theorem Th23:
  (r*p)(#)seq=r(#)(p(#)seq)
proof
  now
    let n be Element of NAT;
    thus ((r*p)(#)seq).n=(r*p)*seq.n by Th9
      .=r*(p*seq.n)
      .=r*(p(#)seq).n by Th9
      .=(r(#)(p(#)seq)).n by Th9;
  end;
  hence thesis by FUNCT_2:63;
end;
