Axiom:Outer Additivity of Equidistance

Axiom

Let $\equiv$ be the relation of equidistance.

Let $\mathsf{B}$ be the relation of betweenness.

This axiom asserts:

$\forall x,y,z,x',y',z' : \left({\mathsf{B}xyz \land \mathsf{B}x'y'z' \land xy \equiv x'y' \land yz \equiv y'z'}\right) \implies xz \equiv x'z'$

where $x, y, z, x', y', z'$ are points.

Intuition

Let $xy$ and $x'y'$ be line segments of the same length

Further let $yz$ and $y'z'$ be line segments of the same length.

If you connect $xy$ to $yz$, and $x'y'$ to $y'z'$, segments $xyz$ and $x'y'z'$ will be the same length.

Also see

• Inner Additivity of Equidistance

Sources

• June 1999: Alfred Tarski and Steven Givant: Tarski's System of Geometry (Bull. Symb. Log. Vol. 5, no. 2: pp. 175 – 214) : p. $188$ : Axiom $23$
  Illustration courtesy of Steven Givant.