theorem Th138:
  (f++F) \ (f++G) c= f ++ (F \ G)
proof
  let i;
  assume
A1: i in (f++F) \ (f++G);
  then consider w such that
A2: i = f+w and
A3: w in F by Th134;
  now
    assume not w in F\G;
    then w in G by A3,XBOOLE_0:def 5;
    then f+w in f++G by Th132;
    hence contradiction by A1,A2,XBOOLE_0:def 5;
  end;
  hence thesis by A2,Th132;
end;
