
theorem Th6:
  for X being non empty set st for a,b being set st a in X & b in X
  ex c being set st a \/ b c= c & c in X holds X is c=directed
proof
  let X be non empty set such that
A1: for a,b being set st a in X & b in X ex c being set st a \/ b c= c &
  c in X;
  set a = the Element of X;
  defpred P[set] means ex a being set st union $1 c= a & a in X;
  let Y be finite Subset of X;
A2: now
    let x,B be set;
    assume that
A3: x in Y and
    B c= Y;
    assume P[B];
    then consider a being set such that
A4: union B c= a and
A5: a in X;
    consider c being set such that
A6: a \/ x c= c & c in X by A1,A3,A5;
    thus P[B \/ {x}]
    proof
      take c;
      union (B \/ {x}) = (union B) \/ union {x} by ZFMISC_1:78
        .= union B \/ x by ZFMISC_1:25;
      then union (B \/ {x}) c= a \/ x by A4,XBOOLE_1:9;
      hence thesis by A6;
    end;
  end;
  union {} c= a by ZFMISC_1:2;
  then
A7: P[ {}];
A8: Y is finite;
  thus P[Y] from FINSET_1:sch 2(A8,A7,A2);
end;
