are the same real number, so that maximum equals p(x,y)p(x,y) by [L3], and (M1') turns step 1.2 into 0p(x,y)0 \le p(x,y), which is claim 2.

step 1.2givenL3
2.3

Adding the supposed inequality of step 1.3 to itself gives p(x,y)+p(x,y)<0+0=0p(x,y) + p(x,y) < 0 + 0 = 0.

step 1.3L2
3.1

Steps 2.1 and 2.3 assert 0p(x,y)+p(x,y)0 \le p(x,y) + p(x,y) and p(x,y)+p(x,y)<0p(x,y) + p(x,y) < 0, which trichotomy forbids; so the supposition of step 1.3 is untenable and p(x,y)0p(x,y) \ge 0, which is claim 1.

step 2.1step 2.2step 2.3L1

Remarks

  • What each claim uses. Claim 1 is the familiar two-line argument 0=p(x,x)p(x,y)+p(y,x)=2p(x,y)0 = p(x,x) \le p(x,y) + p(y,x) = 2p(x,y) followed by the observation that a negative real added to itself stays negative. Claim 2 does not need that second half at all: the strong triangle inequality delivers 0p(x,y)0 \le p(x,y) in one step, because the maximum of a real number with itself is that number.
  • Symmetry is used in both claims and cannot be dropped. Without (M2) the instantiation of step 1.1 only gives 0p(x,y)+p(y,x)0 \le p(x,y) + p(y,x), which leaves the possibility that one of the two values is negative and the other larger and positive. Dropping (M2) instead of weakening (M1) gives the notion usually called a quasimetric, which this library does not treat; for it the argument above is unavailable, so nonnegativity is not redundant there and is imposed as part of the definition (Which metric axiom list this library uses, the live naming fork between semimetric and pseudometric, and why extended metrics are not treated here).

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 results over 9 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