theorem
 dom f misses dom g implies (f +* g) \ g = f
proof assume
A1: dom f misses dom g;
  then
A2: f misses g by RELAT_1:179;
 thus (f +* g) \ g
     = (f \/ g) \ g by A1,Th31
    .= f by A2,XBOOLE_1:88;
end;
