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, it follows at most f i?(w+1) extra cops lie within O c (i, w +1)\O c (i, w) = {O i?(w+1) }. Assume all of them move counterclockwise toward the robber. Since F i?(w+1) contains f i?(w+1) loops, by inductively applying Proposition 3, each of these cops must "stay behind", i.e., become an occupant cop of a loop in F i?(w+1) . Therefore, since there is no free loop between the original sequences O i?w and O i?(w+1) , then O i?w can now be reset to a larger sequence O i?w containing all loops in O i?w, Fj ?Fc(i,w+1) f j extra cops within O c (i, w + 1)
, Hence, strictly fewer than Fj ?Fc(i,w) f j extra cops lie within the vertices in O c (i, w) = O c (i, w) \ {O i?w } ? {O i?w }. By the inductive assumption for w, by only using cops within O c (i, w), at most f i cops can cross L c . Overall, since at most f i cops can cross L c for all possible w i