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;

theorem
  dom I misses dom Reloc(J, card I)
proof
  assume
A1: dom I meets dom Reloc(J, card I);
  dom Reloc(J, card I) = dom Shift(J,card I) by Th20
    .= { l+card I where l is Nat: l in dom J } by VALUED_1:def 12;
  then consider x being object such that
A2: x in dom I and
A3: x in { l+card I where l is Nat: l in dom J } by A1,XBOOLE_0:3;
  consider l being Nat such that
A4: x = l+card I and
  l in dom J by A3;
  l+card I < card I by A2,A4,AFINSQ_1:66;
  hence contradiction by NAT_1:11;
end;
