
theorem
  for L being non empty Poset holds L is with_suprema iff for X being
  finite non empty Subset of L holds ex_sup_of X,L
proof
  let L be non empty Poset;
  hereby
    defpred P[set] means $1 is non empty implies ex_sup_of $1,L;
    assume
A1: L is with_suprema;
    let X be finite non empty Subset of L;
A2: for x,Y being set st x in X & Y c= X & P[Y] holds P[Y \/ {x}]
    proof
      let x,Y be set such that
A3:   x in X and
      Y c= X and
A4:   Y is non empty implies ex_sup_of Y,L;
      reconsider z = x as Element of L by A3;
      assume Y \/ {x} is non empty;
      per cases;
      suppose
        Y is empty;
        then Y \/ {x} = {z};
        hence thesis by Th38;
      end;
      suppose
A5:     Y is non empty;
        thus ex_sup_of Y \/ {x}, L
        proof
          take a = "\/"(Y,L)"\/"z;
A6:       Y is_<=_than "\/"(Y,L) by A4,Def9;
A7:       ex_sup_of {"\/"(Y,L),z},L by A1,Th20;
          then z <= a by Th18;
          then
A8:       {x} is_<=_than a by Th7;
          "\/" (Y,L) <= a by A7,Th18;
          then
A9:       Y is_<=_than a by A6,Th11;
          hence Y \/ {x} is_<=_than a by A8,Th10;
          hereby
            let b be Element of L;
            assume
A10:        Y \/ {x} is_<=_than b;
            Y c= Y \/ {x} by XBOOLE_1:7;
            then Y is_<=_than b by A10;
            then
A11:        "\/"(Y,L) <= b by A4,A5,Def9;
            z in {x} by TARSKI:def 1;
            then z in Y \/ {x} by XBOOLE_0:def 3;
            then z <= b by A10;
            hence b >= a by A7,A11,Th18;
          end;
          let c be Element of L such that
A12:      Y \/ {x} is_<=_than c and
A13:      for b being Element of L st Y \/ {x} is_<=_than b holds b >= c;
          Y c= Y \/ {x} by XBOOLE_1:7;
          then Y is_<=_than c by A12;
          then
A14:      "\/"(Y,L) <= c by A4,A5,Def9;
          z in {x} by TARSKI:def 1;
          then z in Y \/ {x} by XBOOLE_0:def 3;
          then z <= c by A12;
          then
A15:      c >= a by A7,A14,Th18;
          a >= c by A9,A8,A13,Th10;
          hence thesis by A15,ORDERS_2:2;
        end;
      end;
    end;
A16: P[{}];
A17: X is finite;
    P[X] from FINSET_1:sch 2(A17,A16,A2);
    hence ex_sup_of X,L;
  end;
  assume for X being finite non empty Subset of L holds ex_sup_of X,L;
  then for a,b being Element of L holds ex_sup_of {a,b},L;
  hence thesis by Th20;
end;
