Intervals and metric trees are CAT(0)

(i). Every interval $I$ is a convex subset of $\mathbb{R}$; for $x<y$ in $I$ the interval $[x,y]\subseteq I$ with its usual parametrization is a geodesic segment from $x$ to $y$ by [F1], and it is the only one, since a distance-preserving map into $\mathbb{R}$ from an interval is determined by its values at the endpoints and is monotone. A geodesic triangle with vertices in $I$ has its three vertices in a common interval and the sum of two of its side lengths equal to the third, so it is degenerate and by [F5] it is congruent to its Euclidean comparison triangle; hence the CAT(0) inequality holds with equality.

Reduced paths compute the metric. Subdivide edges at the finitely many points under discussion. Subdivision preserves connectedness and cannot create a cycle: a cycle in the subdivided graph would traverse each inserted degree-two vertex straight through and collapse to a cycle in $\mathrm{\Gamma }$. Thus the subdivided graph is still a finite tree, and any two of its vertices $x,y$ have a unique edge path $P$. Traverse $P$ at unit speed, with length $D$ equal to the sum of its edge lengths. Define $f:X_{\mathrm{\Gamma }}\to [0,D]$ to be distance along $P$ on $P$, and constant on each branch attached to $P$. Each component off $P$ attaches at exactly one vertex; two attachment vertices would create a second path between them and hence a cycle. Consequently $f$ is well defined, continuous, and $1$-Lipschitz on every edge. For any chain, summing edgewise inequalities gives $\mathrm{\mid }f(x)-f(y)\mathrm{\mid }=D\le \mathrm{\ell }(\text{chain})$. The path $P$ gives the reverse bound, so $d(x,y)=D$. Taking $x,y$ on a single edge also proves that edge's metric is its interval metric.

Geodesics and uniqueness. Apply step 1.2 after subdividing at any two points $x,y$. Unit-speed traversal of their reduced path is distance preserving on every subinterval, again by the reduced-path formula, so is a geodesic. If $w$ lies off that path, its unique attachment point $v$ satisfies $d(x,w)+d(w,y)=d(x,y)+2d(v,w)>d(x,y)$. Every point of any minimizing segment must instead satisfy equality in this sum. Thus the segment lies on the reduced path, where its distance from $x$ fixes its position, proving uniqueness. The geodesic is a path of length $d(x,y)$, so the space is a length space. Its finite union of compact interval cells is compact: each cell inclusion is $1$-Lipschitz for the chain metric, so the preimages of any open cover have finite subcovers; taking their finite union covers the entire finite gluing. Completeness is [F3].

The median. Subdivide at $x,y,z$. The paths from $x$ to $y$ and from $x$ to $z$ have a common initial path: if they separated and later met, their two portions between the first separation and reunion would contradict unique paths in the tree. Let $o$ be the last vertex of that common initial path. The remaining paths from $o$ to $y$ and from $o$ to $z$ have no vertex in common except $o$, so their concatenation is the unique path from $y$ to $z$. Hence the triple intersection of the three paths is exactly $\{o\}$, including the cases of repeated points, and each side is the union of the corresponding two arms. Every point of an arm lies on its two incident sides.

(ii)(b). Fix a geodesic triangle with vertices $x,y,z$, let $o$ be its median from step 3.1 and put $L:=d(y,z)>0$ if $y\mathrm{\ne }z$ (if two vertices coincide, uniqueness from step 2.1 makes the two nonconstant sides coincide, so [F5] applies). Parametrize the geodesic $[y,z]$ by $\gamma :[0,1]\to X_{\mathrm{\Gamma }}$ with $\gamma (t_o)=o$, so that $d(x,\gamma (t))=h+L\mathrm{\mid }t-t_o\mathrm{\mid }$ with $h:=d(x,o)$, $d(x,y)=h+L{t}_o$ and $d(x,z)=h+L(1-t_o)$. Substituting into the squared right-hand side of [F4], the difference $(1-t)d(x,y{)}^2+td(x,z{)}^2-t(1-t)L^2-d(x,\gamma (t){)}^2$ equals $4hLt(1-t_o)\ge 0$ for $t\le t_o$ and $4hL(1-t)t_o\ge 0$ for $t\ge t_o$; this is a direct expansion, and the two cases are interchanged by $t\leftrightarrow 1-t$, $y\leftrightarrow z$. By [F4] the hinged inequality at the vertex $x$ holds; the same computation with $x$ replaced by $y$ and by $z$, using the median description of step 3.1, gives the hinged inequality at the other two vertices.

(ii)(b), conclusion. Every geodesic triangle in $X_{\mathrm{\Gamma }}$ has all three hinged inequalities at its vertices, and by [F4] (the vertex-opposite-side criterion) it satisfies the CAT(0) inequality for all pairs of its points; hence $X_{\mathrm{\Gamma }}$ is CAT(0), and the Bruhat–Tits inequality is its case $t=1\mathrm{/}2$.