reserve n,k for Element of NAT;

theorem Th11:
  for L be finite distributive LATTICE for x be Element of L holds
  x in Join-IRR L implies ex z be Element of L st z < x & for y be Element of L
  st y < x holds y <= z
proof
  let L be finite distributive LATTICE;
  let x be Element of L;
  assume
A1: x in Join-IRR L;
  then x<>Bottom L by Th10;
  then consider z be Element of L such that
A2: z <(1) x by Th9;
A3: z < x by A2;
  for y be Element of L st y < x holds y <= z
  proof
    let y be Element of L;
    consider Y be set such that
A4: Y={g where g is Element of L : g < x & not (g <= z)};
A5: Y is empty
    proof
A6:   Y c= the carrier of L
      proof
        let f be object;
        assume f in Y;
        then ex g be Element of L st g=f & g<x & not g<=z by A4;
        hence thesis;
      end;
      assume Y is non empty;
      then consider a be Element of L such that
A7:   a in Y and
A8:   for b be Element of L st b in Y holds not a < b by A6,Th8;
A9:   for t be Element of L holds t in Y iff t < x & not t <= z
      proof
        let t be Element of L;
        t in Y implies t < x & not t <= z
        proof
          assume t in Y;
          then ex g be Element of L st g=t & g < x & not g <= z by A4;
          hence thesis;
        end;
        hence thesis by A4;
      end;
      then
A10:  not a<=z by A7;
A11:  z<=x by A3,ORDERS_2:def 6;
      a<x by A9,A7;
      then
A12:  a"\/"z<>x by A1,A3,Th10;
      a<x by A9,A7;
      then a<=x by ORDERS_2:def 6;
      then a"\/"z <= x by A11,YELLOW_5:9;
      then
A13:  a"\/"z<x by A12,ORDERS_2:def 6;
      a"/\"a <= a"\/"z by YELLOW_5:5;
      then
A14:  a <= a"\/"z by YELLOW_5:2;
      a<>a"\/"z
      proof
        assume a=a"\/"z;
        z"/\"z<=a"\/"z by YELLOW_5:5;
        then z<=a"\/"z by YELLOW_5:2;
        then z<a"\/"z by A10,A14,ORDERS_2:def 6;
        hence contradiction by A2,A13;
      end;
      then
A15:  (a"\/"z)>a by A14,ORDERS_2:def 6;
      not a"\/"z <= z by A10,YELLOW_5:3;
      then a"\/"z in Y by A9,A13;
      hence contradiction by A8,A15;
    end;
    assume y < x & not y <= z;
    then y in Y by A4;
    hence contradiction by A5;
  end;
  hence thesis by A3;
end;
