F$ for some $s\in X\cup Y$ or $\{u,w\}\subseteq F$. Then $U\in{{\mathit AS}}(P\cup F)$ since $U$ is a model of $P\cup F$ and each potential model $M\subset U$ of the reduct $P^U$ (see the properties of $\Phi$-reductions) does not satisfy $F\subseteq M$; thus each such $M$ is not a model of $P^U\cup F=(P\cup F)^U$.

Otherwise, we have $F\subseteq M[I,J]$ or $F\subseteq M'[I,J]$ for some $I\subseteq X$, $J\subseteq Y$. In case $F\subseteq M[I,J]$ and $F\not\subseteq O[I]$, we observe that $M[I,J]\in{{\mathit AS}}(P\cup F)$ since $O[I]$ is the only model (being a proper subset of $M[I,J]$) of the reduct $P^{M[I,J]}$. Thus for each such $F$ there cannot be a model $M\subset M[I,J]$ of $P^{M[I,J]}\cup F=(P\cup F)^{M[I,J]}$. As well, in case $F\subseteq M'[I,J]$, where $w\in F$, $M'[I,J]$ can be shown to be an answer set of $P\cup F$. Note that the case $F\subseteq M'[I,J]$ with $w\notin F$ is already taken care of since for such $F$ we have $F\subseteq M[I,J]$ as well.

It remains to consider the case $F\subseteq O[I]$ for each $I\subseteq X$. We show that $M'[I,J]$ is an answer set of $P\cup F$, for some $J\subseteq Y$. Since $\Phi$ is true, we know that, for each $I\subseteq X$, there exists a $J_I\subseteq Y$ 