Slim triangles, the Gromov product, and the four-point condition are equivalent up to constants

If triangles are $\delta$-slim, [F1] proves condition (2) with the same constant $3\delta$ at every basepoint.

Now assume (2) and fix just one point $o\in X$. Let $\kappa ={\delta }_o^{\mathrm{\prime }}$. For any four points $a,b,c,d$, write $p_{uv}=(u\mathrm{\mid }v{)}_o$ and $r_u=d(o,u)$. On these four points define $q_{uv}$ to be the maximum, over all simple edge paths from $u$ to $v$ in the complete graph, of the least $p$-value of an edge on the path; put $q_{uu}=r_u$. There are finitely many paths. The one-edge path gives $q_{uv}\ge p_{uv}$. Along a two-edge path the assumed product inequality gives $p_{uv}\ge \min (p_{us},p_{sv})-\kappa$, and along a three-edge path it gives $p_{uv}\ge \min (p_{us},p_{st},p_{tv})-2\kappa$. Hence $0\le q_{uv}-p_{uv}\le 2\kappa$. Also $q_{uv}\le \min (r_u,r_v)$, since the first and last edges of every path satisfy these respective bounds.

Concatenate paths attaining $q_{uv}$ and $q_{vw}$ and erase any loops. Erasing loops cannot lower the minimum edge value. Thus $q_{uw}\ge \min (q_{uv},q_{vw})$: $q$ is an exact ultrametric similarity on these four labels. For completeness, its positive threshold relations $u{\sim }_{t}v⟺q_{uv}\ge t$ are nested equivalence relations (on labels with $r_u\ge t$). Make a finite rooted tree from these nested clusters, with each leaf $u$ at height $r_u$ and each common ancestor of $u,v$ at height $q_{uv}$. Nonnegative edge lengths follow from the bound in step 1.2. The tree distance between leaves is $D(u,v)=r_u+r_v-2q_{uv}$. Removing the finite subtree spanned by four leaves at its central edge or central vertex shows that the largest two of its three opposite-pair distance sums are equal: each uses the central edge twice, while the third uses it zero times; zero-length edges and repeated leaves follow by the same calculation.

The original metric satisfies $d(u,v)=r_u+r_v-2p_{uv}$, so $0\le d(u,v)-D(u,v)\le 4\kappa$. Each opposite-pair sum therefore differs from its tree counterpart by a number in $[0,8\kappa ]$. Since the two largest tree sums are equal, the largest and second-largest original sums differ by at most $8\kappa$: the two original sums corresponding to those equal tree sums both lie in one interval of length $8\kappa$, while the remaining original sum can only increase the second-largest if it becomes larger. This is the four-point condition with constant $4\kappa$ (additive error $8\kappa$). The bound uses the one fixed basepoint $o$, so condition (2)'s per-basepoint quantifier causes no uniformity gap.

Finally (3) is the four-point condition with constant $\kappa ={\delta }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}2$. By [F2] every triangle is $4\kappa =2{\delta }^{\mathrm{\prime }\mathrm{\prime }}$-slim. This proves (1) and closes the cycle.