theorem
  o1 <= o2 & o2 <= o1 implies o1 = o2
proof
  assume that
A1: o1 <= o2 and
A2: o2 <= o1;
  the_result_sort_of o1 <= the_result_sort_of o2 & the_result_sort_of o2
  <= the_result_sort_of o1 by A1,A2;
  then
A3: the_result_sort_of o1 = the_result_sort_of o2 by ORDERS_2:2;
  (the_arity_of o1) <= (the_arity_of o2) & (the_arity_of o2) <= (
  the_arity_of o1) by A1,A2;
  then
A4: the_arity_of o1 = the_arity_of o2 by Th6;
  o1 ~= o2 by A1;
  hence thesis by A4,A3,Def3;
end;
