Axiom:Outer Additivity of Equidistance
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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
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.