+\[\eqn{Coherence:}{
+ \bigforall_{C,\p} C \haspatch \p \lor C \nothaspatch \p
+}\]
+\[\eqn{Foreign Inclusion:}{
+ \bigforall_{D \text{ s.t. } \patchof{D} = \bot} D \isin C \equiv D \leq C
+}\]
+
+\section{Some lemmas}
+
+\[ \eqn{Exclusive Tip Contents:}{
+ \bigforall_{C \in \py}
+ \neg \Bigl[ D \isin \baseof{C} \land ( D \in \py \land D \le C )
+ \Bigr]
+}\]
+Ie, the two limbs of the RHS of Tip Contents are mutually exclusive.
+
+\proof{
+Let $B = \baseof{C}$ in $D \isin \baseof{C}$. Now $B \in \pn$.
+So by Base Acyclic $D \isin B \implies D \notin \py$.
+}
+\[ \corrolary{
+ \bigforall_{C \in \py} D \isin C \equiv
+ \begin{cases}
+ D \in \py : & D \le C \\
+ D \not\in \py : & D \isin \baseof{C}
+ \end{cases}
+}\]
+
+\[ \eqn{Tip Self Inpatch:}{
+ \bigforall_{C \in \py} C \haspatch \p
+}\]
+Ie, tip commits contain their own patch.
+
+\proof{
+Apply Exclusive Tip Contents to some $D \in \py$:
+$ \bigforall_{C \in \py}\bigforall_{D \in \py}
+ D \isin C \equiv D \le C $
+}
+
+\[ \eqn{Exact Ancestors:}{
+ \bigforall_{ C \hasparents \set{R} }
+ D \le C \equiv
+ ( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R )
+ \lor D = C
+}\]
+
+\[ \eqn{Transitive Ancestors:}{
+ \left[ \bigforall_{ E \in \pendsof{C}{\set P} } E \le M \right] \equiv
+ \left[ \bigforall_{ A \in \pancsof{C}{\set P} } A \le M \right]
+}\]
+
+\proof{
+The implication from right to left is trivial because
+$ \pends() \subset \pancs() $.
+For the implication from left to right:
+by the definition of $\mathcal E$,
+for every such $A$, either $A \in \pends()$ which implies
+$A \le M$ by the LHS directly,
+or $\exists_{A' \in \pancs()} \; A' \neq A \land A \le A' $
+in which case we repeat for $A'$. Since there are finitely many
+commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$
+by the LHS. And $A \le A''$.
+}
+
+\section{Commit annotation}
+
+We annotate each Topbloke commit $C$ with:
+\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
+make it wrong to make plain commits with git because the recorded $\pends$
+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 \true \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{Merge}
+
+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}
+
+\subsection{Conditions}
+
+\[ \eqn{ Tip Merge }{
+ L \in \py \implies
+ \begin{cases}
+ R \in \py : & \baseof{R} \ge \baseof{L}
+ \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
+ R \in \pn : & R \ge \baseof{L}
+ \land M = \baseof{L} \\
+ \text{otherwise} : & \false
+ \end{cases}
+}\]
+
+\subsection{No Replay}
+
+\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK.