p-adic ultrametric balloons I want to show that with p-adic ultrametric : $|.|_{p}=p^{-ord(.)}$ where . is a integer and p is a prime , with $|0|_{p}=0$ if we have balloons $$B(a,R)=\{z\in \mathbb{Z}||z-a|_{p}\le R\}$$  $$B(b,R)=\{z\in \mathbb{Z}||z-b|_{p}\le R\}$$
then they can never overlap. 
Proposition: They can never overlap without being the same. 
Assume we have two different points y and y' in one ball, if $y\in B(a,R)$ with the strong  triangle inequality $|a+b|_{p}\le sup\{|a|_{p} ,|b|_{p}\}$
now set: $a:=z-y'$ and $b:= y'-y$ so we get : $|z-y|_{p}\le sup\{|z-y'|_{p}. |y'-y|_{p} \}$
So that means $|z-y|_{p} \le R \Leftrightarrow |z-y'|_{p}\le R $ e.g. every point of the ball is a center point so the balloons can never overlap without being the same. 
Is this a proof for the proposition? (I never made use of the second balloon...)

Your proof is correct and usually the point where most people would stop, but you can complete the proof explicitly to make it clearer how the second balloon is related:
You have shown that for all $y, y' \in B(a,R)$, you have $ B(a,R) = B(y,R)=B(y',R).$ Now suppose there is some intersection between the two balloons, say at $z$. Then since $z$ is in the first ball, by the previous result $ B(a,R) = B(z,R).$ Similarly, since $z$ is in the second balloon, $B(b,R) = B(z,R)$, and thus $B(a,R) = B(b,R)$ so the balloons are the same.  
