theorem
  (A+)? = A* & (A?)+ = A*
proof
  thus (A+)? = {<%>E} \/ (A+) by FLANG_2:76
    .= A* by Th53;
  <%>E in A? by FLANG_2:78;
  then (A?)+ = (A?)* by Th57;
  hence (A?)+ = A* by FLANG_2:85;
end;
