theorem
  dom f in S & dom g in S implies dom(f+g) in S
proof
  assume that
A1: dom f in S and
A2: dom g in S;
  reconsider E1 = dom f, E2 = dom g as Element of S by A1,A2;
  dom(f+g) = E1 /\ E2 by VALUED_1:def 1;
  hence thesis;
end;
