+We maintain these each time we construct a new commit. \\
+\[ \eqn{No Replay:}{
+ C \has D \implies C \ge D
+}\]
+\[\eqn{Unique Base:}{
+ \bigforall_{C \in \py} \pendsof{C}{\pn} = \{ B \}
+}\]
+\[\eqn{Tip Contents:}{
+ \bigforall_{C \in \py} D \isin C \equiv
+ { D \isin \baseof{C} \lor \atop
+ (D \in \py \land D \le C) }
+}\]
+\[\eqn{Base Acyclic:}{
+ \bigforall_{B \in \pn} D \isin B \implies D \notin \py
+}\]
+\[\eqn{Coherence:}{
+ \bigforall_{C,\p} C \haspatch \p \lor C \nothaspatch \p
+}\]
+\[\eqn{Foreign Inclusion:}{
+ \bigforall_{D \text{ s.t. } \patchof{D} = \bot} D \isin C \equiv D \leq C
+}\]
+
+\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 $A \le A''$.
+}
+\[ \eqn{Calculation Of Ends:}{
+ \bigforall_{C \hasparents \set A}
+ \pendsof{C}{\set P} =
+ \left\{ E \Big|
+ \Bigl[ \Largeexists_{A \in \set A}
+ E \in \pendsof{A}{\set P} \Bigr] \land
+ \Bigl[ \Largenexists_{B \in \set A}
+ E \neq B \land E \le B \Bigr]
+ \right\}
+}\]
+XXX proof TBD.
+
+\subsection{No Replay for Merge Results}
+
+If we are constructing $C$, with,
+\gathbegin
+ \mergeof{C}{L}{M}{R}
+\gathnext
+ L \le C
+\gathnext
+ R \le C
+\end{gather}
+No Replay is preserved. \proofstarts
+
+\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:
+\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{Anticommit}
+
+Given $L, R^+, R^-$ where
+$R^+ \in \pry, R^- = \baseof{R^+}$.
+Construct $C$ which has $\pr$ removed.
+Used for removing a branch dependency.
+\gathbegin
+ C \hasparents \{ L \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ \mergeof{C}{L}{R^+}{R^-}
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Unique Tip }{
+ \pendsof{L}{\pry} = \{ R^+ \}
+}\]
+\[ \eqn{ Currently Included }{
+ L \haspatch \pry
+}\]
+\[ \eqn{ Not Self }{
+ L \not\in \{ R^+ \}
+}\]
+
+\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{Desired Contents}
+
+\[ D \isin C \equiv [ D \notin \pry \land D \isin L ] \lor D = C \]
+\proofstarts
+
+\subsubsection{For $D = C$:}
+
+Trivially $D \isin C$. OK.
+
+\subsubsection{For $D \neq C, D \not\le L$:}
+
+By No Replay $D \not\isin L$. Also $D \not\le R^-$ hence
+$D \not\isin R^-$. Thus $D \not\isin C$. OK.
+
+\subsubsection{For $D \neq C, D \le L, D \in \pry$:}
+
+By Currently Included, $D \isin L$.
+
+By Tip Self Inpatch, $D \isin R^+ \equiv D \le R^+$, but by
+by Unique Tip, $D \le R^+ \equiv D \le L$.
+So $D \isin R^+$.
+
+By Base Acyclic, $D \not\isin R^-$.
+
+Apply $\merge$: $D \not\isin C$. OK.
+
+\subsubsection{For $D \neq C, D \le L, D \notin \pry$:}
+
+By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$.
+
+Apply $\merge$: $D \isin C \equiv D \isin L$. OK.
+
+$\qed$
+
+\subsection{Unique Base}
+
+Need to consider only $C \in \py$, ie $L \in \py$.
+
+xxx tbd
+
+xxx need to finish anticommit
+
+\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
+ \mergeof{C}{L}{M}{R}
+\end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
+
+\subsection{Conditions}
+
+\[ \eqn{ Tip Merge }{
+ L \in \py \implies
+ \begin{cases}
+ R \in \py : & \baseof{R} \ge \baseof{L}
+ \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
+ R \in \pn : & R \ge \baseof{L}
+ \land M = \baseof{L} \\
+ \text{otherwise} : & \false
+ \end{cases}
+}\]
+\[ \eqn{ Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ E \in \pendsof{X}{\py}
+ \implies
+ E \le Y
+}\]
+
+\subsection{No Replay}
+
+See No Replay for Merge Results.
+
+\subsection{Unique Base}
+
+Need to consider only $C \in \py$, ie $L \in \py$,
+and calculate $\pendsof{C}{\pn}$. So we will consider some
+putative ancestor $A \in \pn$ and see whether $A \le C$.
+
+By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$.
+But $C \in py$ and $A \in \pn$ so $A \neq C$.
+Thus $A \le C \equiv A \le L \lor A \le R$.
+
+By Unique Base of L and Transitive Ancestors,
+$A \le L \equiv A \le \baseof{L}$.
+
+\subsubsection{For $R \in \py$:}
+
+By Unique Base of $R$ and Transitive Ancestors,
+$A \le R \equiv A \le \baseof{R}$.
+
+But by Tip Merge condition on $\baseof{R}$,
+$A \le \baseof{L} \implies A \le \baseof{R}$, so
+$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.
+That is, $\baseof{C} = \baseof{R}$.
+
+\subsubsection{For $R \in \pn$:}
+
+By Tip Merge condition on $R$,
+$A \le \baseof{L} \implies A \le R$, so
+$A \le R \lor A \le \baseof{L} \equiv A \le R$.
+Thus $A \le C \equiv A \le R$.
+That is, $\baseof{C} = R$.
+
+$\qed$
+
+\subsection{Coherence and patch inclusion}
+
+Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
+This involves considering $D \in \py$.
+
+\subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
+$D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L
+\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch). So $D \neq C$.
+Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
+
+\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
+$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
+(Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.)
+
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
+
+For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
+ \equiv D \isin L \lor D \isin R$.
+(Likewise $D \le C \equiv D \le X \lor D \le Y$.)
+
+Consider $D \neq C, D \isin X \land D \isin Y$:
+By $\merge$, $D \isin C$. Also $D \le X$
+so $D \le C$. OK for $C \haspatch \p$.
+
+Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
+By $\merge$, $D \not\isin C$.
+And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
+OK for $C \haspatch \p$.
+
+Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
+$D \not\le X$ so $D \not\le M$ so $D \not\isin M$.
+Thus by $\merge$, $D \isin C$. And $D \le Y$ so $D \le C$.
+OK for $C \haspatch \p$.
+
+So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.