 Just as in default logic, some sets of assumptions have no stable extensions while some have multiple stable extensions. We demonstrate the former case with \(\Xi_2 = \{\neg \mathsf{B} \phi \supset \phi\}\). Suppose there is a stable extension \(\Gamma\) of \(\Xi_2\). First note that there is no way to ground \(\phi\) in view of \(\Xi_2\). Hence \(\phi \notin \Gamma\) which means that \(\neg \mathsf{B} \phi \in \Gamma\). But then \(\phi \in \Gamma\) by modus ponens which is a contradiction. 