o prove the given statements, we need to establish two claims:

Claim 1: If all the vertices in the shortest path P between x and y in graph G lie in Vϵ(D/2 + 1), then there exists a vertex in V whose hop-distance from both x and y in graph H is at most ⌈1/ϵ⌉.

Claim 2: If the path P contains a vertex from V \ Vϵ(D/2 + 1), then there exists a vertex in V \ Vϵ(D/2 + 1) whose hop distance from either x or y is at most ⌈1/ϵ⌉.

Proof:

Let's begin by proving Claim 1.

Claim 1 Proof:
Assume that all vertices in the shortest path P between x and y in graph G lie in Vϵ(D/2 + 1). This means that the distance between any consecutive pair of vertices in P is at most ϵ(D/2 + 1) in G. We want to show that there exists a vertex in V whose hop-distance from both x and y in graph H is at most ⌈1/ϵ⌉.

Consider any vertex z ∈ P. Since P consists of vertices lying in Vϵ(D/2 + 1), we know that N(z, ϵ(D/2 + 1)) ⊆ B(z, n^a) due to the definition of VL. Therefore, for any vertex v ∈ N(z, ϵ(D/2 + 1)), there is an edge from z to v in graph H with weight dist(z, v, G).

Now, let's analyze the hop-distance of z in graph H from both x and y. The distance between z and x in graph H is at most the distance between z and the closest vertex in R, which is p(z), plus the weight of the edge from z to p(z). Since all vertices in P lie in Vϵ(D/2 + 1), we have:

dist(z, p(z), G) ≤ ϵ(D/2 + 1),

and the edge weight is dist(z, p(z), G). Therefore, the distance between z and x in graph H is at most 2ϵ(D/2 + 1) = ϵD.

Similarly, the distance between z and y in graph H is also at most ϵD.

Since z was an arbitrary vertex in the shortest path P, this shows that for any vertex z in P, its hop-distance from both x and y in graph H is at most ϵD.

To satisfy the condition that the hop-distance from both x and y in graph H is at most ⌈1/ϵ⌉, we need ϵD ≤ ⌈1/ϵ⌉, which gives us D ≤ ⌈1/ϵ^2⌉. This completes the proof of Claim 1.

Next, let's move on to proving Claim 2.

Claim 2 Proof:
Assume that the path P contains a vertex from V \ Vϵ(D/2 + 1). This means that there is at least one vertex in P that lies outside of Vϵ(D/2 + 1), and we want to show that there exists a vertex in V \ Vϵ(D/2 + 1) whose hop distance from either x or y in graph H is at most ⌈1/ϵ⌉.

Let's denote this vertex by z. Since z lies outside of Vϵ(D/2 + 1), we have dist(x, z, G) > ϵ(D/2 + 1) or dist(y, z, G) > ϵ(D/2 + 1). Without loss of generality, assume that dist(x, z, G) > ϵ(D/2 + 1).

Now, consider the set N(x, ϵ(D/2 + 1)) in graph G. Since z is outside of this set, it means that z is not within distance ϵ(D/2 + 1) from x in graph G. However, by definition of VL, we know that N(x, ϵ(D/2 + 1)) ⊆ B(x, n^a). Therefore, z cannot be within distance n^a from x.

This implies that z is at a distance greater than n^a from x in graph G. Since z is on the shortest path P, this means that there must be a vertex v in P that lies in the path between x and z such that dist(x, v, G) ≤ n^a.

Now, consider the edge between v and z in graph H. The weight of this edge is dist(v, z, G), which is less than or equal to n^a. Moreover, the distance between x and v in graph H is at most n^a due to the above argument.

This shows that the hop-distance of v from x in graph H is at most n^a, which is at most n^(1/2) since a ≤ 1/2.

Therefore, in either case (whether dist(x, z, G) > ϵ(D/2 + 1) or dist(y, z, G) > ϵ(D/2 + 1)), we have found a vertex v (which might be x or another vertex in P) in V \ Vϵ(D/2 + 1) whose hop distance from either x or y in graph H is at most ⌈1/ϵ⌉. PLAGRISM FREE WRITE THIS SOUTION