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 Th27:
  A |^ n = {} iff n > 0 & A = {}
proof
  defpred P[Nat] means A |^ $1 = {};
  thus A |^ n = {} implies n > 0 & A = {}
  proof
    assume that
A1: A |^ n = {} and
A2: n <= 0 or A <> {};
A3: now
      defpred P[Nat] means A |^ $1 <> {};
      assume
A4:   A <> {};
A5:   now
        let n;
        assume P[n];
        then consider a1 such that
A6:     a1 in A |^ n by SUBSET_1:4;
        consider a2 such that
A7:     a2 in A by A4,SUBSET_1:4;
        a1 ^ a2 in A |^ n ^^ A by A6,A7,Def1;
        hence P[n + 1] by Th23;
      end;
      A |^ 0 = {<%>E} by Th24;
      then
A8:   P[0];
      for n holds P[n] from NAT_1:sch 2(A8, A5);
      hence contradiction by A1;
    end;
    now
      assume n <= 0;
      then n = 0;
      then A |^ n = {<%>E} by Th24;
      hence contradiction by A1;
    end;
    hence thesis by A2,A3;
  end;
  assume that
A9: n > 0 and
A10: A = {};
A11: now
    let m be Nat;
    assume that
    1 <= m and
    P[m];
    {}(E^omega) |^ (m + 1) = {}(E^omega) |^ m ^^ {}(E^omega) by Th23
      .= {} by Th12;
    hence P[m + 1] by A10;
  end;
A12: P[1] by A10,Th25;
A13: for m being Nat st m >= 1 holds P[m] from NAT_1:sch 8(A12, A11);
  n >= 0 + 1 by A9,NAT_1:13;
  hence thesis by A13;
end;
