theorem
  the_rank_of A = A
proof
 A c= Rank A by Th38;
  hence the_rank_of A c= A by Th65;
  defpred P[Ordinal] means $1 c= the_rank_of $1;
A1: for A st for B st B in A holds P[B] holds P[A]
  proof
    let A such that
A2: for B st B in A holds B c= the_rank_of B;
 now
      let B such that
A3:   B in A;
      reconsider Y = B as set;
      take Y;
      thus Y in A & B c= the_rank_of Y by A2,A3;
    end;
    hence thesis by Th70;
  end;
 P[B] from ORDINAL1:sch 2(A1);
  hence thesis;
end;
