theorem
  A+ ^^ (A+) = A |^.. 2
proof
  thus A+ ^^ (A+) = A |^.. 1 ^^ (A+) by Th50
    .= A |^.. 1 ^^ (A |^.. 1) by Th50
    .= A |^.. (1 + 1) by Th18
    .= A |^.. 2;
end;
