theorem
  (Cseq1 (#) Cseq2) * seq = Cseq1 * (Cseq2 * seq)
proof
  now
    let n be Element of NAT;
    thus ((Cseq1 (#) Cseq2) * seq).n = (Cseq1 (#) Cseq2).n * seq.n by Def8
      .= (Cseq1.n * Cseq2.n) * seq.n by VALUED_1:5
      .= Cseq1.n * (Cseq2.n * seq.n) by CLVECT_1:def 4
      .= Cseq1.n * (Cseq2 * seq).n by Def8
      .= (Cseq1 * (Cseq2 * seq)).n by Def8;
  end;
  hence thesis by FUNCT_2:63;
end;
