Is it possible to have a single axiom that subsumes axioms 8-10 in this list?

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Think of a totally ordered set as an “order-theoretic line”. Similarly, cyclic orders are “order-theoretic circles”. I want to find the right axioms for an “order-theoretic plane”.

My ultimate goal is to design a data structure that represents a finite set of points on an “order-theoretic plane”, similarly to how binary search trees represent finite sets of points on an “order-theoretic line”.

I will describe the progress I made so far. I started with an axiomatization of oriented triangles:

We postulate a relation $\triangle abc$ (triangle) satisfying the following axioms:

  1. Cyclicity: $\triangle abc \iff \triangle cab$
  2. Asymmetry: $\neg \triangle abc \vee \neg \triangle cba$
  3. Encirclement: $\neg \triangle abc \vee \neg \triangle acd \vee \neg \triangle adb \vee \triangle bcd$

Encirclement is the ternary analogue of a partial order's transitive property. The transitive property of a partial order constructs a large oriented segment by concatenating two smaller ones (in the partial order's Hasse diagram). Similarly, the encirclement property of $\triangle$ constructs a large oriented triangle by concatenating three smaller ones.

Three pairwise distinct points $a,b,c$ are collinear if neither $\triangle abc$ nor $\triangle cba$ holds. If three points are collinear, exactly one of them must be between the other two:

We postulate a relation $\overline {abc}$ (segment) satisfying the following axioms:

  1. Symmetry: $\overline {abc} \iff \overline {cba}$
  2. Acyclicity: $\neg \overline {abc} \vee \neg \overline {cab}$

Triangles and segments satisfy the following axioms:

  1. Mutual exclusion: $\neg \triangle abc \vee \neg \overline {abc}$
  2. Exhaustiveness: $\triangle abc \vee \triangle cba \vee \overline {abc} \vee \overline {bca} \vee \overline {cab} \vee a = b \vee a = c \vee b = c$

This axiomatization is incomplete. I need to postulate that, if $\overline {abc}$, then $\triangle abd$, $\triangle acd$, $\triangle bcd$ are equivalent statements.

Alas, I have not found a neat way to translate this informal statement to formal language. I have to clumsily split it into three statements:

  1. Triangle base translation 1: $\neg \overline {abc} \vee \neg \triangle abd \vee \triangle acd$

  2. Triangle base translation 2: $\neg \overline {abc} \vee \neg \triangle acd \vee \triangle bcd$

  3. Triangle base translation 3: $\neg \overline {abc} \vee \neg \triangle bcd \vee \triangle abd$

Or is there a better way?