%
\label{thm:ultrametric-triangle}
For all $x, y, z \in \tau^3$:
\[
    \boxed{d(x, z)
    \;\leq\;
    \max\bigl(\, d(x, y),\; d(y, z) \,\bigr).}
\]

If any two of $x, y, z$ are equal,
the inequality is immediate.
Assume $x, y, z$ are pairwise distinct.
Let $k$ be any stage with $k \leq \min\bigl(\delta(x,y),\, \delta(y,z)\bigr)$.
Then $\pi_k(x) = \pi_k(y)$ and $\pi_k(y) = \pi_k(z)$,
so by transitivity of equality,
$\pi_k(x) = \pi_k(z)$.
This holds for every such~$k$, hence
\[
    \delta(x,z)
    \;\geq\;
    \min\bigl(\delta(x,y),\, \delta(y,z)\bigr).
\]
Applying the decreasing function $t \mapsto 2^{-t}$:
\[
    2^{-\delta(x,z)}
    \;\leq\;
    2^{-\min(\delta(x,y),\, \delta(y,z))}
    \;=\;
    \max\bigl(2^{-\delta(x,y)},\, 2^{-\delta(y,z)}\bigr),
\]
which is exactly $d(x,z) \leq \max\bigl(d(x,y),\, d(y,z)\bigr)$.