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 Th28:
  A c= B |^.. k & n > 0 implies A |^ n c= B |^.. k
proof
  assume that
A1: A c= B |^.. k and
A2: n > 0;
  defpred P[Nat] means $1 > 0 implies A |^ $1 c= B |^.. k;
A3: now
    let m;
    assume
A4: P[m];
    per cases;
    suppose
      m <= 0;
      then m = 0;
      hence P[m + 1] by A1,FLANG_1:25;
    end;
    suppose
      m > 0;
      then (A |^ m) ^^ A c= (B |^.. k) ^^ (B |^.. k) by A1,A4,FLANG_1:17;
      then A |^ (m + 1) c= (B |^.. k) ^^ (B |^.. k) by FLANG_1:23;
      then
A5:   A |^ (m + 1) c= (B |^.. (k + k)) by Th18;
      B |^.. (k + k) c= B |^.. k by Th5,NAT_1:11;
      hence P[m + 1] by A5,XBOOLE_1:1;
    end;
  end;
A6: P[0];
  for m holds P[m] from NAT_1:sch 2(A6, A3);
  hence thesis by A2;
end;
