Is the relation < necessarily a total order?

Question

Let $M$ be a non-empty set, $+$ a binary operation on $M$, and $<$ a relation on M, satisfying the following:

1. For all $a, b, c \in M$, $(a + b) + c = a + (b + c)$

2. For all $a, b \in M$ with $a \ne b$, exactly one of $a < b$ and $b < a$ is true.

3. For all $a, b \in M$ with $a < b$, at least one of the equations $a + x = b$ and $x + a = b$ has a solution for $x$ in $M$.

4. For all $a, b \in M$ with $a < b$, neither of the equations $b + x = a$ nor $x + b = a$ have a solution for $x$ in $M$.

The question is: is it necessarily true that for all $a, b, c \in M$, that if $a < b$ and $b < c$ then $a < c$?

Answer 1

Consider the oriented graph with vertex set $M$ in which edge $(a,b)$ exists if $a<b$.

Label each edge $(a,b)$ of this graph with the label $L$ if $x+a=b$ has solution.

Label each edge $(a,b)$ of this graph with the label $R$ if $a+x=b$ has solution.

Note each edge has at least one label.

Assume there is a triangle cycle $a\rightarrow b\rightarrow c\rightarrow a$. Such a cycle must contain two consegutive edges with the same label, so we can rotate it so that without loss of generality the first two arrows have the same label. without loss of generality the label is $L$.

We have $b=x+a$ and $c=y+b$ and so $c=y+x+a$ which means $a<c$ which is a contradiction, as we assumed $c<a$