R^+ \in \pry \land R^- = \baseof{R^+}
}\]
\[ \eqn{ Into Base }{
- L \in \pqn
+ L \in \pln
}\]
-\[ \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 Self Inpatch 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 $R^-$, $D \not\isin R^-$.
\subsection{Unique Base}
-Into Base means that $C \in \pqn$, so Unique Base is not
+Into Base means that $C \in \pln$, so Unique Base is not
applicable.
\subsection{Tip Contents}
\subsection{Base Acyclic}
-By Into Base and Base Acyclic for $L$, $D \isin L \implies D \not\in \pqy$.
-And by Into Base $C \not\in \pqy$.
+By Into Base and Base Acyclic for $L$, $D \isin L \implies D \not\in \ply$.
+And by Into Base $C \not\in \ply$.
Now from Desired Contents, above, $D \isin C
\implies D \isin L \lor D = C$, which thus
-$\implies D \not\in \pqy$. $\qed$.
+$\implies D \not\in \ply$. $\qed$.
\subsection{Coherence and Patch Inclusion}
-Need to consider some $D \in \py$. By Into Base, $D \neq C$.
+$$
+\begin{cases}
+ \p = \pr : & C \nothaspatch \p \\
+ \p \neq \pr \land L \nothaspatch \p : & C \nothaspatch \p \\
+ \p \neq \pr \land L \haspatch \p : & C \haspatch \p
+\end{cases}
+$$
+\proofstarts
+~ Need to consider some $D \in \py$. By Into Base, $D \neq C$.
\subsubsection{For $\p = \pr$:}
By Desired Contents, above, $D \not\isin C$.
-So $C \nothaspatch \pr$.
+OK.
\subsubsection{For $\p \neq \pr$:}
By Desired Contents, $D \isin C \equiv D \isin L$
(since $D \in \py$ so $D \not\in \pry$).
If $L \nothaspatch \p$, $D \not\isin L$ so $D \not\isin C$.
-So $L \nothaspatch \p \implies C \nothaspatch \p$.
+OK.
-Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
-so $L \haspatch \p \implies C \haspatch \p$.
+Whereas, if $L \haspatch \p$, $D \isin L \equiv D \le L$,
+so $C \zhaspatch \p$;
+and $\exists_{F \in \py} F \le L$ and this $F \le C$.
+Thus $C \haspatch \p$.
+OK.
$\qed$
+\subsection{Unique Tips:}
+
+Single Parent Unique Tips applies. $\qed$
+
\subsection{Foreign Inclusion}
Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq C$.