$A \le C$ so $D \le C$.
}
+\[ \eqn{Simple Foreign Inclusion:}{
+ \left[
+ C \hasparents \{ L \}
+ \land
+ \bigforall_{D} D \isin C \equiv D \isin L \lor D = C
+ \right]
+ \implies
+ \bigforall_{D \text{ s.t. } \patchof{D} = \bot}
+ D \isin C \equiv D \le C
+}\]
+\proof{
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+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 $D \le L \equiv D \le C$.
+So $D \isin C \equiv D \le C$.
+}
+
\[ \eqn{Totally Foreign Contents:}{
\bigforall_{C \hasparents \set A}
\left[
\subsection{Foreign Inclusion}
-Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq B$
-so $D \isin B \equiv D \isin L$.
-By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
-And by Exact Ancestors $D \le L \equiv D \le B$.
-So $D \isin B \equiv D \le B$. $\qed$
+Simple Foreign Inclusion applies. $\qed$
\subsection{Foreign Contents}