theorem Th24:
  ((f +* g)|(dom f \ dom g)) c= f
proof
A1: for x being object
   st x in dom((f +* g)|(dom f \ dom g)) holds ((f +* g)|(dom f \ dom
  g)).x = f.x
  proof
    let x be object such that
A2: x in dom((f +* g)|(dom f \ dom g));
    dom((f +* g)|(dom f \ dom g)) c= dom f \ dom g by RELAT_1:58;
    then not x in dom g by A2,XBOOLE_0:def 5;
    then (f +* g).x = f.x by Th11;
    hence thesis by A2,FUNCT_1:47;
  end;
  dom((f +* g)|(dom f \ dom g)) c= dom f \ dom g by RELAT_1:58;
  then dom((f +* g)|(dom f \ dom g)) c= dom f by XBOOLE_1:1;
  hence thesis by A1,GRFUNC_1:2;
end;
