theorem
  ||.r(#)H.|| = |.r.| (#) ||.H.||
  proof
    now
      let n be Element of NAT;
      thus ||.r(#)H .||.n=||. ((r(#)H).n) .|| by Def4
      .=||. r(#)(H.n).|| by Def1
      .=|.r.|(#)||. H.n .|| by VFUNCT_1:22
      .=|.r.|(#)(||.H.||).n by Def4
      .=(|.r.|(#)||.H.||).n by SEQFUNC:def 1;
    end;
    hence thesis by FUNCT_2:63;
  end;
