theorem Th63:
  L is convex implies L.x <= 1
 proof
  assume A1: L is convex;
  then sum L=1 by Th62;
  then consider F be FinSequence of V such that
   F is one-to-one and
   A2: rng F=Carrier L and
   A3: 1=Sum(L*F) by Def3;
  assume A4: L.x>1;
  then x in dom L by FUNCT_1:def 2;
  then reconsider v=x as Element of V by FUNCT_2:def 1;
  v in Carrier L by A4;
  then consider n be object such that
   A5: n in dom F and
   A6: F.n=v by A2,FUNCT_1:def 3;
  len(L*F)=len F by FINSEQ_2:33;
  then A7: dom(L*F)=dom F by FINSEQ_3:29;
  A8: now let i be Nat;
   assume i in dom(L*F);
   then (L*F).i=L.(F.i) & F.i=F/.i by A7,FUNCT_1:12,PARTFUN1:def 6;
   hence (L*F).i>=0 by A1,Th62;
  end;
  reconsider n as Nat by A5;
  (L*F).n=L.x by A5,A6,A7,FUNCT_1:12;
  hence contradiction by A3,A4,A5,A7,A8,MATRPROB:5;
 end;
