theorem
  dom f misses dom g implies (f +* g)|(dom f) = f
proof
  assume dom f misses dom g;
  then
A1: dom f \ dom g = dom f by XBOOLE_1:83;
  dom((f +* g)|(dom f)) = dom(f +* g) /\ dom f by RELAT_1:61
    .= (dom f \/ dom g) /\ dom f by Def1
    .= dom f by XBOOLE_1:21;
  hence thesis by A1,Th24,GRFUNC_1:3;
end;
