\[ \eqn{ Into Base }{
L \in \pln
}\]
-\[ \eqn{ Unique Tip }{
+\[ \eqn{ Correct Tip }{
\pendsof{L}{\pry} = \{ R^+ \}
}\]
\[ \eqn{ Currently Included }{
\subsection{Ordering of Ingredients:}
-By Unique Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$
+By Correct Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$
so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$.
$\qed$
By Currently Included, $D \isin L$.
-By Tip Own Contents for $R^+$, $D \isin R^+ \equiv D \le R^+$, but by
-by Unique Tip, $D \le R^+ \equiv D \le L$.
+By Tip Own Contents for $R^+$, $D \isin R^+ \equiv D \le R^+$, but
+by Correct Tip, $D \le R^+ \equiv D \le L$.
So $D \isin R^+$.
By Base Acyclic for $R^-$, $D \not\isin R^-$.
$\qed$
+\subsection{Unique Tips:}
+
+Single Parent Unique Tips applies. $\qed$
+
\subsection{Foreign Inclusion}
-Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq C$.
+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 $L$, $D \isin L \equiv D \le L$.