Verification of a (counter-)example: valuation ring On $\mathbb{Z}\oplus\mathbb{Z}$, we put an order, given by $(a,a')>(b,b')$ if and only if $a>b$ and $b>b'$.

$(0)$ This is a partial order on the abelian group $\mathbb{Z}\oplus \mathbb{Z}$.

Let $R$ be the set of those rational numbers, whose denominator is coprime to $6=2\cdot 3$, when it is written in reduced form.
Now any rational can be written as $\frac{p}{q}=2^a3^b\frac{p'}{q'}$ where $a,b\in\mathbb{Z}$, and $p',q'$ are coprime integers which are also coprime with $6$.
Define $v:\mathbb{Q}\rightarrow \mathbb{Z}\oplus \mathbb{Z}$ by $v(\frac{p}{q})=(a,b)$.
I checked that $v$ satisfies following conditions:

$(1)$ If $v(\frac{p}{q})>\gamma$ and $v(\frac{r}{s})>\gamma$ for some $\gamma\in \mathbb{Z}\oplus\mathbb{Z}$, then $v(\frac{p}{q}+\frac{r}{s})>\gamma$.


$(2)$ $v(\frac{p}{q}\cdot \frac{r}{s})=v(p/q) + v(r/s)$.


$(3)$ $R$ is precisely the set of those rationals whose valuation is $\ge (0,0)$.

However, there are $x\in\mathbb{Q}$ such that neither $x$ not $x^{-1}$ lies in $R$, for example $x=\frac{2}{3}$.

Question: Are the above assertions, namely (0), (1), (2) (3) and the last statement correct?

Why this question and example came to my mind:
In Basic Algebra by P. M. Cohn (p. 307), the author mentions that, if the valuation $v$ on a field has values in totally ordered abelian group, then $R$ has a characterization.
Namely, $R$ is the ring of those elements of the quotient field, such that for every $x$ in the quotient field, either $x$ or $x^{-1}$ lies in $R$ (it is possible that both $x$ and $x^{-1}$ lie in $R$, for example, units).
 A: I think perhaps you might be interested in the Krull-Jaffard-Ohm-Kaplansky theorem, which says every lattice ordered abelian group can be realized as the value group of some Bezout domain.  If $D$ is just a GCD domain, then you know its value group is lattice ordered. You can see all this in

Chapman & Glaz (Eds.) Non-Noetherian Commutative Ring Theory. (2013). United States: Springer US.

The order you chose on $\mathbb Z\times\mathbb Z$ is lattice ordered and not totally ordered, right? And the domain you are talking about is the semilocalization $\mathbb ZS^{-1}$ where $S=((2)\cup (3))^c$, which is a PID (hence Bezout domain, and GCD domain) with exactly two maximal ideals, hence is not a valuation domain.
Maybe part of the problem in the comments is that when you phrase it as "a counterexample" it seems you suggest you disagree with the characterization and that what you're writing has a totally ordered value group but is not a valuation domain.
I think you mean rather that you are giving a non-example of a valuation ring along with its non-totally ordered value group to demonstrate the contrast with how valuation rings and their value groups behave.
