Is the semigroup generated by wellordered positive set wellordered? Let $(A,\leq)$ be a totally ordered abelian group, and $\Gamma\subseteq A$ be a set of nonnegative elements, such that it is wellordered by $\leq$. Is it true then that the semigroup $S$ generated by $\Gamma$ is again wellordered?
Thoughts so far:


*

*Positivity hypothesis is obviously necessary, because for any negative $a\in A$, we have that $n\cdot a$ is an infinite decreasing sequence.

*I can show that for a wellordered $\Gamma$, it is also true that $\Gamma+\Gamma$ is wellordered, which easily extends to sets $\Gamma_n:=\Gamma+\Gamma+\ldots+\Gamma$ ($\Gamma$ is added $n$ times).

*We assume that $0\in \Gamma$, so that $\Gamma_n\subseteq \Gamma_{n+1}$.

*To show that $S$ is wellordered, we have to show that any $B\subseteq S$ has a smallest element; we can assume without loss of generality that $\Gamma\cap B$ is nonempty (extending $\Gamma$ to some $\Gamma_n$ if necessary).

*With the above assumptions, existence of a minimal element of $B$ is equivalent to the statement that the sequence $b_n:=\min (\Gamma_n\cap B)$ stabilizes.


The motivation of the question is to show that the ring of Hahn series $K((X^A))$ (with $K$ a field) is a field, but even if there's a simpler way to show it, I'm curious about the question on its own.
Thoughts? Hints?
 A: Let $(A,\le)$ be a totally ordered abelian group.
Lemma. If $X,Y$ are two wellordered subsets of $A$, then $X+Y$ is wellordered.
Proof.
Let $B$ be a nonempty subset of $X+Y$ and assume it has no minimum.
For $b\in B$ let $$x_b=\min\{x\in X\mid \exists y\in Y\colon x+y\in B, x+y<b\}$$
$$y_b=\min\{y\in Y\mid x_b+y\in B\}$$ and $s(b)=x_b+y_b$.
Then $s(b)<b$. This implies $x_{s(b)}\ge x_b$, hence $y_{s(b)}<y_b$
If $b\in B$ is arbitrary, this produces an infinite descending sequence 
$y_b>y_{s(b)}>y_{s(s(b))}>\ldots $
in $Y$, which is impossible. Therefore $B$ must have a minimal element.$_\blacksquare$
Theorem. If $\Gamma$ a positive wellordered subset of $A$, then $\langle\Gamma\rangle$ is well-ordered.
As a wellordered set, $\Gamma$ is order-isomorphis to some ordinal $\alpha$.
We proceed by induction on $\alpha$, that is: We may assume that $\langle\Gamma'\rangle$ is wellordered for all $\Gamma'$ of the form $\Gamma'=\{g\in \Gamma\mid g<\gamma\}$ for some $\gamma\in\Gamma$.
Let $B\subset\langle\Gamma\rangle$ be a nonempty set and let $b\in B$ be one of its elements.
If $b=0$, it is clearly minimal in the nonnegative set $\langle \Gamma\rangle$.
If $b\ne 0$, write
$$b=g_1+\cdots +g_n$$
with $n\in \mathbb N$, $g_i\in \Gamma$ and let $\gamma=\max\{g_1,\ldots,g_n\}$.
Then $b\le n\gamma$.
Consider $b'\in B$ with $b'\le b$ and write 
$$b'=g'_1+\cdots +g'_m$$
with $m\in \mathbb N$, $g'_i\in \Gamma$.
Then at most $n$ of the $g'_i$ are $\ge \gamma$.
All other summands are in  $\Gamma'=\{g\in \Gamma\mid g<\gamma\}$. By induction hypothesis, $\langle\Gamma'\rangle$ is wellordered. By the lemma,
$$T:=\underbrace{(\Gamma\cup\{0\})+\cdots+(\Gamma\cup\{0\})}_n+\langle\Gamma'\rangle$$
is wellordered.
We have just seen that $\{b'\in B\mid b'\le b\}\subseteq B\cap T$, hence
$$\min B =\min(T\cap B).$$
$_\blacksquare$ 
