theorem Th18:
  Re seq1-Re seq2=Re (seq1-seq2) & Im seq1-Im seq2=Im(seq1-seq2)
proof
  now
    let n be Element of NAT;
    thus (Re seq1-Re seq2).n = Re seq1.n+(-Re seq2).n by SEQ_1:7
      .=Re seq1.n+(Re -seq2).n by Th16
      .=(Re seq1+Re -seq2).n by SEQ_1:7
      .=Re (seq1-seq2).n by Th15;
  end;
  hence Re seq1-Re seq2=Re (seq1-seq2);
  now
    let n be Element of NAT;
    thus (Im seq1-Im seq2).n = Im seq1.n+(-Im seq2).n by SEQ_1:7
      .=Im seq1.n+Im (-seq2).n by Th16
      .=(Im seq1+Im -seq2).n by SEQ_1:7
      .=Im(seq1-seq2).n by Th15;
  end;
  hence thesis;
end;
