theorem Th4:
  a <= c & c <= d & d <= b & ['a,b'] c= dom f & ['a,b'] c= dom g
  implies ['c,d'] c= dom (f+g)
  proof
    assume that
A1: a <= c & c <= d & d <= b and
A2: ['a,b'] c= dom f & ['a,b'] c= dom g;
    dom(f+g) = (dom f) /\ (dom g) by VALUED_2:def 45;
    then ['a,b'] /\ ['a,b'] c= dom(f+g) by A2,XBOOLE_1:27;
    hence thesis by A1,Th2;
  end;
