Axiom:Transitivity of Equidistance

Axiom
Let $a,b,p,q,r,s$ be points.

Let $\equiv$ be the relation of equidistance.

Then the following axiom holds:


 * $\forall a,b,p,q,r,s: \left({ab \equiv pq \land ab \equiv rs}\right) \implies pq \equiv rs$

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

Also see

 * Euclid's Common Notions