
theorem DNF:
  for f be complex-valued Function holds delneg f = delpos (-f)
  proof
    let f be complex-valued Function;
    reconsider g = -f as complex-valued Function;
    (1/2)(#)(f + (abs f)) = (1/2)(#)f + (1/2)(#)(abs f) by RFUNCT_1:16
    .= (1/2)(#)(abs g) + (1/2)(#)(-g) by EUCLID:5
    .= (1/2)(#)(abs g) + -((1/2)(#)g) by VALUED_2:23
    .= (1/2)(#)(abs g) - (1/2)(#)g by VALUED_1:def 9;
    hence thesis by RFUNCT_1:18;
  end;
