Nonshrinkable edge loops of length $<2\pi$ have three edges, and the finite locally CAT(1) large metric flag complex is CAT(1)

Clause (i): a closed edge walk of $K$ with one edge is a loop edge, which does not exist, and with two edges it is an immediate backtrack $u,v,u$; the backtrack loop is short-loop homotopic to the constant loop at $u$ by the family that pulls the second half back along the first, whose loops have length at most the original length, so it is shrinkable [F3]. Hence a nonshrinkable edge loop has at least three edges; if it had four or more, its length would be at least $4\cdot \pi \mathrm{/}2=2\pi$ by [F1], contrary to the hypothesis. Therefore it has exactly three edges.

Clause (ii): let $a,b,c\in [\pi \mathrm{/}2,\pi )$ be the three edge lengths, with $a+b+c<2\pi$, and put $x=\cos a$, $y=\cos b$, $z=\cos c$. By [F7] the determinant of $C$ equals $4\sin s\sin (s-a)\sin (s-b)\sin (s-c)$ with $s=\frac{1}{2}(a+b+c)$: here $0<s<\pi$, and $s-a=\frac{1}{2}(b+c-a)>0$ because $b+c\ge \pi >a$, together with $s-b>0$, $s-c>0$ and $s-a<s$, $s-b<s$, $s-c<s<\pi$; hence all four sines are positive and $\det C>0$. The two leading principal minors are $1>0$ and $1-x^2={\sin }^{2}a>0$, and the principal $2\times 2$ minors are handled the same way as leading minors of submatrices, so by [F6] the matrix $C$ is positive definite.

Clause (iii): by the metric flag condition and step 1.2 the three vertices of the loop span a simplex $\sigma$ of $X$. Its spherical model is $K(C_{\sigma })\cap S^2$; for points $p,q$ in this simplex at distance $\delta <\pi$, the shorter spherical segment has points $(\sin ((1-t)\delta )p+\sin (t\delta )q)\mathrm{/}\sin \delta$ for $0\le t\le 1$. The coefficients are nonnegative, so the whole segment remains in the positive cone $K(C_{\sigma })$ and hence in $\sigma$. Thus the loop is the boundary of a geodesically convex spherical triangle [F1, F2]. At a corner with interior angle $\alpha \in (0,\pi )$, take the points at distance $t>0$ along the two incident edges, with $2t<\pi$; their inner product is ${\cos }^{2}t+{\sin }^{2}t\cos \alpha$, so the spherical segment in $\sigma$ joins them at distance $d(t)=\arccos ({\cos }^{2}t+{\sin }^{2}t\cos \alpha )<2t$, because $\cos d(t)>\cos 2t$ is equivalent to ${\sin }^{2}t(1+\cos \alpha )>0$. Replacing the two boundary subarcs of total length $2t$ by this strictly shorter segment exhibits a strict local shortening, contradicting local geodesicity by its definition. Thus the boundary of $\sigma$, hence $\gamma$, is not locally geodesic.

Clause (iv): assume for contradiction that [assume-contra] $X$ is not CAT(1). Short triangle tests and circles of length $<2\pi$ are unchanged by truncation: their intrinsic side-point or circle distances are below $\pi$. Thus some untruncated intrinsic component is not CAT(1). It is compact geodesic by [F2], so [F4] supplies an attained minimum nonshrinkable circle; across the finitely many components choose one of global minimum length $m<2\pi$. By [F5] some such minimum is a locally geodesic edge loop. Step 1.1 leaves exactly three edges, while step 2.1 excludes that case. This is a contradiction. Therefore each untruncated component is CAT(1), and the same short comparison tests give CAT(1) for the angular truncation of $X$.