theorem Th59:
  A c= B* implies A |^ n c= B*
proof
  defpred P[Nat] means A |^ $1 c= B*;
  assume
A1: A c= B*;
A2: now
    let n;
    assume P[n];
    then (A |^ n) ^^ A c= B* by A1,Th46;
    hence P[n + 1] by Th23;
  end;
  <%>E in B* by Th48;
  then {<%>E} c= B* by ZFMISC_1:31;
  then
A3: P[0] by Th24;
  for n holds P[n] from NAT_1:sch 2(A3, A2);
  hence thesis;
end;
