The all-right triangle must be filled; the disconnected universal-Coxeter nerve is CAT(1) vacuously

Clause (i), the filling. The Coxeter presentation has $s^2=t^2=u^2=1$, and for each distinct pair the relation $(st{)}^2=1$ gives $st=ts$; thus every word reduces to $s^a{t}^b{u}^c$ with $a,b,c\in \{0,1\}$. The map sending $s,t,u$ to the three standard generators of $(\mathbb{Z}\mathrm{/}2{)}^3$ is surjective, so the group has at least eight elements; the word reduction shows it has at most eight, hence $W^{\mathrm{\prime }}\cong (\mathbb{Z}\mathrm{/}2{)}^3$. For the all-right system $B(e_x,e_y)=-\cos (\pi \mathrm{/}2)=0$ for $x\mathrm{\ne }y$, so $C_T$ is the identity matrix of size $\mathrm{\mid }T\mathrm{\mid }$ for every $T\subseteq S^{\mathrm{\prime }}$, in particular positive definite [F6]; by [F3] every subset is spherical, so the nerve $L^{\mathrm{\prime }}$ has all three edges of length $\pi -\pi \mathrm{/}2=\pi \mathrm{/}2$ [F1] and the cell $\mathrm{\Sigma }({\mathrm{i}\mathrm{d}}_3)$ on $S^{\mathrm{\prime }}$ [F1], and no further cells exist; hence $L^{\mathrm{\prime }}$ is exactly the filled all-right triangle $\mathrm{\Sigma }({\mathrm{i}\mathrm{d}}_3)$.

Clause (ii), free product and no edges. Giving a homomorphism from $W^{\mathrm{\prime }\mathrm{\prime }}$ to a group $H$ is exactly giving three involutions in $H$, by the universal property of its Coxeter presentation; these are precisely three homomorphisms $\mathbb{Z}\mathrm{/}2\to H$. Thus $W^{\mathrm{\prime }\mathrm{\prime }}$ has the free-product universal property of three copies of $\mathbb{Z}\mathrm{/}2$ (The free product of an arbitrary family of groups, Coxeter matrices, the presented Coxeter group, reduced words, length, and standard parabolic subgroups). For each two-element subset $\{s_i,s_j\}\subseteq S^{\mathrm{\prime }\mathrm{\prime }}$, [F8] shows that its subgroup contains an element of infinite order, so it is not finite; by [F3] the pair is not spherical and spans no edge of $L^{\mathrm{\prime }\mathrm{\prime }}$ [F1]. Since a cell with two vertices is exactly an edge, the spherical subsets of $S^{\mathrm{\prime }\mathrm{\prime }}$ are $\mathrm{\varnothing }$ and the three singletons, and $L^{\mathrm{\prime }\mathrm{\prime }}$ has no simplex of positive dimension.

Clause (i), the angles. In $\mathrm{\Sigma }({\mathrm{i}\mathrm{d}}_3)$ all three sides equal $\pi \mathrm{/}2$, and the spherical cosine rule with side $c$ opposite the angle $\gamma$ between the two sides of length $\pi \mathrm{/}2$ gives $\cos (\pi \mathrm{/}2)=\cos (\pi \mathrm{/}2)\cos (\pi \mathrm{/}2)+\sin (\pi \mathrm{/}2)\sin (\pi \mathrm{/}2)\cos \gamma$, that is $0=\cos \gamma$ and $\gamma =\pi \mathrm{/}2$ [F7]; the same computation at the other two corners gives three interior angles $\pi \mathrm{/}2$.

Clause (ii), the metric. The truncation sends the value $+\mathrm{\infty }$ between distinct components to $\pi$ [F1], and the components of $L^{\mathrm{\prime }\mathrm{\prime }}$ are its three vertices, so distinct vertices of $L^{\mathrm{\prime }\mathrm{\prime }}$ are at truncated distance $\pi$; therefore $L^{\mathrm{\prime }\mathrm{\prime }}$ has no pair of distinct points at distance $<\pi$, and a triangle with two distinct vertices has a side of length $\pi$ and perimeter at least $2\pi$, while a triangle with a single vertex has perimeter $0$ and satisfies the comparison trivially. Thus all CAT(1) tests of $L^{\mathrm{\prime }\mathrm{\prime }}$ are vacuous and $L^{\mathrm{\prime }\mathrm{\prime }}$ is CAT(1); its components are three one-point CAT(1) spaces at mutual truncated distance $\pi$.

