theorem Th2:
  f(#)(c1+c2) = f(#)c1 + f(#)c2
  proof
A1: dom (f(#)(c1+c2)) = dom f by VALUED_1:def 5;
A2: dom (f(#)c1) = dom f by VALUED_1:def 5;
A3: dom (f(#)c2) = dom f by VALUED_1:def 5;
A4: dom (f(#)c1 + f(#)c2) = dom (f(#)c1) /\ dom (f(#)c2) by VALUED_1:def 1;
    hence dom (f(#)(c1+c2)) = dom (f(#)c1 + f(#)c2) by A2,A3,VALUED_1:def 5;
    let x be object;
    assume
A5: x in dom (f(#)(c1+c2));
    hence (f(#)(c1+c2)).x = f.x * (c1+c2) by VALUED_1:def 5
    .= f.x*c1 + f.x*c2
    .= (f(#)c1).x + f.x*c2 by A1,A2,A5,VALUED_1:def 5
    .= (f(#)c1).x + (f(#)c2).x by A1,A3,A5,VALUED_1:def 5
    .= (f(#)c1 + f(#)c2).x by A1,A2,A3,A4,A5,VALUED_1:def 1;
  end;
