reserve E, x, y, X for set;
reserve A, B, C, D for Subset of E^omega;
reserve a, a1, a2, b, c, c1, c2, d, ab, bc for Element of E^omega;
reserve e for Element of E;
reserve i, j, k, l, n, n1, n2, m for Nat;

theorem
  <%>E in A implies A* = (A*) ^^ (A |^ n) & A* = (A |^ n) ^^ (A*)
proof
  defpred P[Nat] means A* = (A*) ^^ (A |^ $1) & A* = (A |^ $1) ^^ (A*);
A1: (A |^ 0) ^^ (A*) = {<%>E} ^^ (A*) by Th24
    .= A* by Th13;
  assume
A2: <%>E in A;
A3: now
    let n;
    assume
A4: P[n];
A5: (A*) ^^ (A |^ (n + 1)) = (A*) ^^ ((A |^ n) ^^ A) by Th23
      .= (A*) ^^ A by A4,Th18
      .= A* by A2,Th54;
    (A |^ (n + 1)) ^^ (A*) = ((A |^ n) ^^ A) ^^ (A*) by Th23
      .= (A ^^ (A |^ n)) ^^ (A*) by Th32
      .= A ^^ (A*) by A4,Th18
      .= A* by A2,Th54;
    hence P[n + 1] by A5;
  end;
  (A*) ^^ (A |^ 0) = (A*) ^^ {<%>E} by Th24
    .= A* by Th13;
  then
A6: P[0] by A1;
  for n holds P[n] from NAT_1:sch 2(A6, A3);
  hence thesis;
end;
