\renewcommand{\ge}{\geqslant}
\renewcommand{\le}{\leqslant}
+\newcommand{\nge}{\ngeqslant}
+\newcommand{\nle}{\nleqslant}
\newcommand{\has}{\sqsupseteq}
\newcommand{\isin}{\sqsubseteq}
\renewcommand{\implies}{\Rightarrow}
\renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
\renewcommand{\land}{\wedge}
\renewcommand{\lor}{\vee}
\newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
\newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
-\newcommand{\merge}[4]{{\mathcal M}(#1,#2,#3,#4)}
+\newcommand{\merge}{{\mathcal M}}
+\newcommand{\mergeof}[4]{\merge(#1,#2,#3,#4)}
%\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
\newcommand{\patch}{{\mathcal P}}
\newcommand{\patchof}[1]{\patch ( #1 ) }
\newcommand{\baseof}[1]{\base ( #1 ) }
+\newcommand{\eqntag}[2]{ #2 \tag*{\mbox{#1}} }
\newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} }
-\newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} }
%\newcommand{\bigforall}{\mathop{\hbox{\huge$\forall$}}}
\newcommand{\bigforall}{%
\newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}
\newcommand{\qed}{\square}
-\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
+\newcommand{\proofstarts}{{\it Proof:}}
+\newcommand{\proof}[1]{\proofstarts #1 $\qed$}
\newcommand{\gathbegin}{\begin{gather} \tag*{}}
\newcommand{\gathnext}{\\ \tag*{}}
are respectively the base and tip git branches. $\p$ may be used
where the context requires a set, in which case the statement
is to be taken as applying to both $\py$ and $\pn$.
-All these sets are distinct. Hence:
+None of these sets overlap. Hence:
\item[ $ \patchof{ C } $ ]
Either $\p$ s.t. $ C \in \p $, or $\bot$.
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[ $\displaystyle \merge{C}{L}{M}{R} $ ]
+\item[ $\displaystyle \mergeof{C}{L}{M}{R} $ ]
The contents of a git merge result:
$\displaystyle D \isin C \equiv
\end{cases}
$
+Some (overlapping) alternative formulations:
+
+$\displaystyle D \isin C \equiv
+ \begin{cases}
+ D \isin L \equiv D \isin R : & D = C \lor D \isin L \\
+ D \isin L \equiv D \isin R : & D = C \lor D \isin R \\
+ D \isin L \nequiv D \isin R : & D = C \lor D \not\isin M \\
+ D \isin M \equiv D \isin L : & D = C \lor D \isin R \\
+ D \isin M \equiv D \isin R : & D = C \lor D \isin L \\
+ \end{cases}
+$
+
\end{basedescript}
\newpage
\section{Invariants}
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{
+\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
\bigforall_{C \in \py} D \isin C \equiv
\begin{cases}
D \in \py : & D \le C \\
\[ \eqn{Calculation Of Ends:}{
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
- \Bigl\{ E \Big|
+ \left\{ E \Big|
\Bigl[ \Largeexists_{A \in \set A}
E \in \pendsof{A}{\set P} \Bigr] \land
\Bigl[ \Largenexists_{B \in \set A}
E \neq B \land E \le B \Bigr]
- \Bigr\}
+ \right\}
}\]
XXX proof TBD.
\subsection{No Replay for Merge Results}
-If we are constructing $C$ such that $\merge{C}{L}{M}{R}$, No Replay
-is preserved. {\it Proof:}
+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 $D \neq C \land D \not\isin L \land D \not\isin R$:}
$D \not\isin C$. 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
\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
\land D \isin M$:}
-$D \not\isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
-R$ so $D \le C$. OK.
+$D \not\isin C$. OK.
$\qed$
\section{Anticommit}
Given $L, R^+, R^-$ where
-$\patchof{R^+} = \pry, \patchof{R^-} = \prn$.
+$R^+ \in \pry, R^- = \baseof{R^+}$.
Construct $C$ which has $\pr$ removed.
Used for removing a branch dependency.
