\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{>}}}
commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$
by the LHS. And $A \le A''$.
}
+\[ \eqn{Calculation Of Ends:}{
+ \bigforall_{C \hasparents \set A}
+ \pendsof{C}{\set P} =
+ \Bigl\{ E \Big|
+ \Bigl[ \Largeexists_{A \in \set A}
+ E \in \pendsof{A}{\set P} \Bigr] \land
+ \Bigl[ \Largenexists_{B \in \set A}
+ E \neq B \land E \le B \Bigr]
+ \Bigr\}
+}\]
+XXX proof TBD.
\section{Commit annotation}
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 $M$ ($M < L, M < R$):
$\qed$
+\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 $R$,
+$A \le \baseof{L} \implies A \le R$, 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}