theorem
  for f,g being Function , A being set holds dom f misses A implies
  (f +* g)|A = g|A
proof
  let f,g be Function , A be set;
  assume dom f misses A;
  then dom f /\ A = {};
  then dom (f|A) = {} by RELAT_1:61;
  then f|A = {};
  hence (f +* g)|A = {} +* g|A by Th71
    .= g|A;
end;
