
theorem
  for L being with_suprema Poset, F being Filter of L holds F is prime
  iff for A being finite non empty Subset of L st sup A in F ex a being Element
  of L st a in A & a in F
proof
  let L be with_suprema Poset, I be Filter of L;
  thus I is prime implies for A being finite non empty Subset of L st sup A in
  I ex a being Element of L st a in A & a in I
  proof
    defpred P[set] means $1 is non empty & "\/"($1,L) in I implies ex a being
    Element of L st a in $1 & a in I;
    assume
A1: for x,y being Element of L st x"\/"y in I holds x in I or y in I;
    let A be finite non empty Subset of L;
A2: now
      let x,B be set such that
A3:   x in A and
A4:   B c= A and
A5:   P[B];
      thus P[B \/ {x}]
      proof
        reconsider a = x as Element of L by A3;
        reconsider C = B as finite Subset of L by A4,XBOOLE_1:1;
        assume that
        B \/ {x} is non empty and
A6:     "\/"(B \/ {x},L) in I;
        per cases;
        suppose
          B = {};
          then "\/"(B \/ {a},L) = a & a in B \/ {a} by TARSKI:def 1,YELLOW_0:39
;
          hence thesis by A6;
        end;
        suppose
A7:       B <> {};
A8:       sup {a} = a by YELLOW_0:39;
A9:       ex_sup_of {a}, L by YELLOW_0:54;
          ex_sup_of C, L by A7,YELLOW_0:54;
          then
A10:      "\/"(B \/ {x},L) = (sup C)"\/"sup {a} by A9,YELLOW_2:3;
          hereby
            per cases by A1,A6,A10,A8;
            suppose
              sup C in I;
              then consider b being Element of L such that
A11:          b in B & b in I by A5,A7;
              take b;
              thus b in B \/ {x} & b in I by A11,XBOOLE_0:def 3;
            end;
            suppose
A12:          a in I;
              take a;
              a in {a} by TARSKI:def 1;
              hence a in B \/ {x} & a in I by A12,XBOOLE_0:def 3;
            end;
          end;
        end;
      end;
    end;
A13: P[{}];
A14: A is finite;
    P[A] from FINSET_1:sch 2(A14,A13,A2);
    hence thesis;
  end;
  assume
A15: for A being finite non empty Subset of L st sup A in I ex a being
  Element of L st a in A & a in I;
  let a,b be Element of L;
  assume a"\/"b in I;
  then sup {a,b} in I by YELLOW_0:41;
  then ex x being Element of L st x in {a,b} & x in I by A15;
  hence thesis by TARSKI:def 2;
end;
