reserve a,b,c,k,k9,m,n,n9,p,p9 for Nat;
reserve i,i9 for Integer;

theorem Th9:
  for X being set st for m ex n st n >= m & n in X holds X is infinite
proof
  let X be set;
A1: now
    let f be Function;
    defpred P[Nat] means ex m st for n st n >= m holds not n in f.:$1;
A2: for k being Nat st P[k] holds P[k+1]
    proof
      let k be Nat;
      assume ex m st for n st n >= m holds not n in f.:k;
      then consider m such that
A3:   for n st n >= m holds not n in f.:k;
      Segm(k + 1) = Segm k \/ { k } by AFINSQ_1:2;
      then
A4:   f.:(k + 1) = f.:k \/ Im(f,k) by RELAT_1:120;
      per cases;
      suppose
A5:     k in dom f & f.k in NAT;
        then reconsider m9 = f.k as Element of NAT;
        set M = max(m,m9 + 1);
        reconsider M as Element of NAT by ORDINAL1:def 12;
        take M;
        let n;
        assume
A6:     n >= M;
        then
A7:     not n in f.:k by A3,XXREAL_0:30;
        n >= m9 + 1 by A6,XXREAL_0:30;
        then n <> m9 by NAT_1:13;
        then
A8:     not n in { m9 } by TARSKI:def 1;
        f.:(k + 1) = f.:k \/ { m9 } by A4,A5,FUNCT_1:59;
        hence thesis by A7,A8,XBOOLE_0:def 3;
      end;
      suppose
A9:     k in dom f & not f.k in NAT;
        take m;
        set m9 = f.k;
        let n;
        n <> m9 by A9,ORDINAL1:def 12;
        then
A10:    not n in { m9 } by TARSKI:def 1;
        assume n >= m;
        then
A11:    not n in f.:k by A3;
        f.:(k + 1) = f.:k \/ { m9 } by A4,A9,FUNCT_1:59;
        hence thesis by A11,A10,XBOOLE_0:def 3;
      end;
      suppose
        not k in dom f;
        then
A12:    dom f misses { k } by ZFMISC_1:50;
        take m;
        let n;
        assume
A13:    n >= m;
        Im(f,k) = f.:(dom f /\ { k }) by RELAT_1:112
          .= f.:{} by A12,XBOOLE_0:def 7
          .= {};
        hence thesis by A3,A4,A13;
      end;
    end;
A14: P[0]
    proof
      take 0;
      let n such that
      n >= 0;
      thus thesis;
    end;
    thus for k being Nat holds P[k] from NAT_1:sch 2(A14,A2);
  end;
  now
    assume X is finite;
    then consider f being Function such that
A15: rng f = X and
A16: dom f in omega by FINSET_1:def 1;
    reconsider k = dom f as Element of NAT by A16;
    f.:k = X by A15,RELAT_1:113;
    hence ex m st for n st n >= m holds not n in X by A1;
  end;
  hence thesis;
end;
