?. Assume-?v, 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

=. ?. Since-?-i and ?. Pu, 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

M. Concurrency-versus-fairness and .. , 180 6.4.1 Necessary Condition on Concurrency in LRA, p.183

P. Concurrency and .. , 183 6.5.1 Definition, p.184

L. @bullet-t-c, 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)

L. Algorithm and . @bullet-t-c-meets-the, 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

=. Let-e and ?. Lra-@bullet-t-c, 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

. Assume, 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

. Assume-?-r-(-e, ) (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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L. Jane-se-réveille, des capteurs de mouvements détectent son réveil et allument les lampes progressivement ainsi que le chauffage de la salle de bain

Q. Un-place-libre, u se garer grâce aux capteurs qui surveillent l'occupation du parking. Pendant sa journée de travail, JanéJané echange des emails avec ses clientsàclientsclientsà 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 pasàpaspasà 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