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 Th72:
  <%x%> in A* iff <%x%> in A
proof
  thus <%x%> in A* implies <%x%> in A
  proof
    defpred P[Nat] means <%x%> in A |^ $1;
    assume that
A1: <%x%> in A* and
A2: not <%x%> in A;
A3: ex i being Nat st P[i] by A1,Th41;
    ex n1 being Nat st P[n1] & for n2 being Nat st P[n2] holds n1 <= n2
    from NAT_1:sch 5(A3);
    then consider n1 being Nat such that
A4: P[n1] and
A5: for n2 being Nat st P[n2] holds n1 <= n2;
A6: now
      assume n1 = 0;
      then <%x%> in {<%>E} by A4,Th24;
      hence contradiction by TARSKI:def 1;
    end;
A7: now
      assume n1 > 1;
      then consider n2 such that
A8:   n2 + 1 = n1 by NAT_1:6;
      <%x%> in (A |^ n2) ^^ A by A4,A8,Th23;
      then consider a, b such that
A9:   a in (A |^ n2) and
A10:  b in A & <%x%> = a ^ b by Def1;
      now
        reconsider n2 as Element of NAT by ORDINAL1:def 12;
        assume that
A11:    a = <%x%> and
        b = <%>E;
        ex i being Element of NAT st P[i] & n1 > i
        proof
          take n2;
          thus thesis by A8,A9,A11,NAT_1:13;
        end;
        hence contradiction by A5;
      end;
      hence <%x%> in A by A10,Th4;
    end;
    n1 = 1 implies <%x%> in A by A4,Th25;
    hence contradiction by A2,A7,A6,NAT_1:25;
  end;
  assume <%x%> in A;
  then <%x%> in A |^ 1 by Th25;
  hence thesis by Th41;
end;
