theorem Th6:
  for P holds P^^0 = {{}} & for n holds P^^(n+1) = (P^^n)^P
proof
  let P;
  thus P^^0 = {{}} by Def3;
  let n;
  consider Q such that A1: Q = P^^n & P^^(n+1) = Q^P by Def3;
  thus thesis by A1;
end;
