\documentclass[a4paper,leqno]{strayman}
+\errorcontextlines=50
\let\numberwithin=\notdef
\usepackage{amsmath}
\usepackage{mathabx}
\newcommand{\haspatch}{\sqSupset}
\newcommand{\patchisin}{\sqSubset}
-\newcommand{\set}[1]{\mathbb #1}
-\newcommand{\pa}[1]{\varmathbb #1}
+ \newif\ifhidehack\hidehackfalse
+ \DeclareRobustCommand\hidefromedef[2]{%
+ \hidehacktrue\ifhidehack#1\else#2\fi\hidehackfalse}
+ \newcommand{\pa}[1]{\hidefromedef{\varmathbb{#1}}{#1}}
+
+\newcommand{\set}[1]{\mathbb{#1}}
\newcommand{\pay}[1]{\pa{#1}^+}
\newcommand{\pan}[1]{\pa{#1}^-}
\newcommand{\py}{\pay{P}}
\newcommand{\pn}{\pan{P}}
+\newcommand{\pr}{\pa{R}}
+\newcommand{\pry}{\pay{R}}
+\newcommand{\prn}{\pan{R}}
+
%\newcommand{\hasparents}{\underaccent{1}{>}}
%\newcommand{\hasparents}{{%
% \declareslashed{}{_{_1}}{0}{-0.8}{>}\slashed{>}}}
\newcommand{\pancs}{{\mathcal A}}
\newcommand{\pends}{{\mathcal E}}
+\newcommand{\merge}{{\mathcal M}}
\newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
\newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
+\newcommand{\mergeof}[3]{\merge ( #1 , #2, #3 ) }
\newcommand{\patchof}[1]{{\mathcal P} ( #1 ) }
\newcommand{\baseof}[1]{{\mathcal B} ( #1 ) }
the Topbloke patch itself, we hope that git's merge algorithm will
DTRT or that the user will no longer care about the Topbloke patch.
+\item[ $ \mergeof{L}{M}{R} $ ]
+$\displaystyle \left\{ C \middle|
+ \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}
+ \right\} $
+
\end{basedescript}
\newpage
\section{Invariants}
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
+$\patchof{R^+} = \pry, \patchof{R^-} = \prn$.
+Construct $C$ which has $\pr$ removed.
+Used for removing a branch dependency.
+\gathbegin
+ C \hasparents \{ L \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ D \isin C \equiv
+ \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}
+\end{gather}
+
+xxx want to prove $D \isin C \equiv D \not\in pry \land D \isin L$.
+
\section{Merge}
Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
\subsection{Unique Base}
Need to consider only $C \in \py$, ie $L \in \py$,
-and calculate $\pendsof{C}{\pn}$.
+and calculate $\pendsof{C}{\pn}$. So we will consider some
+putative ancestor $A \in \pn$ and see whether $A \le C$.
-%\subsubsection{For $R \in \py$:}
-%foo
+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, $\baseof{C} = R$.
+
+$\qed$
\end{document}