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veriantt1wt2=xt1r0t2y. 100contreditl'hypotheset1wt2=2X.<br />
x0:t1:y0,onat1wt2=x0::u0::y0,donct1wt2=xu0y,soitt1wt22X.Cequi Onadoncjt1wt2j>jxt1j+jt2yj.Onat1wt2=xt1u=vt2y,soitr0l'uniquemot CHAPITRE5.MAXIMALITEETCOMPLETUDEDANSFdif<br />
remarques,nousallonsmontrerquet1r0t22Y. jyj,onat1wt2=2Xett1r0t2=2X+t1A[At2X+.Commelelaissesuggererces {Supposonst1r0t2=2At2Xt1A.Gr^aceauxremarquesprecedentes,nouspouvons endeduirequet1r0t22Cetdoncquet1r0t22Y. Remarquonstoutd'abordquet1r0t2=2X,puisque,parmaximalitedejxjet<br />
{Supposonst1r0t22At2Xt1A.Parconstruction,unprexedet2nepeut^etre suxedet1,onadoncr0t22At2Xt1A.Soitr0lepluscourtmottelquer02<br />
(r0i)i>0tellesque jxj,onat1r0t2=2X+t1A.Enn,onar02r0t2Xt1A,donc,encoreunefoispar onat1r0t2=2At2Xt1A.Dem^emer0=2At2X+.Deplus,parmaximalitede maximalitedejxj,onar0=2X.Onadoncmontrequer02C. r0t2Xt1A.Montronsquet1r0t22C.Puisquelemotr0estdelongueurminimale,<br />
{t1rit22t1r0it2Xt1ri+1t2 {jr0i+1j0et<br />
At2Xt1A.Nousavonsvuplushautquecettederniereconditionimpliqueque rk2C.End'autrestermes,nousavons {t1rit22At2Xt1A<br />
Nousobtenonsdonct1r0t22(CX)CY. Puisquelasuitedesjrij(i>0)decroit,ilexisteunentierk>1telquet1rkt22<br />
t1wt22Y,onadoncw2F(X).L'ensembleYestdonccomplet. Ainsi,commet1wt2=xt1r0t2y,nousavonsmontrequepourtoutmotw,ona t1r0t22t1r0t2Xt1r01t2:::t1r0k1t2Xt1rkt2:<br />
PourconclureilresteaprouverqueYrestetrescoupant. Proposition5.2.9LecodeYesttrescoupant. NousavonsdonccompletelecodeXenuncodecomplet,dem^emedelaiqueX.
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