 However, it does not necessarily satisfy Preservation. To see this suppose that \({\bf Q}=\{H_1, H_2, H_3\}\) and \(H_1 \prec H_2\) but \(H_3\) is not ordered with \(H_1\) or \(H_2\). Then \(\cap \Bel = H_1\cup H_3.\) However \(\cap\Bel_{\neg H_1} = H_2 \cup H_3\nsubseteq \cap \Bel\) even though \(\cap \Bel \cap \neg H_1 \neq \varnothing.\)