\renewcommand{\implies}{\Rightarrow}
\renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
\renewcommand{\land}{\wedge}
\renewcommand{\lor}{\vee}
\end{cases}
$
+Some (overlapping) alternative formulations:
+
+$\displaystyle D \isin C \equiv
+ \begin{cases}
+ D \isin L \equiv D \isin R : & D = C \lor D \isin L \\
+ D \isin L \equiv D \isin R : & D = C \lor D \isin R \\
+ D \isin L \nequiv D \isin R : & D = C \lor D \not\isin M \\
+ D \isin M \equiv D \isin L : & D = C \lor D \isin R \\
+ D \isin M \equiv D \isin R : & D = C \lor D \isin L \\
+ \end{cases}
+$
+
\end{basedescript}
\newpage
\section{Invariants}
\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} \\
+ R \in \pn : & M = \baseof{L} \\
\text{otherwise} : & \false
\end{cases}
}\]
\subsubsection{For $R \in \pn$:}
-By Tip Merge condition on $R$,
+By Tip Merge condition on $R$ and since $M \le R$,
$A \le \baseof{L} \implies A \le R$, so
$A \le R \lor A \le \baseof{L} \equiv A \le R$.
Thus $A \le C \equiv A \le R$.
$\qed$
-\subsection{Coherence and patch inclusion}
+\subsection{Coherence and Patch Inclusion}
Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
This involves considering $D \in \py$.
We need worry only about $C \in \py$.
And $\patchof{C} = \patchof{L}$
-so $L \in \py$ so $L \haspatch \p$. We will use the coherence and
-patch inclusion of $C$ as just proved.
+so $L \in \py$ so $L \haspatch \p$. We will use the Unique Base
+of $C$, and its Coherence and Patch Inclusion, as just proved.
-Firstly we prove $C \haspatch \p$: If $R \in \py$, then $R \haspatch
-\p$ and by coherence/inclusion $C \haspatch \p$ . If $R \not\in \py$
+Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+\p$ and by Coherence/Inclusion $C \haspatch \p$ . If $R \not\in \py$
then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
-of $\nothaspatch$, $M \nothaspatch \p$. So by coherence/inclusion $C
+of $\nothaspatch$, $M \nothaspatch \p$. So by Coherence/Inclusion $C
\haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
-We will consider some $D$ and prove the Exclusive Tip Contents form.
+We will consider an arbitrary commit $D$
+and prove the Exclusive Tip Contents form.
\subsubsection{For $D \in \py$:}
$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
\le C$. OK.
-xxx up to here
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
-\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+$D \neq C$. By Tip Contents of $L$,
+$D \isin L \equiv D \isin \baseof{L}$, and by Tip Merge condition,
+$D \isin L \equiv D \isin M$. So by definition of $\merge$, $D \isin
+C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
+Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
+\subsubsection{For $D \not\in \py, R \in \py$:}
+
+xxx up to here
%D \in \py$:}