theorem
  r<>0 implies (r(#)H)" = r" (#) H"
proof
  assume
A1: r<>0;
  now
    let n be Element of NAT;
    thus (r(#)H)".n = ((r(#)H).n)^ by Def2
      .= (r (#) H.n)^ by Def1
      .= r" (#) ((H.n)^) by A1,RFUNCT_1:28
      .= r" (#) (H".n) by Def2
      .= (r" (#) H").n by Def1;
  end;
  hence thesis by FUNCT_2:63;
end;
