
theorem Th33:  :: Proposition 3 1L 4L 8LH
  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= (L.(X`))`) &
    (for X, Y being Subset of A holds L.(X /\ Y) = L.X /\ L.Y) holds
  ex R being non empty finite serial 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= (L.(X`))` and
A3: for X, Y being Subset of A holds L.(X /\ Y) = L.X /\ L.Y;
    consider R being non empty finite RelStr such that
A4: the carrier of R = A & L = LAp R by Th30,A1,A3;
    set XX = {}A;
A5: (L.(XX`)) = A by A1; then
A6: L.{} c= ([#]A)` by A2;
    for x be object st x in the carrier of R
    ex y being object st
      y in the carrier of R & [x,y] in the InternalRel of R
    proof
      let x be object;
      assume
A7:   x in the carrier of R;
      reconsider Z = [#]A as Subset of R by A4;
A8:   (LAp R).{} = LAp {}R by Def10
                .= { y where y is Element of R :
         Class (the InternalRel of R, y) c= {} };
      for y being Element of R holds Class (the InternalRel of R, y) <> {}
      proof
        let y be Element of R;
        assume Class (the InternalRel of R, y) = {}; then
        y in {z where z is Element of R :
          Class (the InternalRel of R, z) c= {}};
        hence contradiction by A8,A4,A6,A5;
      end; then
      Class (the InternalRel of R, x) <> {} by A7; then
      consider t be object such that
A9:   t in Im(the InternalRel of R,x) by XBOOLE_0:def 1;
A10:   [x,t] in the InternalRel of R by A9,RELAT_1:169; then
      t in rng the InternalRel of R by RELSET_1:25,A7;
      hence thesis by A10;
    end; then
    R is serial by Def1;
    hence thesis by A4;
  end;
