theorem Th17:
  for P, Q st P c= Q for n holds P^^n c= Q^^n
proof
  let P, Q;
  assume A1: P c= Q;
  defpred S[ Nat ] means P^^$1 c= Q^^$1;
  P^^0 = {{}} by Th6 .= Q^^0 by Th6;
  then A2: S[ 0 ];
  A3: for n holds S[ n ] implies S[ n + 1 ]
  proof
    let n;
    assume A4: S[ n ];
    A5: P^^(n+1) = (P^^n)^P by Th6;
    Q^^(n+1) = (Q^^n)^Q by Th6;
    hence thesis by A1, A4, A5, Th16;
  end;
  for n holds S[ n ] from NAT_1:sch 2(A2, A3);
  hence thesis;
end;
