theorem
  A+ ^^ (A |^.. k) = A |^.. k ^^ (A+)
proof
  thus A+ ^^ (A |^.. k) = A |^.. (k + 1) by Th90
    .= (A |^.. k) ^^ (A |^.. 1) by Th18
    .= A |^.. k ^^ (A+) by Th50;
end;
