Let $R$ be a discrete valuation ring. We have a valuation $v$ on the fraction field $K$ of $R$. Let $x,y \in K$ with $v(x)<v(y)$. I want to prove that $v(x+y) = \min(v(x), v(y))$.
So, we have $v(x) = v((x+y)-y) \geq \min(v(x+y),v(y)) \geq \min(v(x),v(y)) = v(x)$.
Questions about the proof:
(1) What justification makes it valid that $\min(v(x+y),v(y)) \geq \min(v(x),v(y))$?
(2) I thought of adding a little details by saying $$\min(v(x+y),v(y)) \geq \min(\min(v(x),v(y)),v(y)) \geq \min(v(x),v(y)),$$ but it seems to me that I am assuming what I want to prove is true in the proof.
I need help for some clarification. Thanks.