theorem Th4:
  f(/)c + g(/)c = (f+g)(/)c
  proof
A1: dom (f(/)c + g(/)c) = dom (f(/)c) /\ dom (g(/)c) by VALUED_1:def 1;
A2: dom (f(/)c) = dom f by VALUED_2:28;
A3: dom (g(/)c) = dom g by VALUED_2:28;
A4: dom ((f+g)(/)c) = dom (f+g) by VALUED_2:28;
A5: dom (f+g) = dom f /\ dom g by VALUED_1:def 1;
    thus dom (f(/)c + g(/)c) = dom ((f+g)(/)c) by A1,A2,A3,A4,VALUED_1:def 1;
    let x be object;
    assume
A6: x in dom (f(/)c + g(/)c);
    thus ((f+g)(/)c).x = (f+g).x/c by VALUED_2:29
    .= (f.x+g.x)/c by A1,A2,A3,A6,A5,VALUED_1:def 1
    .= f.x/c + g.x/c
    .= f.x/c + (g(/)c).x by VALUED_2:29
    .= (f(/)c).x + (g(/)c).x by VALUED_2:29
    .= (f(/)c + g(/)c).x by A6,VALUED_1:def 1;
  end;
