 If \(\Xi\) were an extension, then the justification \(\theta\) of the default above would either be consistent with \(\Xi\) or not, and either case is impossible. For if \(\theta\) were consistent with \(\Xi\), then the default would be applied to derive \(\neg\theta\) in contradiction the the consistency of \(\Xi\) with \(\theta\). Similarly, if \(\Xi\) were inconsistent with \(\theta\) then \(\Xi \vDash \neg \theta\) and hence, by deductive closure, \(\neg\theta \in \Theta\). Our default would not be applicable, in which case \(\Xi = \Xi_1 = \mathrm{Cn}(\emptyset)\). But then \(\neg\theta \notin \Xi\),—a contradiction. 