+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}
+
+\[ \eqn{ Into Base }{
+ L \in \pn
+}\]
+\[ \eqn{ Unique Tip }{
+ \pendsof{L}{\pry} = \{ R^+ \}
+}\]
+\[ \eqn{ Currently Included }{
+ L \haspatch \pry
+}\]
+
+\subsection{Ordering of ${L, R^+, R^-}$:}
+
+By Unique Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$
+so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$.
+
+(Note that the merge base $R^+ \not\le R^-$, i.e. the merge base is
+later than one of the branches to be merged.)
+
+\subsection{No Replay}
+
+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}
+
+Into Base means that $C \in \pn$, so Unique Base is not
+applicable. $\qed$
+
+\subsection{Tip Contents}
+
+Again, not applicable. $\qed$
+
+\subsection{Base Acyclic}
+
+By Base Acyclic for $L$, $D \isin L \implies D \not\in \py$.
+And by Into Base $C \not\in \py$.
+Now from Desired Contents, above, $D \isin C
+\implies D \isin L \lor D = C$, which thus
+$\implies D \not\in \py$. $\qed$.
+
+\subsection{Coherence and Patch Inclusion}
+
+Need to consider some $D \in \py$. By Into Base, $D \neq C$.
+
+\subsubsection{For $\p = \pr$:}
+By Desired Contents, above, $D \not\isin C$.
+So $C \nothaspatch \pr$.
+
+\subsubsection{For $\p \neq \pr$:}
+By Desired Contents, $D \isin C \equiv D \isin L$
+(since $D \in \py$ so $D \not\in \pry$).
+
+If $L \nothaspatch \p$, $D \not\isin L$ so $D \not\isin C$.
+So $L \nothaspatch \p \implies C \nothaspatch \p$.
+
+Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
+so $L \haspatch \p \implies C \haspatch \p$.
+
+\section{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq C$.
+So by Desired Contents $D \isin C \equiv D \isin L$.
+By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+
+And $D \le C \equiv D \le L$.
+Thus $D \isin C \equiv D \le C$. $\qed$
+
+\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
+ \mergeof{C}{L}{M}{R}
+\end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
+
+\subsection{Conditions}
+
+\[ \eqn{ Tip Merge }{
+ L \in \py \implies
+ \begin{cases}
+ R \in \py : & \baseof{R} \ge \baseof{L}
+ \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
+ R \in \pn : & M = \baseof{L} \\
+ \text{otherwise} : & \false
+ \end{cases}
+}\]
+\[ \eqn{ Merge Acyclic }{
+ L \in \pn
+ \implies
+ R \nothaspatch \p
+}\]
+\[ \eqn{ Removal Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \haspatch \p
+ \implies
+ \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p \land
+ M \nothaspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+ \right]
+}\]
+
+\subsection{Non-Topbloke merges}
+
+We require both $\patchof{L} = \bot$ and $\patchof{R} = \bot$.
+I.e. not only is it forbidden to merge into a Topbloke-controlled
+branch without Topbloke's assistance, it is also forbidden to
+merge any Topbloke-controlled branch into any plain git branch.
+
+Given those conditions, Tip Merge and Merge Acyclic do not apply.
+And $Y \not\in \py$ so $\neg [ Y \haspatch \p ]$ so neither
+Merge Ends condition applies. Good.
+
+\subsection{No Replay}
+
+No Replay for Merge Results applies. $\qed$
+
+\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 $R$ 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 $A \le C \equiv A \le R$.
+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 $C \haspatch \p$.