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A. .. Annexe-générale-sommaire, 1 Preuve de correction du calcul de la grammaire de W I(L(G)), p.187

. Démonstration, Nous supposons un arbre XML T qui est complet par rapport à IP

]. and .. .. P-n-[-e-n-]-}-?-p-i, La preuve sera faite par contradiction. Supposons que T |= f . D'après la Définition 4.16, nous déduisons qu'il existe deux projections ? 1 and ? 2 pour la branche

C. Pour-chaque and . Ip, = 2 sauf dans les cas suivants : (a) ou (c) en considérant l'égalité par valeur, |Instances(C/P, t)| = 2 provoque la violation de la condition 1 pour un autre chemin C, IP tel que P ? Q, ou (d) Last(P) ? ? att et Parent(C/P) vérifie les conditions 2a, ou 2b

A. Lemme, F. Soit, and X. Un-ensemble-de, type, value) un document XML où t est un arbre bi-instance. Soit IP l'ensemble de chemins associés à T et soit C/X ? IP et E les types d'égalités associés à X. Les propriétés suivantes sont vérifiées pour t : ? pour tous les chemins R i nous Cette situation est similaire à la situation 1b. Par conséquent nous pouvons construire un arbre avec deux instances de P qui soit una arbre bi-instance respectant les contraintes sur les chemins R i

@. Il-existe-un-chemin, R. Tel-que, and C. /. , Cette situation est similaire à la situation 1c. Par conséquent nous ne pouvons pas construire un arbre avec deux instances de P qui soit un arbre bi-instance respectant les contraintes sur le chemin R j . Ainsi nous concluons que

. Et-comme-l-'arbre-t-est-un-arbre and . Complet, chacun des deux instances de Parent(P) doit avoir un noeud attribut Last(P) et donc |Instances(C/P, t)| = 2. Nous obtenons une contradiction avec l'hypothèse |Instances(C/P, t)| = 1. En conclusion l'arbre t ne possède qu'une seule instance pour le chemin Parent(P), Notons que puisque Last(Parent(P)) ? ? ele , nous avons déjà prouvé que C/Parent(P)[E ? ] ? (C, X[ E]) +

. Démonstration, Nous allons d'abord prouver que l'algorithme termine, ensuite prouver qu'il est correct et enfin prouver qu'il est complet. Pour plus de simplicité, nous allons omettre le contexte c'est à dire écrire

C. Terminaison:-l-'algorithme-termine-parce-que-ip-est-fini-et-comme, X (i) est un sousensemble de IP, nous allons atteindre la valeur j pour i telle que X (j) = X (j+1) Par conséquent nous avons la garantie que la boucle termine et l'ensemble X (j) est retourné par l'algorithme. Nous allons maintenant prouver que X + est égal à X (j)

?. Cas-de-base, A la ligne 5 de l'Algorithme 2, l'ensemble X (0) contient des chemins P qui sont dans S, ou T, ou V ou W. 1

P. Sinon-si, . Att, Q. [. Il-existe-un-chemin, Q. Tel-que, and . Parent, Comme nous en appliquant l'axiome A6 (Unicité de l'attribut)

P. [. Sinon-si, Comme nous avons X[ E] ? P[N], en appliquant l'axiome A9 (De l'identité de noeud à l'égalité par valeur) nous avons P

P. Sinon-si-le-chemin, est dans Y alors il existe une XFD f dans F telle que l'une des conditions (a)-(b) de la ligne 7 soit satisfaite, Lorsque nous posons Z =, issue.1 1