theorem
  (A |^ k) ^^ (A+) = (A+) ^^ (A |^ k)
proof
  thus (A |^ k) ^^ (A+) = (A |^ k) ^^ (A |^.. 1) by Th50
    .= (A |^.. 1) ^^ (A |^ k) by Th24
    .= A+ ^^ (A |^ k) by Th50;
end;
