Question: Prove that for any link (u, v), d[u, v]=Max {d(u, y) : y T [vu]} = 1+ Max{d(v, y) : y T [uv]} = Max{d[v,

Prove that for any link (u, v), d[u, v]=Max {d(u, y) : y∈ T [v−u]} =

1+ Max{d(v, y) : y∈ T [u−v]} = Max{d[v, z] : z = u ∈ N(v)}.

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