theorem Th32:
  (A |^.. m) ^^ (A*) = A* ^^ (A |^.. m)
proof
  thus (A |^.. m) ^^ (A*) = (A |^.. m) ^^ (A |^.. 0) by Th11
    .= (A |^.. 0) ^^ (A |^.. m) by Th27
    .= A* ^^ (A |^.. m) by Th11;
end;
