reserve x, x1, x2, y, X, D for set,
  i, j, k, l, m, n, N for Nat,
  p, q for XFinSequence of NAT,
  q9 for XFinSequence,
  pd, qd for XFinSequence of D;

theorem Th5:
  n >= m implies (n-->0)^(m-->1) is dominated_by_0
proof
  assume
A1: n>=m;
  set p=(n-->0)^(m-->1);
  rng (m-->1) c= {1} & {1} c= {0,1} by FUNCOP_1:13,ZFMISC_1:7;
  then
A2: rng (m-->1) c= {0,1};
  rng (n-->0) c={0} & {0} c= {0,1} by FUNCOP_1:13,ZFMISC_1:7;
  then rng (n-->0) c= {0,1};
  then rng (n-->0) \/ rng (m-->1) c= {0,1} by A2,XBOOLE_1:8;
  hence rng p c= {0,1} by AFINSQ_1:26;
  let k such that
A3: k <= dom p;
  now
    per cases;
    suppose
A4:   k <= dom (n-->0);
A5:   (n-->0)|k = (k-->0) by A4,Lm1;
A6:   Sum (k-->0)=0*k by AFINSQ_2:58;
      p|k=(n-->0)|k by A4,AFINSQ_1:58;
      hence thesis by A5,A6;
    end;
    suppose
      k > dom (n-->0);
      then reconsider kd=k-dom (n-->0) as Nat by NAT_1:21;
      k<=len (n-->0)+len (m-->1) by A3,AFINSQ_1:17;
      then k-len (n-->0)<=len (m-->1)+len (n-->0)-len (n-->0) by XREAL_1:9;
      then kd <= m;
      then
A8:   (m-->1)|kd = (kd-->1) by Lm1;
    reconsider m1 = m-->1 as XFinSequence of NAT;
      k=kd+dom (n-->0);
      then p|k = (n-->0)^(m1|kd) by AFINSQ_1:59;
      then
A9:   Sum(p|k)=Sum (n-->0) + Sum (kd-->1) by A8,AFINSQ_2:55;
      dom p= len (n-->0) +len (m-->1) & dom (m-->1)=m by AFINSQ_1:def 3
;
      then k-n <= m+n-n by A3,XREAL_1:9;
      then k-n <= n by A1,XXREAL_0:2;
      then
A10:  k-n+(k-n)<=n+(k-n) by XREAL_1:6;
      Sum (n-->0) =n*0 & Sum (kd-->1)=kd*1 by AFINSQ_2:58;
      hence thesis by A9,A10;
    end;
  end;
  hence thesis;
end;
