\newcommand{\merge}[4]{{\mathcal M}(#1,#2,#3,#4)}
%\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
-\newcommand{\patchof}[1]{{\mathcal P} ( #1 ) }
-\newcommand{\baseof}[1]{{\mathcal B} ( #1 ) }
+\newcommand{\patch}{{\mathcal P}}
+\newcommand{\base}{{\mathcal B}}
+
+\newcommand{\patchof}[1]{\patch ( #1 ) }
+\newcommand{\baseof}[1]{\base ( #1 ) }
\newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} }
\newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} }
}\]
XXX proof TBD.
+\subsection{No Replay for Merge Results}
+
+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 $D \isin L \land D \isin R$:}
+$D \isin C$. And $D \isin L \implies D \le L \implies D \le C$. OK.
+
+\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
+$D \not\isin C$. OK.
+
+\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
+ \land D \not\isin M$:}
+$D \isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
+R$ so $D \le C$. OK.
+
+\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
+ \land D \isin M$:}
+$D \not\isin C$. OK.
+
+$\qed$
+
\section{Commit annotation}
We annotate each Topbloke commit $C$ with:
\end{cases}
}\]
-\subsection{No Replay}
-
-\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK.
-
-\subsubsection{For $D \isin L \land D \isin R$:}
-$D \isin C$. And $D \isin L \implies D \le L \implies D \le C$. 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 $D \le L \lor D \le
-R$ so $D \le C$. OK.
-
-\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.
+\subsection{Merge Results}
-$\qed$
+As above.
\subsection{Unique Base}