theorem Th3:
  (g(#)seq)*Ns = g(#)(seq*Ns)
proof
  now
    let n be Element of NAT;
    thus ((g(#)seq)*Ns).n = (g(#)seq).(Ns.n) by FUNCT_2:15
      .= g*(seq.(Ns.n)) by VALUED_1:6
      .= g*((seq*Ns).n) by FUNCT_2:15
      .= (g(#)(seq*Ns)).n by VALUED_1:6;
  end;
  hence thesis by FUNCT_2:63;
end;
