Equidistance is Independent of Betweenness

Theorem
Let $\mathcal{G}$ be a formal systematic treatment of geometry containing only:


 * The language and axioms of first-order logic, and the disciplines preceding it


 * The undefined terms of Tarski's Geometry (excluding equidistance)


 * Some or all of Tarski's Axioms of Geometry

In $\mathcal{G}$, equidistance $\equiv$ is necessarily an undefined term with respect to betweenness $\mathsf{B}$.

Proof
Our assertion is that $\equiv$ cannot be defined in terms of $\mathsf{B}$.

Seeking a contradiction, assume that it can. Call this assumption $\left({A}\right)$

If $\left({A}\right)$ holds, it must hold in all systems.

Let one such system be $\left({\R^2, \mathsf{B}_1, \equiv_1}\right)$ where:


 * $\R^2$ is the cartesian product of the reals with itself


 * $\mathsf{B}_1$ is a ternary relation of betweenness


 * $\equiv_1$ is a quaternary relation of equidistance

Let $\mathcal{G}$ be a discipline preceding the given discipline, where $\mathcal{G}$ is as defined above (excluding both $\equiv$ and $\mathsf{B}$

Also see

 * Betweenness Not Independent of Equidistance, which states that there are models where one can define $\mathsf{B}$ in terms of $\equiv$.