\subsection{No Replay for Merge Results}
-If we are constructing $C$ such that $\merge{C}{L}{M}{R}$, No Replay
-is preserved. {\it Proof:}
+If we are constructing $C$, given
+\gathbegin
+ \merge{C}{L}{M}{R}
+\gathnext
+ L \le C
+\gathnext
+ R \le C
+\end{gather}
+No Replay is preserved. {\it Proof:}
\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK.
\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
$D \not\isin C$. OK.
-\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
-$D \not\isin C$. OK.
-
\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
\land D \not\isin M$:}
$D \isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
\land D \isin M$:}
-$D \not\isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
-R$ so $D \le C$. OK.
+$D \not\isin C$. OK.
$\qed$
\section{Anticommit}
Given $L, R^+, R^-$ where
-$\patchof{R^+} = \pry, \patchof{R^-} = \prn$.
+$R^+ \in \pry, R^- = \baseof{R^+}$.
Construct $C$ which has $\pr$ removed.
Used for removing a branch dependency.
\gathbegin
\[ \eqn{ Unique Tip }{
\pendsof{L}{\pry} = \{ R^+ \}
}\]
-\[ \eqn{ Correct Base }{
- \baseof{R^+} = R^-
-}\]
\[ \eqn{ Currently Included }{
L \haspatch \pry
}\]
+\subsection{Desired Contents}
+
+xxx need to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+
+\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{Unique Base}
+Need to consider only $C \in \py$, ie $L \in \py$.
-xxx want to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+xxx tbd
\section{Merge}
\end{cases}
}\]
-\subsection{Merge Results}
+\subsection{No Replay}
-As above.
+See No Replay for Merge Results.
\subsection{Unique Base}