{\hbox{\scriptsize$\forall$}}}%
}
+\newcommand{\Largeexists}{\mathop{\hbox{\Large$\exists$}}}
+\newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}
\newcommand{\qed}{\square}
\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
\newcommand{\gathbegin}{\begin{gather} \tag*{}}
\newcommand{\gathnext}{\\ \tag*{}}
+\newcommand{\true}{t}
+\newcommand{\false}{f}
+
\begin{document}
\section{Notation}
$ D \le C $.
Contents of $C$:
-$ D \isin C \equiv \ldots \lor t \text{ so } D \haspatch C $.
+$ D \isin C \equiv \ldots \lor \true \text{ so } D \haspatch C $.
\subsubsection{For $A \haspatch P, D \neq C$:}
Ancestors: $ D \le C \equiv D \le A $.
If $D = C$, trivial. For $D \neq C$:
$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$
-\section{Test more symbols}
+\section{Merge}
-$ C \haspatch \p $
+Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
+\gathbegin
+ C \hasparents \{ L, R \}
+\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}
+\end{gather}
-$ C \nothaspatch \p $
+\subsection{Conditions}
+
+\[ \eqn{ Tip Merge }{
+ L \in \py \implies
+ \begin{cases}
+ R \in \py : & \baseof{R} \ge \baseof{L}
+ \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
+ R \in \pn : & R \ge \baseof{L}
+ \land M = \baseof{L} \\
+ \text{otherwise} : & \false
+ \end{cases}
+}\]
-$ \p \patchisin C $
+\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.
-$ \p \notpatchisin C $
+\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
+$D \not\isin C$. OK.
-$ \{ B \} \areparents C $
+\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.
+
+$\qed$
\end{document}