{\hbox{\scriptsize$\forall$}}}%
}
-\newcommand{\proof}[1]{{\it Proof.} #1 $\square$}
+
+\newcommand{\qed}{\square}
+\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
+
+\newcommand{\gathbegin}{\begin{gather} \tag*{}}
+\newcommand{\gathnext}{\\ \tag*{}}
\begin{document}
\item[ $ \baseof{C} $ ]
$ \pendsof{C}{\pn} = \{ \baseof{C} \} $ where $ C \in \py $.
A partial function from commits to commits.
-See ``unique base'', below.
+See Unique Base, below.
\item[ $ C \haspatch \p $ ]
$\displaystyle \bigforall_{D \in \py} D \isin C \equiv D \le C $.
\section{Commit annotation}
We annotate each Topbloke commit $C$ with:
-\begin{gather}
-\tag*{} \patchof{C} \\
-\tag*{} \baseof{C}, \text{ if } C \in \py \\
-\tag*{} \bigforall_{\pa{Q}}
- \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q} \\
-\tag*{} \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
+\gathbegin
+ \patchof{C}
+\gathnext
+ \baseof{C}, \text{ if } C \in \py
+\gathnext
+ \bigforall_{\pa{Q}}
+ \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q}
+\gathnext
+ \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
\end{gather}
We do not annotate $\pendsof{C}{\py}$ for $C \in \py$. Doing so would
would have to be updated. The annotation is not needed because
$\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$.
+\section{Simple commit}
+
+A simple single-parent forward commit $C$ as made by git-commit.
+\begin{gather}
+\tag*{} C \hasparents \{ A \} \\
+\tag*{} \patchof{C} = \patchof{A} \\
+\tag*{} D \isin C \equiv D \isin A \lor D = C
+\end{gather}
+
+\subsection{No Replay}
+Trivial.
+
+\subsection{Unique Base}
+If $A, C \in \py$ then $\baseof{C} = \baseof{A}$. $\qed$
+
+\subsection{Tip Contents}
+We need to consider only $A, C \in \py$. From Tip Contents for $A$:
+\[ D \isin A \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A ) \]
+Substitute into the contents of $C$:
+\[ D \isin C \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A )
+ \lor D = C \]
+Since $D = C \implies D \in \py$,
+and substituting in $\baseof{C}$, this gives:
+\[ D \isin C \equiv D \isin \baseof{C} \lor
+ (D \in \py \land D \le A) \lor
+ (D = C \land D \in \py) \]
+\[ \equiv D \isin \baseof{C} \lor
+ [ D \in \py \land ( D \le A \lor D = C ) ] \]
+So by Exact Ancestors:
+\[ D \isin C \equiv D \isin \baseof{C} \lor ( D \in \py \land D \le C
+) \]
+$\qed$
+
+\subsection{Base Acyclic}
+
+Need to consider only $A, C \in \pn$.
+
+For $D = C$: $D \in \pn$ so $D \not\in \py$. OK.
+
+For $D \neq C$: $D \isin C \equiv D \isin A$, so by Base Acyclic for
+$A$, $D \isin C \implies D \not\in \py$. $\qed$
+
+\subsection{Coherence and patch inclusion}
+
+Need to consider $D \in \py$
+
+\subsubsection{For $A \haspatch P, D = C$:}
+
+Ancestors of $C$:
+$ D \le C $.
+
+Contents of $C$:
+$ D \isin C \equiv \ldots \lor t \text{ so } D \haspatch C $.
+
+\subsubsection{For $A \haspatch P, D \neq C$:}
+Ancestors: $ D \le C \equiv D \le A $.
+
+Contents: $ D \isin C \equiv D \isin A \lor f $
+so $ D \isin C \equiv D \isin A $.
+
+So:
+\[ A \haspatch P \implies C \haspatch P \]
+
+\subsubsection{For $A \nothaspatch P$:}
+
+Firstly, $C \not\in \py$ since if it were, $A \in \py$.
+Thus $D \neq C$.
+
+Now by contents of $A$, $D \notin A$, so $D \notin C$.
+
+So:
+\[ A \nothaspatch P \implies C \nothaspatch P \]
+$\qed$
+
+\subsection{Foreign inclusion:}
+
+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}
$ C \haspatch \p $