I was studying from some book and I came across something I haven't been able to justify. Let's suppose se have a field $K$ and an ordered group $G$ (with multiplicative notation) with an extra element called $0$ satisfying $$0x=0$$ and $$0<x$$ for all $x\in G$.
We then define a valuation as a function $$v:K\to G$$ satisfying $$v(x)=0\Leftrightarrow x=0$$ $$(\forall x,y),v(xy)=v(x)v(y)$$ $$v(x+y)\leq \max(v(x),v(y))$$
We can prove that $v(1)=1,v(- x)=v(x)$ for all $x$. Then in the book the proposition
If $v(x)<v(y)$ then $v(x+y)=v(y)$
is proved like this: $$v(y)=v(y+x-x)\leq \max(v(y+x),v(x))=v(y+x)\leq \max(v(y),v(x))=v(y)$$
Then step I've trouble with is:
$$\max(v(y+x),v(x))=v(y+x)$$
Why can we say that? I mean, why can we say that $v(y+x)>v(x)$ if $v(x)<v(y)$?
If $\max(v(y+x),v(x))=v(x)$, then plugging it into that inequality you would get $v(y)\leq v(x)$, which is a contradiction.