. Proof, First notice that since G does not contain infinite isometric rays and all cells of G are Cartesian products of cycles and edges, all cells of G are finite. Since Z ? (G) is an intersection of convex subgraphs, Z ? (G) is convex. Since every cell X of G is a Cartesian product of edges and even cycles, any proper halfspace of X is maximal by inclusion, hence F (X) = ?

Z. Therefore, Since H is convex, H is hypercellular Since there are no infinite isometric rays, Z(H) = H if and only if for every edge ab of H the halfspaces W (a, b) and W (b, a) are maximal, which is equivalent to the condition that for every edge ab the carrier N (E ab ) is the whole graph H. Let X be the intersection of all maximal cells of H; consequently, X is a cell of H. As in the proof of Theorem 2, if there exist two disjoint maximal cells of H, then for every edge f on a shortest path between them, the carrier N (E f ) is not the whole graph H. Hence the maximal cells of H pairwise intersect. Since they are finite and gated, by the Helly property for gated sets, X is nonempty. If X = H, by Claim 10, there exists an edge of H whose carrier does not include all maximal cells, a contradiction, Hence H = X, i.e., H is a finite cell. Consequently, Z ? (G) is a finite cell of G. We continue with the proof of assertion (i) of Theorem F

. Proof, Every automorphism ? of G maps maximal halfspaces to maximal halfspaces By Lemma 21, Z ? (G) is a finite cell

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