Is $x-x = 0x$ true in wheel theory? https://en.wikipedia.org/wiki/Wheel_theory
On the wikipedia page about wheel theory, it says that you can derive that $x-x = 0x^2$. You can also derive from the rules that $/0*/0=/0$ and $0/0*0/0=0/0$. So if $x=/0$ or $x=0/0$, then $0x^2 = 0x$
There are 3 examples of wheels on the wikipedia page, in all of them it's true that if $x≠/0$ and $x≠0/0$ then $x-x=0$. As far as I know, this is true for any wheel. So why doesn't the wikipedia page just say that $x-x = 0x$? What's the point of x being squared?
These are my guesses:

*

*There are wheels were $x≠/0$ and $x≠0/0$ does not imply $x-x=0$. If this is the case, are there any examples?


*$x-x = 0x$ is true, but can't be derived by only using algebraic manipulation.


*You can derive that $x-x = 0x$, wikipedia just didn't.
 A: The following is the fifth example (not highlighted by an example environment) found on page 6 in this pdf from the references in the Wikipedia article.
Consider the ring $R = \Bbb{Z} / 4\Bbb{Z}$, and define the following equivalence relation on $R \times R$:
\begin{align*}
(a, b) \sim (c, d) &\iff \exists e, f \in \{1, 3\} \subseteq R : (ea, eb) = (fc, fd) \\
&\iff \exists e \in \{1, 3\} \subseteq R : (ea, eb) = (c, d).
\end{align*}
The pointless equivalence keeps the example within the discussion of the ring of fractions sketched in 1.2.
As in the pdf, we denote $[a, b]$ to be the equivalence class of $(a, b)$. We further denote by $a$ the equivalence class $[a, 1]$, $/a$ the equivalence class $[1, a]$, and $a/b$ the equivalence class $[a,b]$. We let $W$ be the wheel of all equivalence classes; it is a wheel of fractions with respect to the multiplicative submonoid $\{1, 3\} \subseteq R$.
Addition on $W$ is defined by
$$a/b + c/d = (ad + bc)/bd,$$
where the addition and multiplication on the right are considered modulo $4$. Similarly, multiplication is defined by
$$(a/b)(c/d) = (ac)/(bd).$$
You can prove, if you like, that these operations are well-defined and define a wheel. Or you can look at the proof in the pdf.
Note that $1$ has additive inverse $3$, so $x - x \equiv x + 3x$, by the definition given in the Wikipedia article. We then have
$$/2 - /2 = /2 + 3/2 = (1\cdot 2 + 3 \cdot 2)/(2 \cdot 2) = 0/0.$$
On the other hand,
$$0 \cdot /2 = 0/2.$$
So, in order to have $/2 - /2 = 0 \cdot /2$, we must have $0/0 = 0/2$. This would only be true if $(0, 0) \sim (0, 2)$, i.e. there exist $e \in \{1, 3\}$ such that $0 \cdot e = 0$ and $0 \cdot e = 2$. This is clearly not the case, so we do not have $x - x = 0x$ in general for this ring, even if we exclude the $x = /0$ and $x = 0/0$ cases.
As a sanity check, let's verify that $0(/2)^2 = /2 - /2 = 0/0$. We have
$$(/2)^2 = (1 \cdot 2 + 1 \cdot 2)/(2 \cdot 2) = 0/0.$$
But then
$$0(/2)^2 = 0(0/0) = (0 \cdot 0)/(1 \cdot 0) = 0/0,$$
as required.
