
theorem  :: Proposition 5 1L 4L 3L
  for A being non empty finite set,
      L being Function of bool A, bool A st
    L.A = A &
    (for X being Subset of A holds L.X c= X) &
    (for X,Y being Subset of A holds L.(X /\ Y) = L.X /\ L.Y) holds
  ex R being non empty finite reflexive RelStr st
    the carrier of R = A & L = LAp R
  proof
    let A be non empty finite set;
    let L be Function of bool A, bool A;
    assume that
A1: L.A = A and
A2: for X being Subset of A holds L.X c= X and
A3: for X,Y being Subset of A holds L.(X /\ Y) = L.X /\ L.Y;
    set U = Flip L;
A4: U.{} = {} by A1,Th19;
A5: for X being Subset of A holds X c= U.X
    proof
      let X be Subset of A;
      X`` c= (L.X`)` by A2,SUBSET_1:12;
      hence thesis by Def14;
    end;
    for X,Y being Subset of A holds U.(X \/ Y) = U.X \/ U.Y
      by A3,Th22; then
    consider R being non empty finite reflexive RelStr such that
A6: the carrier of R = A & U = UAp R by Th37,A4,A5;
    L = Flip UAp R by Th23,A6; then
    L = LAp R by Th27;
    hence thesis by A6;
  end;
