Axiom:Transitivity of Equidistance
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Axiom
Let $\equiv$ be the relation of equidistance.
Then the following axiom holds:
- $\forall a,b,p,q,r,s: \left({ab \equiv pq \land ab \equiv rs}\right) \implies pq \equiv rs$
where $a, b, p, q, r, s$ are points.
Intuition
If a line segment has the same length as two others, those two others have the same length as each other.
Also see
Sources
- June 1999: Alfred Tarski and Steven Givant: Tarski's System of Geometry (Bull. Symb. Log. Vol. 5, no. 2: pp. 175 – 214) : p. $177$ : Axiom $2$