theorem
  seq1/"seq + seq19/"seq = (seq1 + seq19) /" seq & seq1/"seq - seq19/"
  seq = (seq1 - seq19) /" seq
proof
  thus seq1/"seq + seq19/"seq = (seq1 + seq19) (#) (seq") by Th15
    .= (seq1 + seq19) /" seq;
  thus seq1/"seq - seq19/"seq = (seq1 - seq19) (#) (seq") by Th20
    .= (seq1 - seq19) /" seq;
end;
