theorem Th10:
  (seq1 + seq2) + seq3 = seq1 + (seq2 + seq3)
proof
  let n be Element of NAT;
  thus ((seq1+seq2)+seq3).n = (seq1+seq2).n+ seq3.n by Th4
    .= seq1.n+seq2.n+seq3.n by Th4
    .=seq1.n+(seq2.n+seq3.n) by RLVECT_1:def 3
    .=seq1.n+(seq2+seq3).n by Th4
    .=(seq1+(seq2+seq3)).n by Th4;
end;
