What are mathematically definable/ useful properties of ternary relations? ( How to define for example symmetry or transitivity?) My question may seem gratuitous. Here are my presuppositions. What makes a relation mathematically interesting is that one can define abstract properties of this relation , such as reflexivity, symmetry, transitivity. These three are, it seems to me, the basic ones. So, knowing that ternary relations exist and are sometimes used  in mathematics  ( for example the relation of colinearity in geometry), I presuppose that abstract properties can be defined for these relations that make them interesting. So I began to ask myself how to define these properties for ternary relations. And I started  with reflexivity, symmetry and transitivity. But maybe I did not take the right way. What other properties are interesting for ternary relations? 

As an attenpt to define reflexivity for a ternary relation , I would say that 
R  is reflexive iff ( for all a, the triple ( a,a,a) belongs to R ). 
But I cannot manage to find a plausible definition of symmetry or of transitivity for ternary relations. 
Symmetry requires the notion of , so to say, " converse" n-tuple. The " converse" couple of (a,b) is (b,a). What would be the " converse" triple of 
(a,b,c)? Should one  say ( a, c, b)? ( c, a ,b) ? 
If I analyse (a,b,c) as ( (a,b), c); may I say that ( c, (a,b) ) is its " converse" triple? 
The problem, is, apparently, that if ( (a,b), c) ( with a, b and c elements of some set A) belongs to a relation R, this relation is from A² to A . But the pair ( c, (a,b)) is not a possible element of a relation from A² to A. So the second pair cannot belong to the same relation as the first one . But this is required to define symmetry: a pair and it's "converse" pair have to belong to the same relation in order this relation to be called symmetric. 
I think the same problem would occur with transitivity. 
 A: You could try defining things like so:
$1)$ Reflexivity: $(x,y,x)\in R\;$ for all $\;x,y\in A$ 
$2)$ Symmetry: $(x,y,z)\Leftrightarrow(x,z,y)\Leftrightarrow(y,x,z)\Leftrightarrow(y,z,x)\Leftrightarrow(z,x,y)\Leftrightarrow(z,y,x)$ (all permutations are equivalent, essentially) for all $\;x,y,z\in A$ 
$3)$ Transitivity: $(x,y,z)\wedge(y,z,w)\wedge(z,w,a)\implies (x,z,a)\;$ for all $\; x,y,z,w,a\in A$ 
This equivalence relation produces a natural partition over $A^2$, although its usefulness, of course, depends on where the "ternary equivalence relation" is being used.
Define the equivalence class of any element, $i\in A$, under a ternary equivalence relation as $[i]=\{(j,k)\;|\;(i,j,k)\}$ 
(here, the first set of parentheses denotes an ordered pair while the second represents the ternary relation- in the remainder of the answer, take parentheses around two elements to mean an ordered pair and parentheses around three elements to mean that the three elements satisfy the ternary equivalence relation). 
By the reflexive property, we will have $(i,i)\in [i]$ for all $i$ so that each equivalence class is nonempty.  
Also, by reflexivity, we have $(k,j,k)$ for any pair of $j,k\in A$ so each element in $A^2$ belongs to some equivalence class. 
Suppose for some $x,y\in A$ $[x]\cap[y]\neq \varnothing$.  
There must then exist some $j,k\in A$ s.t. $(x,j,k)$ and $(y,j,k)$. 
Now, let $(b,c)$ be any element in $[y]$. 
Step 1: By symmetry we have $(j,k,y)$. By reflexivity we have $(y,k,y)$ and, then, symmetry we get $(k,y,y)$. Overall, we have $(x,j,k)\wedge (j,k,y)\wedge (k,y,y)$ which, by transitivity, gives us $(x,k,y)$. This was the first application of transitivity.  
Step 2: We have shown $(x,k,y)$. $(k,y,y)$ still holds (by reflexivity+symmetry). By analogous reasoning we have $(y,y,b)$ (recall $(b,c)$ was chosen to be an arbitrary element in $[y]$). The net result is $(x,k,y)\wedge (k,y,y)\wedge (y,y,b)\implies(x,y,b)$. 
Step 3: $(x,y,b)$ has been shown and $(y,b,c)$ holds by assumption. $(b,c,c)$ is obvious. A final application of transitivity yields $(x,b,c)$, i.e. $(b,c)\in [x]$.  
Since $(b,c)$ was an arbitrary member of $[y]$, we have $[y]\subset [x]$. We can then reverse the roles of $x$ and $y$ to similarly get $[x]\subset [y]$ so that $[x]=[y]$ whenever $[x]$ and $[y]$ are not disjoint, meaning that the ternary equivalence relation partitions $A^2$. 
One thing worth noting is that the condition for symmetry may be stronger than required. It was assumed that any permutation implied all the others simply to simplify the argument. If we were more careful, we could note that, treating the permutations of $(x,y,z)$ as members of the permutation group $S_3$, we require only three cycles of $S_3$ to prove that the equivalence relation produces a partition: $(132), (12)$ OR $(213)$, and $(32)$ OR $(231)$.   
So one could weaken the symmetry condition, if need be, while still retaining the partition property.
A: For the relation $C$ that's the set of all triples of colinear points you know


*

*$(x, y , z) \in C$ implies $(p,q,r) \in C$ for any of the six
permutations of the three points.

*$(x, x , z) \in C$ .

*If $(x, y , z) \in C$ , $(x, y , w) \in C$ and $x \ne y$ then $(x, w
   , z) \in C$ .


You could call these properties "symmetric", "reflexive" and "transitive" if you wanted to. You could look for other ternary relationships that share these properties, and try to prove theorems.
If you start from another interesting ternary relationship you might be led to other definitions for your abstraction.
For fun, how might you begin to write down an abstract generalization of the ternary relation $B$ on points on a line given by $(x,y,z) \in B$ just when $y$ is between $x$ and $z$?
