reserve E, x, y, X for set;
reserve A, B, C for Subset of E^omega;
reserve a, a1, a2, b for Element of E^omega;
reserve i, k, l, m, n for Nat;

theorem Th25:
  (A |^ (k, l)) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^ (k, l))
proof
  defpred P[Nat] means (A |^ (k, l)) ^^ (A |^.. $1) = (A |^.. $1) ^^ (A |^ (k,
  l));
A1: now
    let n;
    assume
A2: P[n];
    (A |^ (k, l)) ^^ (A |^.. (n + 1)) = (A |^ (k, l)) ^^ ((A |^.. n) ^^ A)
    by Th16
      .= (A |^.. n) ^^ (A |^ (k, l)) ^^ A by A2,FLANG_1:18
      .= (A |^.. n) ^^ ((A |^ (k, l)) ^^ A) by FLANG_1:18
      .= (A |^.. n) ^^ (A ^^ (A |^ (k, l))) by FLANG_2:36
      .= (A |^.. n) ^^ A ^^ (A |^ (k, l)) by FLANG_1:18
      .= (A |^.. (n + 1)) ^^ (A |^ (k, l)) by Th16;
    hence P[n + 1];
  end;
  (A |^ (k, l)) ^^ (A |^.. 0) = (A |^ (k, l)) ^^ (A*) by Th11
    .= A* ^^ (A |^ (k, l)) by FLANG_2:66
    .= (A |^.. 0) ^^ (A |^ (k, l)) by Th11;
  then
A3: P[0];
  for n holds P[n] from NAT_1:sch 2(A3, A1);
  hence thesis;
end;
