consistè a vérifier que pos' n'est pas consommée. Si c'est le cas on indique que Correction de la combinatorisation (?s ,
´ evaluation de M dans s ne termine pas Dans le second cas, l'´ evaluation de M dans s bloque dans l'´ etat mémoire s ? 1 . Dans letroisì eme cas on a : if M 1 then M ,
l'expression M 1 est réduite en premier (IF est une valeur) Nous retrouvons les trois issues possibles pour l'´ evaluation de M 1 . Si l'´ evaluation de M 1 ne termine pas, alors l'´ evaluation de [M ] ne termine pas 1 . Si l'´ evaluation de M 1 bloque dans l'´ etat s ? 1 alors l'´ evaluation de [M ] bloqué egalement dans cetétatcetétat. Dans le cas restant, l'´ evaluation de M 1 donne V 1 dans l'´ etat s 1 et on a donc ,
) (fun x ? M 3 ) / s 1 ? E 2 (fun x ? M 3 ) / s 1 (5) o` u E 2 est fun z ? if V 1 then ,
s 1 ? if V 1 then (fun x ? M 2 ) ( ) else (fun x ? M 3 ) ( ) / s 1 (6) Or on a (fun x ? M 2 ) ( ) ? M 2 {x ? ( )} et comme x n'apparait pas dans M 2, {x ? ( )} vaut M 2 . De même, (fun x ? M 3 ) ( ) ? M 3 . Donc, en remplaçant d'´ egaì a ´ egal (r` egles (cond- comp-2) et (cond-comp-3)) on a : if V 1 then (fun x ? M 2 ) ( ) else (fun x ? M 3 ) ( ) ? if V 1 then M 2 else M ,
Or, on a vu que par (2) nous avons ,
N 2 )) X : (fun x ? (N 1 N 2 )) X / s ? (N 1 N 2 ){x ? X} ,
){x ? X} estégaìestégaì a (N 1 {x ? X}) (N 2 {x ? X}) (par la définition de l'application d'une substitution), Par conséquent on a : (fun x ? (N 1 N 2 )) X ? (N 1 {x ? X}) (N 2 {x ? X}) ,
M est de la forme fun x ? fun y ? N . Par définition, [fun x ? fun y ? N ] = [fun x ? [fun y ? N ]]. Or, [fun y ? N ] est soit un opérateur (I), soit une application (de la forme K C ou S N 1 N 2 ) On peut donc appliquer l'hypothèse d'inductionàinductionà fun x ? ,
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