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 ,
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 ? ,
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 ? ? ,
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 ? ? ,
?}, q (i+L,j) = p (i,j) . The ID of q (i+L,j) becomes (i + L ? 1)? + j and q (i+L,j) .level = i + L. The value of variables color and done do not change, We increase the level and the ID of the L? processes of E ??1 as follows ,
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