theorem Th67:
  seq1 is_compared_to seq2 implies seq2 is_compared_to seq1
proof
  assume
A1: seq1 is_compared_to seq2;
  let r;
  assume r > 0;
  then consider k such that
A2: for n st n >= k holds dist((seq1.n), (seq2.n)) < r by A1;
  take m = k;
  let n;
  assume n >= m;
  hence thesis by A2;
end;
