reserve
  X for non empty set,
  FX for Filter of X,
  SFX for Subset-Family of X;

theorem Th26:
  for F be sequence of bool NAT st
  for x be Element of NAT holds F.x={y where y is Element of NAT:x <= y} holds
  #(Tails OrderedNAT)=rng F
  proof
    let F be sequence of bool NAT such that
A1: for x be Element of NAT holds F.x =
    {y where y is Element of NAT:x <= y};
    set F1=rng F;
    set F2=the set of all uparrow p where p is Element of OrderedNAT;
    now
      hereby
        let t be object;
        assume t in F1;
        then consider x0 be object such that
A2:     x0 in dom F & t=F.x0 by FUNCT_1:def 3;
        reconsider x1=x0 as Element of NAT by A2;
        reconsider x2=x0 as Element of OrderedNAT by A2;
        t={y where y is Element of NAT:x1 <= y} by A1,A2;
        then t={x where x is Element of OrderedNAT: x2 <= x} by Lm3;
        then t={x where x is Element of NAT:ex p0 be Element of NAT st
        x2=p0 & p0 <= x} by Lm4;
        then t=uparrow x2 by Th23;
        hence t in F2;
      end;
      let t be object;
      assume t in F2;
      then consider p0 be Element of OrderedNAT such that
A3:   t=uparrow p0;
      t={x where x is Element of NAT:ex p1 be Element of NAT st
      p0=p1 & p1 <= x} by A3,Th23;
      then
A4:   t={y where y is Element of OrderedNAT:p0<=y} by Lm4;
      reconsider p2=p0 as Element of NAT;
      t={x where x is Element of NAT:p2 <= x} by A4,Lm3;
      then
A5:   t = F.p2 by A1;
      dom F=NAT by FUNCT_2:def 1;
      hence t in F1 by A5,FUNCT_1:3;
    end;
    then F1 c= F2 & F2 c= F1;
    hence thesis;
  end;
