Proof: Let C be a shortest cycle in G. If g(G)>= 2diam(G) + 2, then C has two vertices whose distance in C is at least diam(G)+1. Question: why is it at least diam(G)+1? cont. of proof: In G, these vertices have a lesser distance; any shortest path P between them is therefore not a subgraph of C. how i understood: A path "around" the cycle creates a longer path than the shortest path between x and y in G. Therefore, this shortest path P is not a subgraph of the shortest cycle C. rest of the proof: Thus, P contains a C-path xPy. Together with the shorter of the two x-y paths in C, this path xPy forms a shorter cycle than C, a contradiction. Question: why does P contain a C-path xPy if it is not a subgraph of C?