There is a bounded set containing a set, a set that properly contains that set, a third set that properly contains the second set, and so forth. In short, there is at least one bounded set of infinite cardinality.Notes:
- Source: KIF Version 3.0 Specification
(Exists (?U) (And (Bounded ?U) (Not (Empty ?U)) (Forall (?X) (=> (Member ?X ?U) (Exists (?Y) (And (Member ?Y ?U) (Proper-Subset ?X ?Y)))))))