
theorem Th34:  :: Proposition 4 2H 4H 8LH
  for A being non empty finite set,
      U being Function of bool A, bool A st
    U.{} = {} &
    (for X being Subset of A holds (U.(X`))` c= U.X) &
    (for X, Y being Subset of A holds U.(X \/ Y) = U.X \/ U.Y) holds
  ex R being non empty serial RelStr st
  the carrier of R = A & U = UAp R
  proof
    let A be non empty finite set;
    let U be Function of bool A,bool A;
    assume that
A1: U.{} = {} and
A2: for X being Subset of A holds (U.(X`))` c= U.X and
A3: for X, Y being Subset of A holds U.(X \/ Y) = U.X \/ U.Y;
    consider R being non empty finite RelStr such that
A4: the carrier of R = A & U = UAp R by Th29,A1,A3;
    for x being 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
A5:   x in the carrier of R;
      reconsider Z = [#]A as Subset of R by A4;
      set XX = [#]A;
      (U.(XX`))` c= U.XX by A2; then
      ({}A)` c= U.XX by A1; then
A6:   (Flip UAp R).{} = {} by Th19,A4,XBOOLE_0:def 10;
A7:   (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 A7,A6,Th27;
      end; then
      Class (the InternalRel of R, x) <> {} by A5; then
      consider t being object such that
A8:   t in Im(the InternalRel of R,x) by XBOOLE_0:def 1;
A9:   [x,t] in the InternalRel of R by A8,RELAT_1:169; then
      t in rng the InternalRel of R by RELSET_1:25,A5;
      hence thesis by A9;
    end; then
    R is serial by Def1;
    hence thesis by A4;
  end;