\gathbegin
\gathnext
\patchof{C} = \patchof{L}
\gathnext
- \merge{C}{L}{R^+}{R^-}
+ \mergeof{C}{L}{R^+}{R^-}
\end{gather}
\subsection{Conditions}
\[ \eqn{ Unique Tip }{
\pendsof{L}{\pry} = \{ R^+ \}
}\]
-\[ \eqn{ Correct Base }{
- \baseof{R^+} = R^-
-}\]
\[ \eqn{ Currently Included }{
L \haspatch \pry
}\]
+\[ \eqn{ Not Self }{
+ L \not\in \{ R^+ \}
+}\]
+
+\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.
-xxx want to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+\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}
\gathnext
\patchof{C} = \patchof{L}
\gathnext
- \merge{C}{L}{M}{R}
+ \mergeof{C}{L}{M}{R}
\end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
\subsection{Conditions}
\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{Merge Results}
+\subsection{No Replay}
-As above.
+See No Replay for Merge Results.
\subsection{Unique Base}
$\qed$
+\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 $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
+
+\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
+
+$C \haspatch \p \equiv M \nothaspatch \p$.
+
+\proofstarts
+
+One of the Merge Ends conditions applies.
+Recall that we are considering $D \in \py$.
+$D \isin Y \equiv D \le Y$. $D \not\isin X$.
+We will show for each of
+various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
+(which suffices by definition of $\haspatch$ and $\nothaspatch$).
+
+Consider $D = C$: Thus $C \in \py, L \in \py$, and by Tip
+Self Inpatch $L \haspatch \p$, so $L=Y, R=X$. By Tip Merge,
+$M=\baseof{L}$. So by Base Acyclic $D \not\isin M$, i.e.
+$M \nothaspatch \p$. And indeed $D \isin C$ and $D \le C$. OK.
+
+Consider $D \neq C, M \nothaspatch P, D \isin Y$:
+$D \le Y$ so $D \le C$.
+$D \not\isin M$ so by $\merge$, $D \isin C$. OK.
+
+Consider $D \neq C, M \nothaspatch P, D \not\isin Y$:
+$D \not\le Y$. If $D \le X$ then
+$D \in \pancsof{X}{\py}$, so by Addition Merge Ends and
+Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
+Thus $D \not\le C$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \isin Y$:
+$D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
+and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
+Thus $D \isin M$. By $\merge$, $D \not\isin C$. OK.
+
+Consider $D \neq C, M \haspatch P, D \not\isin Y$:
+By $\merge$, $D \not\isin C$. OK.
+
+$\qed$
+
+\subsection{Base Acyclic}
+
+This applies when $C \in \pn$.
+$C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
+
+Consider some $D \in \py$.
+
+By Base Acyclic of $L$, $D \not\isin L$. By the above, $D \not\isin
+R$. And $D \neq C$. So $D \not\isin C$. $\qed$
+
+\subsection{Tip Contents}
+
+We need worry only about $C \in \py$.
+And $\patchof{C} = \patchof{L}$
+so $L \in \py$ so $L \haspatch \p$. We will use the coherence and
+patch inclusion of $C$ as just proved.
+
+Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+\p$ and by coherence/inclusion $C \haspatch \p$ . If $R \not\in \py$
+then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
+of $\nothaspatch$, $M \nothaspatch \p$. So by coherence/inclusion $C
+\haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
+
+We will consider some $D$ and prove the Exclusive Tip Contents form.
+
+\subsubsection{For $D \in \py$:}
+$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
+\le C$. OK.
+
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
+
+$D \neq C$. By Tip Contents of $L$,
+$D \isin L \equiv D \isin \baseof{L}$, and by Tip Merge condition,
+$D \isin L \equiv D \isin M$. xxx up to here
+
+
+\subsubsection{For $D \not\in \py, R \in \py$:}
+
+
+%D \in \py$:}
+
+
+
+xxx the coherence is not that useful ?
+
+$L \haspatch \p$ by
+
+xxx need to recheck this
+
+$C \in \py$ $C \haspatch P$ so $D \isin C \equiv D \le C$. OK.
+
+\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+
+Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$.
+
+Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$.
+
+But by Tip Merge, $\baseof{R} \ge \baseof{L}$
+
\end{document}