
theorem VAL:
  for f be complex-valued Function, x be object holds
  f.x = (delneg f).x - (delpos f).x
  proof
    let f be complex-valued Function, x be object;
K1: dom(delneg f) = dom f & dom(delpos f) = dom f by DMN;
    per cases;
    suppose not x in dom f;
      then f.x = {} & (delneg f).x = {} & (delpos f).x = {}
      by K1,FUNCT_1:def 2;
      hence thesis;
    end;
    suppose
    A1: x in dom f;
    per cases;
    suppose
      B1: f is empty; then
      (1/2)(#)(f + abs f) is empty & (1/2)(#)((abs f) - f) is empty;
      hence thesis by B1;
    end;
    suppose not f is empty; then
      reconsider X = dom f as non empty set;
      reconsider f as total X-defined complex-valued Function
        by RELAT_1:def 18,PARTFUN1:def 2;
      A2: dom ((delneg f)-(delpos f)) = (dom (delneg f))/\(dom (delpos f))
        by VALUED_1:12
    .= (dom f) /\ (dom (delpos f)) by DMN
    .= (dom f) /\ (dom f) by DMN
    .= dom f;
    f.x = ((1/2 + (1/2))(#)f).x
    .= ((1/2)(#)f + ((1/2)(#)(abs f) - (1/2)(#)(abs f)) + (1/2)(#)f ).x
      by TOPREALC:2
    .= ((1/2)(#)f  + (1/2)(#)(abs f) - (1/2)(#)(abs f) + (1/2)(#)f).x
      by RFUNCT_1:23
    .= ((1/2)(#)f + (1/2)(#)(abs f) - ((1/2)(#)(abs f) - (1/2)(#)f)).x
      by RFUNCT_1:22
    .= (delneg f - ((1/2)(#)(abs f) - (1/2)(#)f)).x by RFUNCT_1:16
    .= ((delneg f) - (delpos f)).x   by RFUNCT_1:18
    .= (delneg f).x - (delpos f).x by A1,A2,VALUED_1:13;
    hence thesis;
  end;
  end;
end;
