theorem Th27:
  (A |^.. m) ^^ (A |^.. n) = (A |^.. n) ^^ (A |^.. m)
proof
  thus (A |^.. m) ^^ (A |^.. n) = A |^.. (m + n) by Th18
    .= (A |^.. n) ^^ (A |^.. m) by Th18;
end;
