\renewcommand{\ge}{\geqslant}
\renewcommand{\le}{\leqslant}
+\newcommand{\nge}{\ngeqslant}
+\newcommand{\nle}{\nleqslant}
\newcommand{\has}{\sqsupseteq}
\newcommand{\isin}{\sqsubseteq}
\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{\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*{}}
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
\[ \eqn{Calculation Of Ends:}{
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
- \Bigl\{ E \Big|
+ \left\{ E \Big|
\Bigl[ \Largeexists_{A \in \set A}
E \in \pendsof{A}{\set P} \Bigr] \land
\Bigl[ \Largenexists_{B \in \set A}
E \neq B \land E \le B \Bigr]
- \Bigr\}
+ \right\}
}\]
XXX proof TBD.
\subsection{No Replay for Merge Results}
-If we are constructing $C$, given
+If we are constructing $C$, with,
\gathbegin
- \merge{C}{L}{M}{R}
+ \mergeof{C}{L}{M}{R}
\gathnext
L \le C
\gathnext
R \le C
\end{gather}
-No Replay is preserved. {\it Proof:}
+No Replay is preserved. \proofstarts
\subsubsection{For $D=C$:} $D \isin C, D \le C$. OK.
\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.
+\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 want to prove $D \isin C \equiv D \not\in \pry \land D \isin L$.
+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}
}\]
+\[ \eqn{ Merge Ends }{
+ X \not\haspatch \p \land
+ Y \haspatch \p
+ \implies \left[
+ \bigforall_{E \in \pendsof{X}{\py}}
+ E \le Y
+ \right]
+}\]
-\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
+
+Merge Ends 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 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.
+
\end{document}
~ian / topbloke-formulae.git / blobdiff