The rule of three steps for a cyclically ordered group

Question

Is this rule correct?
If yes, is there a better way to prove it?
If not, what would be an example that does not satisfy the rule?

Theorem. For any element $a$ of a group with a non-linear cyclic order there is a combination of not more than three non-negative elements $b$, $c$, $d$ such that $a + b + c + d = 0$.
If the cyclic order is Archimedean or it has an apex, then there is a combination of not more than two non-negative elements $b$, $c$ such that $a + b + c = 0$.

Corollary. For any positive element $a$ of a group with a non-linear cyclic order there is a combination of not more than three positive elements $b$, $c$, $d$ such that $a + b + c + d = 0$.

(Positive and negative elements of a cyclically ordered group)

Proof:

Lemma 1. For any non-negative element $x$ of an Archimedean cyclically ordered group there is a non-negative element $y$ such that $x + y$ is non-positive (A property of an Archimedean cyclically ordered group).

Lemma 2. If $z$ is an apex, then $x + z$ is negative (positive) for any positive (negative) $x$ (Apex of a cyclically ordered group).

• Any cyclically ordered group $G$ has a representation $K \times L$ where $K$ is an Archimedean cyclically ordered group, and $L$ is a linearly ordered group (Swierczkowski, Jakubik, Pringerova);
• The cyclic order on $G$ is not linear $\iff$ $K$ is not trivial;
• Checking all possible cases for an element $g = (k, x), k \in K, x \in L$:
  1. Case $L$ is trivial: $G = K$, $G$ is Archimedean;
     • Case $g$ is negative:
       • $g' = –g$ is positive,
       • $g + g' = 0$;
     • Case $g$ is not negative:
       • Applying Lemma 1 to $G$: there is a non-negative $g'$: $g + g'$ is not positive;
       • $g'' = -(g + g')$ is not negative;
       • $g + g' + g'' = 0$;
  2. Case $L$ is not trivial: $G$ is not Archimedean since $[(0, 0), n(0, x), (0, -x)]$ for any $x > 0, n \in \mathbb N$;
     • Case $g = 0$;
     • Case $g$ is negative or an apex:
       • $g' = –g$ is not negative;
       • $g + g' = 0$;
     • Case $g$ is positive:
       • Case $k$ is an apex in $K$ $(x < 0)$:
         • An element $g' = (k, 0)$ is an apex in $G$;
         • Applying Lemma 2: $g + g’$ is negative;
         • $g'' = -(g + g')$ is positive;
         • $g + g' + g'' = 0$;
       • Case $k$ is positive in $K$ $(\forall x \in L)$. Applying Lemma 1 to $K$:
         there is a non-negative $k'$ such that $k + k'$ is not positive.
         • Case $k’ = 0$ is not applicable: $k + 0$ is positive;
         • Case $k’$ is an apex in $K$: $g' = (k', 0)$ is an apex in $G$, reviewed;
         • Case $k'$ is positive in $K$:
           • Case $k + k' = 0$ is not applicable: $k' = -k$ is negative;
           • Case $k + k'$ is an apex in $K$: reviewed;
           • Case $k + k'$ is negative in $K$:
             • $g' = (k', 0)$ is positive;
             • $g + g' = (k + k', x)$ is negative;
             • $g'' = -(g + g')$ is positive;
             • $g + g' + g'' = 0$;
       • Case $k = 0$ $(x > 0)$. $K$ is non-trivial; therefore, there is an element $k' ≠ 0$;
         • Case $k'$ is an apex in $K$: reviewed;
         • Case $k'$ is positive in $K$:
           • An element $g' = (k', 0)$ is positive;
           • $g + g' = (k', x)$ is positive;
           • Applying the case for a positive element $(k', x)$ with a positive $k'$:
             there is a combination of not more than two non-negative elements $g'', g'''$ such that $g + g' + g'' + g''' = 0$;
         • Case $k'$ is negative in $K$: applying the previous case with $k'' = -k'$.