theorem
  Cseq * (seq1 + seq2) = Cseq * seq1 + Cseq * seq2
proof
  now
    let n be Element of NAT;
    thus (Cseq * (seq1 + seq2)).n = Cseq.n * (seq1 + seq2).n by Def8
      .= Cseq.n * (seq1.n + seq2.n) by NORMSP_1:def 2
      .= Cseq.n * seq1.n + Cseq.n * seq2.n by CLVECT_1:def 2
      .= (Cseq * seq1).n + Cseq.n * seq2.n by Def8
      .= (Cseq * seq1).n + (Cseq * seq2).n by Def8
      .= (Cseq * seq1 + Cseq * seq2).n by NORMSP_1:def 2;
  end;
  hence thesis by FUNCT_2:63;
end;
