+\[ \eqn{Ingredients Prevent Replay:}{
+ \left[
+ {C \hasparents \set A} \land
+ \\
+ \left(
+ D \isin C \implies
+ D = C \lor
+ \Largeexists_{A \in \set A} D \isin A
+ \right)
+ \right] \implies \left[
+ D \isin C \implies D \le C
+ \right]
+}\]
+\proof{
+ Trivial for $D = C$. Consider some $D \neq C$, $D \isin C$.
+ By the preconditions, there is some $A$ s.t. $D \in \set A$
+ and $D \isin A$. By No Replay for $A$, $D \le A$. And
+ $A \le C$ 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 $D \isin C \equiv D \le C$.
+}
+