Let $(A, +, \preceq)$ be an ordered group, namely $(A, +)$ is a group and $\preceq$ is a total order on $A$ such that $x + z \prec y + z$ and $z + x \prec z + y$ for all $x,y,z \in A$ with $x \prec y$.

Q. If $x,y,z \in A$ and $x \preceq z$, is it true that $x+y \ne y + z$ unless $x = z$?

Some observations: The answer is yes if (i) $y$ is in the commutator of either of $x$ or $z$, (ii) $x \preceq y \preceq z$, (iii) $x+y \preceq y+x$, (iv) $z+y \preceq y + z$, or (v) the subgroup, $S$, generated by $x$, $y$ and $z$ is Archimedean (this follows from a 1902 theorem by O. Hölder).

My feeling is that the answer is positive in general (here is an informal reasoning): For any $n \in \mathbb{Z}$ we have $-y+nx+y = nz$, so the smallest subgroup, say $T$, containing $z$ and $-y+x+y$ is torsion-free, abelian, and finitely generated. In my fantasy, this shoud be in contradiction to the fact that for $x \prec z$ the "gap" between $nx$ and $nz$ "grows" larger and larger as $n \to \infty$, while the "distance" between $nx$ and $-y+nx+y$ is "uniformly bounded". The point is that I don't see how to turn this into rigorous arguments, and particularly how to give a formal meaning to words or expressions like "gap", "grows", "distance", and "uniformly bounded". I had thought of introducing a group norm on $T$, but it didn't help much.

Motivation. I was led to Q while working at the proof of some estimates on sumsets in ordered monoids.