theorem Th16:
  g (#) c1 (#) c2 = g (#) (c1 * c2)
proof
  dom((g(#)c1)(#)c2) = dom(g(#)c1) & dom(g(#)c1) = dom g by VALUED_1:def 5;
  hence dom((g(#)c1)(#)c2) = dom(g(#)(c1*c2)) by VALUED_1:def 5;
  let x be object;
  assume x in dom((g(#)c1)(#)c2);
  thus ((g(#)c1)(#)c2).x = (g(#)c1).x*c2 by VALUED_1:6
    .= g.x*c1*c2 by VALUED_1:6
    .= g.x*(c1*c2)
    .= (g(#)(c1*c2)).x by VALUED_1:6;
end;
