Let $K$ be a complete local division ring (note $v$ its valuation). For $x,y\in K$ ($y\ne0$), one puts $x^y=yxy^{-1}$. Let $r\in\mathbb N$. Consider $x,y\in K$ and $a,b\in K^*$ such that $v(x-y)\ge r$ and $v(a-b)\ge r$. Do we have $v(x^a-y^b)\ge r$? In the commutative case, it is obvious but in the non-commutative case, I can not see the answer.
Thanks in advance
$\newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$ Not necessarily.
Take $K = \Q_3 + \Q_3 i + \Q_3 j + \Q_3 ij$ with $i^2=-1$ and $j^2=3$, $r=2$, $x=y=j$, $a=3i$ and $b=3(1+i)$. Then $v(j)=1$, and the maximal order of $K$ is $\Z_3+\Z_3 i+\Z_3 j + \Z_3 ij$. We have $a\equiv b \bmod 3$ so that $v(a-b) = 2$, but $x^a = -j$ and $y^b=ij$, which are not congruent modulo $3$.
Basically the problem is that $v(a-b)\ge r$ gives you no information since from arbitrary $a,b$ you can always ensure that this condition holds by multiplying by a high power of the uniformiser of the center of $K$, which does not change $x^a$ and $y^b$.
