Then, since ? i (y).c = ?, ? i (q).c = 0. Then, ? i+1 (p).c = ? i+1 (q).c = 0 (p necessarily moves in ? i ? s ? i+1 and gets clock 0 owing the fact that the c-clock values of all its non-? neighbors are well computed according to their t-clock values, by Definition 5.10.1; moreover q is disabled in ? i since Locked(q) hold because of y), and we are done. Otherwise, ?v ? V i , (? i (v).c = ? ? (?w ? ? i (v).N , ? i (w).c = ?)). Then, ?x ? ? i (p).N such that c? i (x ) = ? ? t? i (x ) = M inT ime(p) in ? i . By Definition 5.10.2c, d ? (t? i (x ), ? i (q).t) ? µ because they have neighbors whose c-variables equal ? (p and y, respectively). Moreover, q is disabled in ? i because of y: ? i+1 (q), ).t. Finally, ? i+1 (p).t = ? i (x ).t since p executes DSU-J-action. So, d ? (? i+1 (p).t, ? i+1 (q).t) ? µ ,
10.2a. If ?v ? V ic = ? i+1 (q).c = 0 (owing the fact that the c-clock values of all their non-? neighbors are well computed according to their t-clock values, by Definition 5.10.1), and we are done. Otherwise, ?v ? V i , (? i (v).c = ? ? (?w ? ? i (v).N , ? i (w).c = ?)). So, ?x ? ? i (p).N such that ? i (x ).c = ? ? ? i (x ).t = M inT ime(p) in ? i and ?y ? ? i (q).N such that ? i (y ).c = ? ? ? i (y ).t = M inT ime(q) in ? i . By Definition 5.10.2c, d ? (? i (x ).t, ? i (y ).t) ? µ because they have neighbors whose c-variables equal ? (p and q, respectively), i+1 (p).t = ? i (x ).t and ? i+1 (q).t = ? i (y ).t since p and q execute DSU-J-action. So d ? (? i+1 (p).t, ? i+1 (q).t) ? µ ,
180 6.4.1 Necessary Condition on Concurrency in LRA, p.183 ,
183 6.5.1 Definition, p.184 ,
Assume Result(? k . . . ? k , p, r) Assume ?l < k such that Start(? l , ? l+1 , p, r) In particular, ? l (p).status = In. Hence, N oConf lict(? l , p) trivially holds, Proof : Let e = (? i ) i?0 be an execution, p.p) ,
No Deadlock property of strong concurrency: for every subset of processes X ? V , for every configuration ?, if P strong (X, ?) holds and P F ree (?) ? X, there exists a configuration ? and a step ? ? ? such that, ContinuousCS ,
0 such that T C is stabilized in ? i (i is defined by Lemma 6.5), and t ? V the unique tokenholder in ? i . If R(e, i, 6) exists, R(e, i, 6) ? reqU p(e, i), and ?j ? {i + 1, . . . , R(e, i, 6)}, P assT oken(t) is not executed in step ? j?1 ? ? j , then for every k ? {R(e, i, 4), . . . , reqU p(e, i)}: ? ? k (t), req = ?, ? k (t).status = Blocked, and ? ?q ? ? k (t).CN such that ? k (q).status = In and ?p ? P F ree (? k ) ? ? k (t).CN , p / ? {q} ? ? k (q).CN ,
then RsT-action (the highest priority action) is continuously enabled at p until p executes it. Now, in this case, p executes it within at most 2 rounds (Remark 6.1), hence, there is a configuration between ? i and ? R(e,i,2) , where p.token = T okenReady(p), a contradiction, ) (t).status = Blocked ,
) (t).status = Out. If t.req = ?, then t.req = ? holds in all configurations between ? i and ? R(e,i,6) , by hypothesis Similarly to ,
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des capteurs de mouvements détectent son réveil et allument les lampes progressivement ainsi que le chauffage de la salle de bain ,
u se garer grâce aux capteurs qui surveillent l'occupation du parking. Pendant sa journée de travail, JanéJané echange des emails avec ses clientsàclients`clientsà l'autre bout du monde Elle participè a une réunion par visioconférence avec une autre filiale etéchangeetéchange des données avec sescolì egues via le réseau local de l'entreprise. Pendant qu'elle n'est pasàpas`pasà la maison, les capteurs de ses panneaux solaires détectent une grande production d'´ electricitéelectricité`electricitéà la mi-journée. Ils allument donc le chauffe-eau et le lave-vaisselle, Jane vérifie lesdernì eres informations sur Internet et partage les photos du weekend dernier avec sa famille grâcegrâcè a un service de partage de fichiers sur le " cloud ,