reserve n,k for Element of NAT;

theorem
  for L be finite LATTICE for x be Element of L holds height (x) = 1 iff
  x = Bottom L
proof
  let L be finite LATTICE;
  let x be Element of L;
A1: x = Bottom L implies height (x) = 1
  proof
A2: for B be Chain of Bottom L,Bottom L holds B = {Bottom L}
    proof
      let B be Chain of Bottom L,Bottom L;
A3:   B c= {Bottom L}
      proof
        let r be set;
        r in B implies r in {Bottom L}
        proof
          assume r in B;
          then reconsider r as Element of B;
          Bottom L <= r & r <= Bottom L by Def2;
          then Bottom L = r by ORDERS_2:2;
          hence thesis by TARSKI:def 1;
        end;
        hence thesis;
      end;
      {Bottom L} c= B
      proof
        let r be set;
        r in {Bottom L} implies r in B
        proof
          assume r in {Bottom L};
          then r = Bottom L by TARSKI:def 1;
          hence thesis by Def2;
        end;
        hence thesis;
      end;
      hence thesis by A3,XBOOLE_0:def 10;
    end;
    assume
A4: x = Bottom L;
    ( ex A be Chain of Bottom L,Bottom L st height(Bottom L)= card A)&
    card { Bottom L} = 1 by Def3,CARD_1:30;
    hence thesis by A4,A2;
  end;
  height (x) = 1 implies x = Bottom L
  proof
A5: for z be Element of L st z in {Bottom L,x} holds Bottom L <= z & z <= x
    proof
      let z be Element of L;
      assume
A6:   z in {Bottom L,x};
      per cases by A6,TARSKI:def 2;
      suppose
        z=Bottom L;
        hence thesis by YELLOW_0:44;
      end;
      suppose
        z=x;
        hence thesis by YELLOW_0:44;
      end;
    end;
A7: x in {Bottom L,x} & Bottom L in {Bottom L,x} by TARSKI:def 2;
A8: Bottom L<=x by YELLOW_0:44;
    then {Bottom L,x} is Chain of L by ORDERS_2:9;
    then
A9: {Bottom L,x} is Chain of Bottom L,x by A8,A7,A5,Def2;
    assume that
A10: height (x) = 1 and
A11: x <> Bottom L;
    card {Bottom L,x} = 2 by A11,CARD_2:57;
    hence contradiction by A10,A9,Def3;
  end;
  hence thesis by A1;
end;
