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

Define $\mathsf{B}_1$ as follows.

Define $\equiv_1$ as follows.

Now, define the isomorphism $\phi$ on $\left({\R^2, \mathsf{B}_2, \equiv_2}\right)$ as:

$\phi: \R^2 \to \R^2$ on $\left({\R^2, \mathsf{B}_1, \equiv_1}\right), \left({x_1,x_2}\right) \mapsto \left({x_1,2x_2}\right)$

Now consider the system $\left({\R^2, \mathsf{B}_2, \equiv_2}\right)$

Where $\mathsf{B}_2$ and $\equiv_2$ are the relations defined as above, but on the elements in the images of $\mathsf{B}_1$ and $\equiv_1$, respectively.

Observe that $\mathsf{B_1}$ and $\mathsf{B_2}$ coincide, because in:


 * $\left({x_1-y_1}\right) \cdot \left({2y_2-2z_2}\right) = \left({2x_2-2y_2}\right) \cdot \left({y_1-z_1}\right) \land$


 * $\left({0 \le \left({x_1-y_1}\right) \cdot \left({y_1-z_1}\right)}\right) \land \left({0 \le \left({2x_2-2y_2}\right) \cdot \left({2y_2-2z_2}\right)}\right)$

we can simply factor out the $2$ and divide both sides of the equality of inequality by $2$.

But consider the elements:


 * $p_1 = \left({0,0}\right)$


 * $p_2 = \left({0,1}\right)$


 * $p_3 = \left({1,0}\right)$

Observe that $p_1 p_2 \equiv_1 p_1 p_3$:


 * $\left({0-0}\right)^2 + \left({0 - 1}\right)^2 = \left({0-1}\right)^2 + \left({0-0}\right)^2$

But $\neg\left({p_1 p_2 \equiv_2 p_1 p_3}\right)$:


 * $\left({0-0}\right)^2 + \left({0 - 2}\right)^2 \ne \left({0-1}\right)^2 + \left({0-0}\right)^2$

But both $\left({\R^2, \mathsf{B}_1, \equiv_1}\right)$ and $\left({\R^2, \mathsf{B}_2, \equiv_2}\right)$ are both models of $\mathcal{G}$.

Recall that if $\left({A}\right)$ holds, it must hold in all systems. But it does not.

Hence $\left({A}\right)$ is false, from Proof by Contradiction

Also see

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