All value spaces are partially ordered.

Dave is right.  The null relation, null(a, b) = false for all values of a 
and b:



For no a in the value space null(a,a).



For all a and b in the value space null(a, b) implies not(null(b,a)).



For all a, b, and c in the value space null(a, b) and null(b, a) implies 
null(a, c).



(Note that the second point is just f -> ~f == ~f \/ ~f == t, and the third is

(f /\ f) -> f == ~(f /\ f) \/ f == t \/ f == t)



Matthew