theorem
  (-seq)*Ns = -(seq*Ns) & (abs(seq))*Ns = abs((seq*Ns))
proof
  thus (-seq)*Ns = ((-1)(#)seq)*Ns .= (-1)(#)(seq*Ns) by Th3
    .= -(seq*Ns);
  now
    let n be Element of NAT;
    thus ((abs(seq))*Ns).n = (abs(seq)).(Ns.n) by FUNCT_2:15
      .= |.seq.(Ns.n).| by SEQ_1:12
      .= |.(seq*Ns).n.| by FUNCT_2:15
      .= (abs(seq*Ns)).n by SEQ_1:12;
  end;
  hence thesis by FUNCT_2:63;
end;
