\subsection{Foreign Inclusion}
-Consider some $D$ s.t. $\isforeign{D}$. $D \neq C$.
+Consider some $D \in \foreign$. $D \neq C$.
So by Desired Contents $D \isin C \equiv D \isin L$.
By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
\bigforall_{C,\p} C \haspatch \p \implies \pendsof{C}{\py} = \{ T \}
}\]
\[\eqn{Foreign Inclusion}{
- \bigforall_{D \text{ s.t. } \isforeign{D}} D \isin C \equiv D \leq C
+ \bigforall_{D \in \foreign} D \isin C \equiv D \leq C
}\]
\[\eqn{Foreign Contents}{
- \bigforall_{C \text{ s.t. } \isforeign{C}}
+ \bigforall_{C \in \foreign}
D \le C \implies \isforeign{D}
}\]
\right]
\implies
\left[
- \bigforall_{D \text{ s.t. } \isforeign{D}}
+ \bigforall_{D \in \foreign}
D \isin C \equiv D \le C
\right]
$$
\proof{
-Consider some $D$ s.t. $\isforeign{D}$.
+Consider some $D \in \foreign$.
If $D = C$, trivially true. For $D \neq C$,
by Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
And by Exact Ancestors $D \le L \equiv D \le C$.
\subsection{Foreign Inclusion}
-Consider some $D$ s.t. $\isforeign{D}$.
+Consider some $D \in \foreign$.
By Foreign Inclusion of $L, M, R$:
$D \isin L \equiv D \le L$;
$D \isin M \equiv D \le M$;
}\]
\[ \eqn{ Foreign Unaffected }{
- \bigforall_{ D \text{ s.t. } \isforeign{D} }
+ \bigforall_{ D \in \foreign }
\left[ \bigexists_{A \in \set A} D \le A \right]
\implies
D \le L