Axiom:Outer Connectivity of Betweenness

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

Let $=$ be the relation of equality.

This axiom asserts that:
 * $\forall a, b, c, d: \left({\mathsf{B}abc \land \mathsf{B}abd \land \neg \left({a = b}\right) }\right) \implies \left({\mathsf{B}acd \lor \mathsf{B}adc}\right)$

where $a, b, c, d$ are points.

Intuition
Let $abc$ and $abd$ be line segments.

They exist in one of the following configurations:

Case 1

 * Outer Connectivity of Betweenness Case1.png

Point $c$ is between $a$ and $d$.

Case 2

 * Outer Connectivity of Betweenness Case2.png

Point $d$ is between $a$ and $c$.

Note that this axiom does not assert that exactly one of the cases happens.

This axiom still holds in the degenerate cases where not all the points are not (pairwise) distinct.

For example, if we are dealing with exactly three points, this axiom could be interpreted as "three points on a line are between each other".

Also see

 * Inner Connectivity of Betweenness