\newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
\newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
-\newcommand{\patchof}[1]{{\mathcal P} ( #1 ) }
-\newcommand{\baseof}[1]{{\mathcal B} ( #1 ) }
+\newcommand{\merge}[4]{{\mathcal M}(#1,#2,#3,#4)}
+%\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
+
+\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.}} }
the Topbloke patch itself, we hope that git's merge algorithm will
DTRT or that the user will no longer care about the Topbloke patch.
+\item[ $\displaystyle \merge{C}{L}{M}{R} $ ]
+The contents of a git merge result:
+
+$\displaystyle D \isin C \equiv
+ \begin{cases}
+ (D \isin L \land D \isin R) \lor D = C : & \true \\
+ (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
+ \text{otherwise} : & D \not\isin M
+ \end{cases}
+$
+
\end{basedescript}
\newpage
\section{Invariants}
\[ \eqn{Calculation Of Ends:}{
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
- \Bigl\{ E \Big|
+ \left\{ E \Big|
\Bigl[ \Largeexists_{A \in \set A}
E \in \pendsof{A}{\set P} \Bigr] \land
\Bigl[ \Largenexists_{B \in \set A}
E \neq B \land E \le B \Bigr]
- \Bigr\}
+ \right\}
}\]
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:
\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
\gathnext
\patchof{C} = \patchof{L}
\gathnext
- D \isin C \equiv
- \begin{cases}
- R \in \py : & \baseof{R} \ge \baseof{L}
- \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
- R \in \pn : & R \ge \baseof{L}
- \land M = \baseof{L} \\
- \text{otherwise} : & \false
- \end{cases}
+ \merge{C}{L}{R^+}{R^-}
\end{gather}
-xxx want to prove $D \isin C \equiv D \not\in pry \land D \isin L$.
+\subsection{Conditions}
+
+\[ \eqn{ Unique Tip }{
+ \pendsof{L}{\pry} = \{ R^+ \}
+}\]
+\[ \eqn{ Currently Included }{
+ L \haspatch \pry
+}\]
+
+\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 tbd
\section{Merge}
\gathnext
\patchof{C} = \patchof{L}
\gathnext
- D \isin C \equiv
- \begin{cases}
- (D \isin L \land D \isin R) \lor D = C : & \true \\
- (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\
- \text{otherwise} : & D \not\isin M
- \end{cases}
+ \merge{C}{L}{M}{R}
\end{gather}
\subsection{Conditions}
\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.
-
-$\qed$
+See No Replay for Merge Results.
\subsection{Unique Base}