\subsubsection{For $\p = \pq$:}
By Base Acyclic, $D \not\isin B$. So $D \isin C \equiv D = C$.
-By No Sneak, $D \le B \equiv D = C$. Thus $C \haspatch \pq$.
+By No Sneak, $D \not\le B$ so $D \le C \equiv D = C$. Thus $C \haspatch \pq$.
\subsubsection{For $\p \neq \pq$:}
Not applicable.
-\section{Anticommit}
+\section{Dependency Removal}
-Given $L$ and $\pr$ as represented by $R^+, R^-$.
-Construct $C$ which has $\pr$ removed.
+Given $L$ which contains $\pr$ as represented by $R^+, R^-$.
+Construct $C$ which has $\pr$ removed by applying a single
+commit which is the anticommit of $\pr$.
Used for removing a branch dependency.
\gathbegin
C \hasparents \{ L \}
R^+ \in \pry \land R^- = \baseof{R^+}
}\]
\[ \eqn{ Into Base }{
- L \in \pn
+ L \in \pqn
}\]
\[ \eqn{ Unique Tip }{
\pendsof{L}{\pry} = \{ R^+ \}
\subsection{No Replay}
-By definition of $\merge$,
+By $\merge$,
$D \isin C \implies D \isin L \lor D \isin R^- \lor D = C$.
So, by Ordering of Ingredients,
Ingredients Prevent Replay applies. $\qed$
\subsubsection{For $D \neq C, D \not\le L$:}
-By No Replay $D \not\isin L$. Also $D \not\le R^-$ hence
+By No Replay for $L$, $D \not\isin L$.
+Also, by Ordering of Ingredients, $D \not\le R^-$ hence
$D \not\isin R^-$. Thus $D \not\isin C$. OK.
\subsubsection{For $D \neq C, D \le L, D \in \pry$:}
by Unique Tip, $D \le R^+ \equiv D \le L$.
So $D \isin R^+$.
-By Base Acyclic, $D \not\isin R^-$.
+By Base Acyclic for $R^-$, $D \not\isin R^-$.
Apply $\merge$: $D \not\isin C$. OK.
\subsection{Unique Base}
-Into Base means that $C \in \pn$, so Unique Base is not
+Into Base means that $C \in \pqn$, so Unique Base is not
applicable. $\qed$
\subsection{Tip Contents}
\subsection{Base Acyclic}
-By Base Acyclic for $L$, $D \isin L \implies D \not\in \py$.
-And by Into Base $C \not\in \py$.
+By Into Base and Base Acyclic for $L$, $D \isin L \implies D \not\in \pqy$.
+And by Into Base $C \not\in \pqy$.
Now from Desired Contents, above, $D \isin C
\implies D \isin L \lor D = C$, which thus
-$\implies D \not\in \py$. $\qed$.
+$\implies D \not\in \pqy$. $\qed$.
\subsection{Coherence and Patch Inclusion}
Not applicable.
+\section{Dependency Insertion}
+
+Given $L$ construct $C$ which additionally
+contains $\pr$ as represented by $R^+$ and $R^-$.
+This may even be used for reintroducing a previous-removed branch
+dependency.
+\gathbegin
+ C \hasparents \{ L, R^+ \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ \mergeof{C}{L}{R^-}{R^+}
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ R^- = \baseof{R^+}
+}\]
+\[ \eqn{ Into Base }{
+ L \in \pqn
+}\]
+\[ \eqn{ Currently Excluded }{
+ L \nothaspatch \pr
+}\]
+\[ \eqn{ Inserted's Ends }{
+ E \in \pendsof{L}{\pry} \implies E \le R^+
+}\]
+\[ \eqn{ Others' Ends }{
+ \bigforall_{\p \patchisin \L}
+ E \in \pendsof{R^+}{\py} \implies E \le L
+}\]
+\[ \eqn{ Insertion Acyclic }{
+ R^+ \nothaspatch \pq
+}\]
+
+\subsection{No Replay}
+
+By $\merge$,
+$D \isin C \implies D \isin L \lor D \isin R^+ \lor D = C$.
+So Ingredients Prevent Replay applies. $\qed$
+
+\subsection{Unique Base}
+
+Not applicable.
+
+\subsection{Tip Contents}
+
+Not applicable.
+
+\subsection{Base Acyclic}
+
+Consider some $D \isin C$. We will show that $D \not\in \pqy$.
+By $\merge$, $D \isin L \lor D \isin R^+ \lor D = C$.
+
+For $D \isin L$, Base Acyclic for L suffices. For $D \isin R^+$,
+Insertion Acyclic suffices. For $D = C$, trivial. $\qed$.
+
+\subsection{Coherence and Patch Inclusion}
+
+$$
+\begin{cases}
+ \p = \pr \lor L \haspatch \p : & C \haspatch \p \\
+ \p \neq \pr \land L \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+
+\proofstarts
+~ Consider some $D \in \py$.
+$D \neq C$ so $D \le C \equiv D \le L \lor D \le R^+$.
+
+\subsubsection{For $\p = \pr$:}
+
+$D \not\isin L$ by Currently Excluded.
+$D \not\isin R^-$ by Base Acyclic.
+So by $\merge$, $D \isin C \equiv D \isin R^+$,
+which by Tip Self Inpatch of $R^+$, $\equiv D \le R^+$.
+
+And by definition of $\pancs$,
+$D \le L \equiv D \in \pancsof{L}{R^+}$.
+Applying Transitive Ancestors to Inserted's Ends gives
+$A \in \pancsof{L}{R^+} \implies A \le R^+$.
+So $D \le L \implies D \le R^+$.
+Thus $D \le C \equiv D \le R^+$.
+
+So $D \isin C \equiv D \le C$, i.e. $C \haspatch \pr$.
+OK.
+
+\subsubsection{For $\p \neq \pr$:}
+
+By Exclusive Tip Contents for $R^+$ ($D \not\in \pry$ case)
+$D \isin R^+ \equiv D \isin R^-$.
+So by $\merge$, $D \isin C \equiv D \isin L$.
+
+If $L \nothaspatch \p$, $D \not\isin L$ so $C \nothaspatch \p$. OK.
+
+If $L \haspatch \p$, Others' Ends applies; by Transitive
+Ancestors, $A \in \pancsof{R^+}{\py} \implies A \le L$.
+So $D \le R^+ \implies D \le L$,
+since $D \le R^+ \equiv D \in \pancsof{R^+}{\py}$.
+Thus $D \le C \equiv D \le L$.
+And by $\haspatch$, $D \le L \equiv D \isin L$ so
+$D \isin C \equiv D \le C$. Thus $C \haspatch \p$.
+OK.
+
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+
+By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$.
+So by $\merge$, $D \isin C \equiv D \isin L$.
+
+xxx up to here, need new condition
+
+$D \neq C$.
+
+
+
\section{Merge}
Merge commits $L$ and $R$ using merge base $M$: