+\section{Anticommit}
+
+Given $L, R^+, R^-$ where
+$R^+ \in \pry, R^- = \baseof{R^+}$.
+Construct $C$ which has $\pr$ removed.
+Used for removing a branch dependency.
+\gathbegin
+ C \hasparents \{ L \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ \merge{C}{L}{R^+}{R^-}
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Unique Tip }{
+ \pendsof{L}{\pry} = \{ R^+ \}
+}\]
+\[ \eqn{ Currently Included }{
+ L \haspatch \pry
+}\]
+\[ \eqn{ Not Self }{
+ L \not\in \{ R^+ \}
+}\]
+
+\subsection{No Replay}
+
+By Unique Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$
+so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$ and No Replay for
+Merge Results applies. $\qed$
+
+\subsection{Desired Contents}
+
+\[ $D \isin C \equiv [ D \not\in \pry \land D \isin L$ ] \lor D = C \]
+{\it Proof.}
+
+\subsubsection{For $D = C$:}
+
+Trivially $D \isin C$. OK.
+
+\subsubsection{For $D \not\le C$:}
+
+
+
+\subsubsection{For $D \in R^+$:}
+By Currently Included,
+
+\subsection{Unique Base}
+
+Need to consider only $C \in \py$, ie $L \in \py$.
+
+xxx tbd
+