theorem
  abs(r(#)H) = |.r.| (#) abs(H)
proof
  now
    let n be Element of NAT;
    thus abs(r(#)H).n=abs((r(#)H).n) by Def4
      .=abs(r(#)(H.n)) by Def1
      .=|.r.|(#)abs(H.n) by RFUNCT_1:25
      .=|.r.|(#)(abs(H)).n by Def4
      .=(|.r.|(#)abs(H)).n by Def1;
  end;
  hence thesis by FUNCT_2:63;
end;
