theorem
  seq1 is_compared_to seq2 implies (seq1 ^\k) is_compared_to (seq2 ^\k)
proof
  assume
A1: seq1 is_compared_to seq2;
  let r;
  assume r > 0;
  then consider m1 be Nat such that
A2: for n st n >= m1 holds dist((seq1.n), (seq2.n)) < r by A1;
  take m = m1;
  let n such that
A3: n >= m;
  n + k >= n by NAT_1:11;
  then n + k >= m by A3,XXREAL_0:2;
  then dist((seq1.(n + k)), (seq2.(n + k))) < r by A2;
  then dist((seq1 ^\k).n, (seq2.(n + k))) < r by NAT_1:def 3;
  hence thesis by NAT_1:def 3;
end;
