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)))))))