Axiom:Outer Connectivity of Betweenness
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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
Point $c$ is between $a$ and $d$.
Case 2
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".
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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. $186$ : Axiom $18$