theorem Th2:
  seq1+seq2-seq2=seq1
proof
  for n being Element of NAT holds (seq1+seq2-seq2).n = seq1.n
  proof
    let n be Element of NAT;
    (seq1+seq2-seq2).n = (seq1+(seq2-seq2)).n by SEQ_1:31
      .= seq1.n + (seq2+-seq2).n by SEQ_1:7
      .= seq1.n + (seq2.n +(-seq2).n) by SEQ_1:7
      .= seq1.n + (seq2.n +-seq2.n) by SEQ_1:10
      .= seq1.n;
    hence thesis;
  end;
  hence thesis by FUNCT_2:63;
end;
