theorem Th7:
  (seq1+seq2)+seq3=seq1+(seq2+seq3)
proof
  now
    let n;
    thus ((seq1+seq2)+seq3).n=(seq1+seq2 qua Complex_Sequence).n+ seq3.n by
VALUED_1:1
      .=seq1.n+seq2.n+seq3.n by VALUED_1:1
      .=seq1.n+(seq2.n+seq3.n)
      .=seq1.n+(seq2+seq3).n by VALUED_1:1
      .=(seq1+(seq2+seq3)).n by VALUED_1:1;
  end;
  hence thesis by FUNCT_2:63;
end;
