theorem
  A div^ 1 = A & A mod^ 1 = {}
proof
A1: A = A*^1 by ORDINAL2:39;
A2: A = A+^{} by ORDINAL2:27;
  1 = succ 0 .= { 0 };
  then
A3: {} in 1 by Th8;
  hence A div^ 1 = A by A1,A2,Def6;
  thus A mod^ 1 = A-^A by A1,A2,A3,Def6
    .= {} by Th54;
end;
