theorem Th3:
  (p(#)seq)*Ns = p(#)(seq*Ns)
proof
  now
    let n be Element of NAT;
    thus ((p(#)seq)*Ns).n = (p(#)seq).(Ns.n) by FUNCT_2:15
      .= p*(seq.(Ns.n)) by SEQ_1:9
      .= p*((seq*Ns).n) by FUNCT_2:15
      .= (p(#)(seq*Ns)).n by SEQ_1:9;
  end;
  hence thesis by FUNCT_2:63;
end;
