theorem Th63:
  seq is Cauchy implies seq + x is Cauchy
proof
  assume
A1: seq is Cauchy;
  let r;
  assume r > 0;
  then consider m1 be Nat such that
A2: for n, m st n >= m1 & m >= m1 holds dist((seq.n), (seq.m)) < r by A1;
  take k = m1;
  let n, m;
  dist((seq.n) + x, (seq.m) + x) <= dist((seq.n), (seq.m)) + dist(x, x) by
CSSPACE:56;
  then
A3: dist((seq.n) + x, (seq.m) + x) <= dist((seq.n), (seq.m)) + 0 by CSSPACE:50;
  assume n >= k & m >= k;
  then dist((seq.n), (seq.m)) < r by A2;
  then dist((seq.n) + x, (seq.m) + x) < r by A3,XXREAL_0:2;
  then dist((seq + x).n, (seq.m) + x) < r by BHSP_1:def 6;
  hence thesis by BHSP_1:def 6;
end;
