let A be non empty finite set ; :: thesis: for 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 )
let L be Function of (bool A),(bool A); :: thesis: ( 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) ) implies ex R being non empty finite reflexive RelStr st
( the carrier of R = A & L = LAp R ) )
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) ; :: thesis: ex R being non empty finite reflexive RelStr st
( the carrier of R = A & L = LAp R )
set U = Flip L;
A4: (Flip L) . {} = {} by A1, Th19;
A5: for X being Subset of A holds X c= (Flip L) . X

proof

let X be Subset of A; :: thesis: X c= (Flip L) . X
(X `) ` c= (L . (X `)) ` by A2, SUBSET_1:12;
hence X c= (Flip L) . X by Def14; :: thesis: verum

end;

for X, Y being Subset of A holds (Flip L) . (X \/ Y) = ((Flip L) . X) \/ ((Flip L) . Y) by A3, Th22;
then consider R being non empty finite reflexive RelStr such that
A6: ( the carrier of R = A & Flip L = UAp R ) by Th37, A4, A5;
L = Flip (UAp R) by Th23, A6;
then L = LAp R by Th27;
hence ex R being non empty finite reflexive RelStr st
( the carrier of R = A & L = LAp R ) by A6; :: thesis: verum