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

theorem Th9:
  <%x%> in A |^ n iff <%x%> in A & ( <%>E in A & n > 1 or n = 1)
proof
  thus <%x%> in A |^ n implies <%x%> in A & ( <%>E in A & n > 1 or n = 1)
  proof
    assume
A1: <%x%> in A |^ n;
    A |^ n c= A* by FLANG_1:42;
    hence <%x%> in A by A1,FLANG_1:72;
    assume
A2: ( not <%>E in A or n <= 1)& n <> 1;
    per cases by A2;
    suppose
A3:   not <%>E in A & n <> 1;
      per cases by A3,NAT_1:25;
      suppose
        n = 0;
        then A |^ n = {<%>E} by FLANG_1:24;
        hence contradiction by A1,TARSKI:def 1;
      end;
      suppose
A4:     n > 1;
        then consider m such that
A5:     m + 1 = n by NAT_1:6;
        <%x%> in (A |^ m) ^^ A by A1,A5,FLANG_1:23;
        then consider a, b such that
A6:     a in (A |^ m) and
A7:     b in A and
A8:     <%x%> = a ^ b by FLANG_1:def 1;
        per cases by A8,FLANG_1:4;
        suppose
A9:       a = <%>E & b = <%x%>;
          m + 1 > 0 + 1 by A4,A5;
          then m > 0;
          hence contradiction by A3,A6,A9,FLANG_1:31;
        end;
        suppose
          b = <%>E & a = <%x%>;
          hence contradiction by A3,A7;
        end;
      end;
    end;
    suppose
      n <= 1 & n <> 1;
      then n = 0 by NAT_1:25;
      then A |^ n = {<%>E} by FLANG_1:24;
      hence contradiction by A1,TARSKI:def 1;
    end;
  end;
  assume that
A10: <%x%> in A and
A11: <%>E in A & n > 1 or n = 1;
  per cases by A11;
  suppose
    <%>E in A & n > 1;
    then A c= A |^ n by FLANG_1:35;
    hence thesis by A10;
  end;
  suppose
    n = 1;
    hence thesis by A10,FLANG_1:25;
  end;
end;
