. With-d-=-n and ?. V. We-have-?i-?-r-n, EBL(p, k) is false in ? i : hence, in at most n rounds, there is no more process of status EB in abnormal trees, those ones being dead. So, all processes (and in particular the abnormal roots) in abnormal trees have status EF

R. As, status = ? ? (p).status = EF If Self Root(p) ? ¬Self RootOk(p) holds in ?, it also holds in ? ? because p does not execute an action between ? and ? ? and these predicates only depends on the local state of p. Otherwise ¬Self Root(p) ? ¬KinshipOk(p, p.par) holds in ?. Let q = p.par. If q does not execute an action between ? and ? ? , p is still an abnormal root, Otherwise, three cases are possible: ? ¬GoodIdR(p, q) holds in ?

@. ¬goodlevel, idR = ?(q).idR but ?(p).level = ?(q).level + 1. First, if q executes EB-action or EF -action, its idR and its level do not change, so ? ? (p).idR = ? ? (q).idR and ? ? (p).level = ? ? (q).level+1, so AbRoot(p) holds in ? ? . Otherwise, q executes R-action or J-action and consequently ? ? (q), .status = C. So ¬GoodStatus(p, q) and AbRoot(p) holds in ? ?

@. ¬goodstatus, Then ?(q).status = C, and q can only execute EB-action or J-action between ? and ? ? . If q executes EB-action, then EBroadcast(q) holds in ?, so q is in an abnormal tree (Lemma 8). But, by hypothesis, all abnormal trees are dead in ?, so ?(q).status = C, a contradiction, If q executes J-action then ? ? (q).status = C, so ¬GoodStatus(p, q) and AbRoot(p) holds in ? ?

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