theorem
  a * (seq1 + seq2) = a * seq1 + a * seq2
proof
  now
    let n be Element of NAT;
    thus (a * (seq1 + seq2)).n = a * (seq1 + seq2).n by CLVECT_1:def 14
      .= a * (seq1.n + seq2.n) by NORMSP_1:def 2
      .= a * seq1.n + a * seq2.n by CLVECT_1:def 2
      .= (a * seq1).n + a * seq2.n by CLVECT_1:def 14
      .= (a * seq1).n + (a * seq2).n by CLVECT_1:def 14
      .= (a * seq1 + a * seq2).n by NORMSP_1:def 2;
  end;
  hence thesis by FUNCT_2:63;
end;
