reserve x,A for set, i,j,k,m,n, l, l1, l2 for Nat;
reserve D for non empty set, z for Nat;
reserve S for COM-Struct;
reserve ins for Element of the InstructionsF of S;
reserve k, m for Nat,
  x, x1, x2, x3, y, y1, y2, y3, X,Y,Z for set;
reserve i, j, k for Nat,
  n for Nat,
  l,il for Nat;
reserve
  i,j,k for Instruction of S,
  I,J,K for Program of S;
reserve k1,k2 for Integer;
reserve l,l1,loc for Nat;
reserve i1,i2 for Instruction of S;
reserve
  i,j,k for Instruction of S,
  I,J,K for Program of S;
reserve m for Nat;
reserve I,J for Program of S;
reserve i for Instruction of S,
        I for Program of S;
reserve loc for Nat;

theorem
  loc in dom stop I & (stop I).loc <> halt S implies loc in dom I
proof
  assume that
A1: loc in dom stop I and
A2: (stop I).loc <> halt S;
  set SS=Stop S, S2=Shift(SS, card I);
A3: stop I = I +* S2 by AFINSQ_1:77;
  assume not loc in dom I;
  then loc in dom S2 by A1,A3,FUNCT_4:12;
  then loc in {l1+ card I where l1 is Nat : l1 in dom SS} by VALUED_1:def 12;
  then consider l1 being Nat such that
A4: loc=l1+ card I and
A5: l1 in dom SS;
A6:  0 in dom Stop S by Th2;
A7: (Stop S). 0 = halt S;
  l1= 0 by A5,TARSKI:def 1;
  hence contradiction by A2,A4,A7,A6,AFINSQ_1:def 3;
end;
