Equidistance is Independent of Betweenness

Theorem
Let $\GG$ 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 $\GG$, 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 $\mathsf B$.

that it can.

Call this assumption $\paren A$.

If $\paren A$ holds, it must hold in all systems.

Let one such system be $\tuple {\R^2, \mathsf B_1, \equiv_1}$ where:


 * $\R^2$ is the cartesian product of the set of real numbers with itself


 * $\mathsf B_1$ is a ternary relation of betweenness


 * $\equiv_1$ is a quaternary relation of equidistance

Let $\GG$ be the discipline preceding the given discipline, where $\GG$ 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 $\struct {\R^2, \mathsf B_2, \equiv_2}$ as:


 * $\phi: \R^2 \to \R^2$ on $\struct {\R^2, \mathsf B_1, \equiv_1}, \tuple {x_1, x_2} \mapsto \tuple {x_1, 2 x_2}$

Now consider the system:
 * $\struct {\R^2, \mathsf B_2, \equiv_2}$

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:


 * $\paren {x_1 - y_1} \cdot \paren {2 y_2 - 2 z_2} = \paren {2 x_2 - 2 y_2} \cdot \paren {y_1 - z_1} \land$


 * $\paren {0 \le \paren {x_1 - y_1} \cdot \paren {y_1 - z_1} } \land \paren {0 \le \paren {2 x_2 - 2 y_2} \cdot \paren {2 y_2 - 2 z_2} }$

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

But consider the elements:


 * $p_1 = \tuple {0, 0}$


 * $p_2 = \tuple {0, 1}$


 * $p_3 = \tuple {1, 0}$

Observe that $p_1 p_2 \equiv_1 p_1 p_3$:


 * $\paren {0 - 0}^2 + \paren {0 - 1}^2 = \paren {0 - 1}^2 + \paren {0 - 0}^2$

But $\map \neg {p_1 p_2 \equiv_2 p_1 p_3}$:


 * $\paren {0 - 0}^2 + \paren {0 - 2}^2 \ne \paren {0 - 1}^2 + \paren {0 - 0}^2$

But both $\struct {\R^2, \mathsf B_1, \equiv_1}$ and $\struct {\R^2, \mathsf B_2, \equiv_2}$ are both models of $\GG$.

Recall that if $\paren A$ holds, it must hold in all systems.

But it does not.

Hence $\paren A$ 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 $\equiv$.