theorem Th17:
  (r*p)(#)seq=r(#)(p(#)seq)
proof
  now
    let n;
    thus ((r*p)(#)seq).n=(r*p)*seq.n by VALUED_1:6
      .=r*(p*seq.n)
      .=r*(p(#)seq).n by VALUED_1:6
      .=(r(#)(p(#)seq)).n by VALUED_1:6;
  end;
  hence thesis by FUNCT_2:63;
end;
