Axiom:Transitivity of Equidistance

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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.


If a line segment has the same length as two others, those two others have the same length as each other.

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