
theorem
  for P being complete non empty Poset, V being non empty Subset of P
  holds uparrow sup V = meet {uparrow u where u is Element of P : u in V}
proof
  let P be complete non empty Poset, V be non empty Subset of P;
  set F = {uparrow u where u is Element of P : u in V};
  consider u being object such that
A1: u in V by XBOOLE_0:def 1;
A2: F c= bool the carrier of P
  proof
    let X be object;
    assume X in F;
    then ex u being Element of P st X = uparrow u & u in V;
    hence thesis;
  end;
  reconsider u as Element of P by A1;
A3: uparrow u in F by A1;
  reconsider F as Subset-Family of P by A2;
  reconsider F as Subset-Family of P;
  now
    let x be object;
    hereby
      assume
A4:   x in uparrow sup V;
      then reconsider d = x as Element of P;
A5:   d >= sup V by A4,WAYBEL_0:18;
      now
        let Y be set;
        assume Y in F;
        then consider u being Element of P such that
A6:     Y = uparrow u and
A7:     u in V;
        sup V is_>=_than V by YELLOW_0:32;
        then sup V >= u by A7,LATTICE3:def 9;
        then d >= u by A5,ORDERS_2:3;
        hence x in Y by A6,WAYBEL_0:18;
      end;
      hence x in meet F by A3,SETFAM_1:def 1;
    end;
    assume
A8: x in meet F;
    then reconsider d = x as Element of P;
    now
      let b be Element of P;
      assume b in V;
      then uparrow b in F;
      then d in uparrow b by A8,SETFAM_1:def 1;
      hence d >= b by WAYBEL_0:18;
    end;
    then d is_>=_than V by LATTICE3:def 9;
    then d >= sup V by YELLOW_0:32;
    hence x in uparrow sup V by WAYBEL_0:18;
  end;
  hence thesis by TARSKI:2;
end;
