Axiom:Inner Additivity of Equidistance

Axiom

Let $x,y,z,x',y',z'$ be points.

Let $\equiv$ be the relation of equidistance.

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

This axiom asserts:

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

Intuition

Let $xz$ and $x'z'$ be line segments of the same length

Cut off segments $yz$ and $y'z'$ from lines $xz$ and $x'z'$, respectively.

If $yz$ is the same length as $y'z'$, then the remaining line segments $xy$ and $x'y'$ are the same length as each other.

Also see

• Outer 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 $24$
  Illustration courtesy of Steven Givant.