strict subset of $O[I]$ does not satisfy some rule of the form $x \vee \overline{x} \leftarrow$, and thus it is not a model of $P^{M[I,J]}$. Similarly, any interpretation $W$ such that $O[I] \subset W \subset M[I,J]$ does not satisfy some rule in $P^{M[I,J]}$ (note that rules of the form $u \leftarrow z$ and $u \leftarrow \overline{z}$ occur in $P^{M[I,J]}$ because $w \not\in M[I,J]$; such rules are obtained by rules in (\[eq:red:third\])).

Let us now consider $P^{M'[I,J]}$ and let $W \subseteq M'[I,J]$ be one of its models. We shall show that either $W = M'[I,J]$, or $W = N[I,J,K]$ for some $K \subseteq Z$ such that $I \cup J \cup K \not\models \phi$. Note that $v$ is a fact in $P^{M'[I,J]}$, hence $v$ must belong to $W$. By (\[eq:red:first\]) and (\[eq:red:second\]), since $v \in W$ and $W \subseteq M'[I,J]$, we can conclude that all of the atoms in $I\cup \overline{(X\setminus I)}\cup J\cup \overline{(Y\setminus J)}$ belong to $W$. Consider now the atom $w$. If $w$ belongs to $W$, by the rules in (\[eq:red:third\]) we conclude that all of the atoms in $Z \cup \overline{Z}$ belong to $W$, and thus $W = M'[I,J]$. Otherwise, if $w \not\in W$, by the rules of the 