theorem Th14:
  abs(H)"=abs(H")
proof
  now
    let n be Element of NAT;
    thus (abs(H")).n=abs(H".n) by Def4
      .=abs((H.n)^) by Def2
      .=(abs((H.n)))^ by RFUNCT_1:30
      .=(abs(H).n)^ by Def4
      .=(abs(H)").n by Def2;
  end;
  hence thesis by FUNCT_2:63;
end;
