Three Points in Ultrametric Space have Two Equal Distances/Corollary 2

Theorem
Let $\struct {R, \norm{\,\cdot\,}}$ be a normed division ring with non-Archimedean norm $\norm{\,\cdot\,}$,

Let $x, y \in R$ and $\norm x \neq \norm y$.

Then:
 * $\norm {x+y} = \max \set {\norm x, \norm y}$.

Proof
Let $d$ be the metric induced by the norm $\norm{\,\cdot\,}$.

By Non-Archimedean Norm iff Non-Archimedean Metric then $d$ is a non-Archimedean metric and $\struct {R,d}$ is an ultrametric space.

Let $x, y \in R$ and $\norm x \neq \norm y$.

By the definition of the non-Archimedean metric $d$ then:
 * $\norm x = \norm {x - 0} = d \tuple{x, 0}$

and:
 * $\norm y = \norm {0 - \paren {-y}} = d \tuple{0, -y} = d \tuple{-y, 0}$

By assumption then:
 * $d \tuple{x, 0} \neq d \tuple{-y, 0}$

By Three Points in Ultrametric Space has Two Equal Distances then:
 * $d \tuple{x, -y} = \max \set {d \tuple{x, 0}, d \tuple{-y, 0}} = \max \set { \norm x, \norm y}$

By the definition of the non-Archimedean metric $d$ then:
 * $d \tuple{x, -y} = \norm {x - \paren {-y}} = \norm {x + y}$

The result follows.