
theorem DMN:
  for f be complex-valued Function holds dom f = dom (delpos f)
  & dom f = dom (delneg f) & dom f = dom (delall f)
  proof
    let f be complex-valued Function;
    A1: dom ((1/2)(#)(f + abs f)) = dom (f + abs f) by VALUED_1:def 5
    .= (dom f) /\ (dom (abs f)) by VALUED_1:def 1
    .= dom f by VALUED_1:def 11;
    dom ((1/2)(#)((abs f) -  f)) = dom ((abs f) - f) by VALUED_1:def 5
    .= dom ((abs f) + (-f)) by VALUED_1:def 9
    .= (dom (-f)) /\ (dom (abs f)) by VALUED_1:def 1
    .= (dom ((-1)(#)f))/\(dom (abs f)) by VALUED_1:def 6
    .= (dom f) /\ (dom (abs f)) by VALUED_1:def 5
    .= dom f by VALUED_1:def 11;
    hence thesis by A1,VALUED_1:def 5;
  end;
