theorem Th5:
  (seq*Ns)" = (seq")*Ns
proof
  now
    let n be Element of NAT;
    thus ((seq*Ns)").n = ((seq*Ns).n)" by VALUED_1:10
      .= (seq.(Ns.n))" by FUNCT_2:15
      .= (seq").(Ns.n) by VALUED_1:10
      .= ((seq")*Ns).n by FUNCT_2:15;
  end;
  hence thesis by FUNCT_2:63;
end;