Clause (iii), the filled triangle has no circle. Since $C_{S^{\mathrm{\prime }}}={\mathrm{i}\mathrm{d}}_3$, its Cholesky realization has the standard basis as vertices, so $\mathrm{\Sigma }({\mathrm{i}\mathrm{d}}_3)=\{x\in S^2:x_1,x_2,x_3\ge 0\}$ [F7]. For any $x,y$ in this octant, $x\cdot y\ge 0$, so their round distance $\delta \le \pi \mathrm{/}2$. If $0<\delta <\pi$, the shorter great-circle segment has points $(\sin ((1-t)\delta )x+\sin (t\delta )y)\mathrm{/}\sin \delta$ for $0\le t\le 1$; its coefficients are nonnegative, so it stays in the octant. Thus $\mathrm{\Sigma }({\mathrm{i}\mathrm{d}}_3)$ is geodesically convex and its chain metric is the round metric, with diameter at most $\pi \mathrm{/}2$. If an isometric circle $S_{\mathrm{\ell }}^1$ embedded in $L^{\mathrm{\prime }}$, its diameter $\mathrm{\ell }\mathrm{/}2$ would be at most $\pi \mathrm{/}2$, so $\mathrm{\ell }\le \pi <2\pi$. The two semicircles between opposite points of $S_{\mathrm{\ell }}^1$ would give distinct geodesic segments of length $\mathrm{\ell }\mathrm{/}2<\pi$ in $L^{\mathrm{\prime }}$, contradicting uniqueness of the shorter great-circle arc in the round sphere [F7]. Hence $L^{\mathrm{\prime }}$ contains no isometrically embedded circle.

Clause (i), the boundary loop. Let $\mathrm{\Gamma }$ be the boundary $3$-cycle, of length $3\cdot \pi \mathrm{/}2<2\pi$. Fix a corner and let $x_t,y_t$ be the points at distance $t\in (0,\pi \mathrm{/}4)$ from it on the two incident edges: their inner product is ${\cos }^{2}t+{\sin }^{2}t\cos (\pi \mathrm{/}2)={\cos }^{2}t$, so their distance is $d(t)=\arccos ({\cos }^{2}t)<2t$ because $\cos$ is strictly decreasing on $[0,\pi ]$ and ${\cos }^{2}t>\cos 2t$ for $t\in (0,\pi \mathrm{/}2)$ [F7]; replacing the two boundary subarcs of total length $2t$ by the geodesic segment of length $d(t)$ inside the convex triangle is a strict shortening of $\mathrm{\Gamma }$ near the corner, so $\mathrm{\Gamma }$ is not locally geodesic and hence not an isometrically embedded circle, whose sufficiently short arcs would be minimizing [F5].

Clause (iii), the discrete nerve and the comparison. Since $L^{\mathrm{\prime }\mathrm{\prime }}$ is discrete [step 2.2], every continuous map from the connected space $\mathbb{R}\mathrm{/}\mathrm{\ell }\mathbb{Z}$ into $L^{\mathrm{\prime }\mathrm{\prime }}$ is constant, so $L^{\mathrm{\prime }\mathrm{\prime }}$ contains no isometrically embedded circle. Together with step 2.3 this proves both girth values are $+\mathrm{\infty }$, hence at least $2\pi$. For the comparison: the value $0$ of the all-right cosine matrix lies at the boundary of the large hypothesis and makes the metric flag test the ordinary flag test, which fills the triangle of 1.1; the value $-1$ of a non-edge pair of $L^{\mathrm{\prime }\mathrm{\prime }}$ would make the $2\times 2$ principal block $\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$ have the nonzero kernel vector $(1,1)$, hence not positive definite [F6], so it is never the cosine of an edge and no set of two or more vertices triggers the metric flag test.