reserve n,k for Element of NAT;

theorem Th8:
  for L be finite LATTICE for X be non empty Subset of L holds ex x
  be Element of L st x in X & for y be Element of L st y in X holds not x < y
proof
  let L be finite LATTICE;
  let X be non empty Subset of L;
  defpred P[Nat] means ex x be Element of L st x in X & $1=height(x);
  ex k st P[k] & for n st P[n] holds n <= k
  proof
A1: for k be Nat st P[k] holds k <= card the carrier of L
    proof
      let k be Nat;
      assume P[k];
      then consider x be Element of L such that
      x in X and
A2:   k=height(x);
      ex B be Chain of Bottom L,x st k=card B by A2,Def3;
      hence thesis by NAT_1:43;
    end;
A3: ex k be Nat st P[k]
    proof
      consider x be object such that
A4:   x in X by XBOOLE_0:def 1;
      reconsider x as Element of L by A4;
      ex B be Chain of Bottom L,x st height(x)= card B by Def3;
      hence thesis by A4;
    end;
    consider k being Nat such that
A5: P[k] and
A6: for n being Nat st P[n] holds n <= k from NAT_1:sch 6 (A1,A3);
    for n being Element of NAT st P[n] holds n <= k by A6;
    hence thesis by A5;
  end;
  then consider k such that
A7: ( P[k])& for n st P[n] holds n <= k;
  consider x be Element of L such that
A8: x in X and
A9: k=height(x) & for n st ex y be Element of L st y in X & n=height(y)
  holds n <= k by A7;
  for y be Element of L st y in X holds not x < y
  proof
    let y be Element of L;
    assume that
A10: y in X and
A11: x < y;
    height(x)<height(y) by A11,Th2;
    hence contradiction by A9,A10;
  end;
  hence thesis by A8;
end;
