theorem
  a * (seq1 - seq2) = a * seq1 - a * seq2
proof
  let n be Element of NAT;
  thus (a * (seq1 - seq2)).n = a * (seq1 - seq2).n by NORMSP_1:def 5
    .= a * (seq1.n - seq2.n) by NORMSP_1:def 3
    .= a * seq1.n - a * seq2.n by RLVECT_1:34
    .= (a * seq1).n - a * seq2.n by NORMSP_1:def 5
    .= (a * seq1).n - (a * seq2).n by NORMSP_1:def 5
    .= (a * seq1 - a * seq2).n by NORMSP_1:def 3;
end;
