reserve r for Real,
  X for set,
  f, g, h for real-valued Function;

theorem Th2:
  f c= g implies h-f c= h-g
proof
A1: dom(h-g) = dom h /\ dom g by VALUED_1:12;
A2: dom(h-f) = dom h /\ dom f by VALUED_1:12;
  assume
A3: f c= g;
  then dom f c= dom g by GRFUNC_1:2;
  then
A4: dom(h-f) c= dom(h-g) by A1,A2,XBOOLE_1:27;
  now
    let x be object;
    assume
A5: x in dom(h-f);
    then
A6: x in dom f by A2,XBOOLE_0:def 4;
    thus (h-f).x = h.x - f.x by A5,VALUED_1:13
      .= h.x - g.x by A3,A6,GRFUNC_1:2
      .= (h-g).x by A4,A5,VALUED_1:13;
  end;
  hence thesis by A4,GRFUNC_1:2;
end;
