for each $I\neq{{\mathcal I}}$, $M[I,J]$ and $M'[I,J]$ cannot be models of $P\cup F_{{\mathcal I}}$ because $F_{{\mathcal I}}\not\subseteq M[I,J]$, resp. $F_{{\mathcal I}}\not\subseteq M'[I,J]$. We now analyze these three types of potential candidates:

-   ${{\mathcal M}}=U$: Then, for instance, $M[{{\mathcal I}},J]\subset U$ is a model of $(P\cup F_{{\mathcal I}})^{{\mathcal M}}=P^{{\mathcal M}}\cup F_{{\mathcal I}}$ for any $J \subseteq Y$. Thus, ${{\mathcal M}}\notin{{\mathit AS}}(P\cup F_{{\mathcal I}})$.

-   ${{\mathcal M}}=M[{{\mathcal I}},J]$ for some $J\subseteq Y$. Then, by the properties of $\Phi$-reductions, $O[{{\mathcal I}}]\subset {{\mathcal M}}$ is a model of $(P\cup F_{{\mathcal I}})^{{\mathcal M}}=P^{{\mathcal M}}\cup F_{{\mathcal I}}$. Thus, ${{\mathcal M}}\notin{{\mathit AS}}(P\cup F_{{\mathcal I}})$.

-   ${{\mathcal M}}=M'[{{\mathcal I}},J]$ for some $J\subseteq Y$. By the initial assumption, there exists a $K_Y\subseteq Z$ with ${{\mathcal I}}\cup J \cup K_Y\not\models \phi$. Then, by the properties of $\Phi$-reductions, $N[{{\mathcal I}},J,K]\subset {{\mathcal M}}$ is a model of $P^{{\mathcal M}}$ and thus also of $(P\cup F_{{\mathcal I}})^{{\mathcal M}}$. Hence, ${{\mathcal M}}\notin{{\mathit AS}}(P\cup F_{{\mathcal I}})$.

In each of the cases we have obtained ${{\mathcal M}}\notin{{\mathit AS}}(P\cup F_{{\mathcal I}})$, hence ${{\mathit AS}}(P\cup F_{{\mathcal I}})=\emptyset$ and $P$ is not super-coherent.

Suppose that $\Phi$ is true. It is sufficient to show that for each $F\subseteq U$, ${{\mathit AS}}(P\cup F)\neq \emptyset$. We have the following cases:

If $\{s,\overline{s}\}\subseteq 