\renewcommand{\implies}{\Rightarrow}
\renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
\renewcommand{\land}{\wedge}
\renewcommand{\lor}{\vee}
$\displaystyle \bigforall_{D \in \py} D \not\isin C $.
~ Informally, $C$ has none of the contents of $\p$.
-Non-Topbloke commits are $\nothaspatch \p$ for all $\p$; if a Topbloke
+Non-Topbloke commits are $\nothaspatch \p$ for all $\p$. This
+includes commits on plain git branches made by applying a Topbloke
+patch. If a Topbloke
patch is applied to a non-Topbloke branch and then bubbles back to
-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.
+the relevant Topbloke branches, we hope that
+if the user still cares about the Topbloke patch,
+git's merge algorithm will DTRT when trying to re-apply the changes.
\item[ $\displaystyle \mergeof{C}{L}{M}{R} $ ]
The contents of a git merge result:
\section{Some lemmas}
+\[ \eqn{Alternative (overlapping) formulations defining
+ $\mergeof{C}{L}{M}{R}$:}{
+ D \isin C \equiv
+ \begin{cases}
+ D \isin L \equiv D \isin R : & D = C \lor D \isin L \\
+ D \isin L \nequiv D \isin R : & D = C \lor D \not\isin M \\
+ D \isin L \equiv D \isin M : & D = C \lor D \isin R \\
+ D \isin L \nequiv D \isin M : & D = C \lor D \isin L \\
+ \text{as above with L and R exchanged}
+ \end{cases}
+}\]
+\proof{
+ Truth table xxx
+
+ Original definition is symmetrical in $L$ and $R$.
+}
+
\[ \eqn{Exclusive Tip Contents:}{
\bigforall_{C \in \py}
\neg \Bigl[ D \isin \baseof{C} \land ( D \in \py \land D \le C )
\tag*{} \patchof{C} = \patchof{A} \\
\tag*{} D \isin C \equiv D \isin A \lor D = C
\end{gather}
+This also covers Topbloke-generated commits on plain git branches:
+Topbloke strips the metadata when exporting.
\subsection{No Replay}
Trivial.
\subsection{Conditions}
+\[ \eqn{ From Base }{
+ L \in \pn
+}\]
\[ \eqn{ Unique Tip }{
\pendsof{L}{\pry} = \{ R^+ \}
}\]
\[ \eqn{ Currently Included }{
L \haspatch \pry
}\]
-\[ \eqn{ Not Self }{
- L \not\in \{ R^+ \}
-}\]
\subsection{No Replay}
\subsection{Unique Base}
-Need to consider only $C \in \py$, ie $L \in \py$.
+From Base means that $C \in \pn$, so Unique Base is not
+applicable. $\qed$
+
+\subsection{Tip Contents}
+
+Again, not applicable. $\qed$
xxx tbd
\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} \\
+ R \in \pn : & M = \baseof{L} \\
\text{otherwise} : & \false
\end{cases}
}\]
\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}
See No Replay for Merge Results.
\subsubsection{For $R \in \pn$:}
-By Tip Merge condition on $R$,
+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 $A \le C \equiv A \le R$.
$\qed$
-\subsection{Coherence and patch inclusion}
+\subsection{Coherence and Patch Inclusion}
Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
This involves considering $D \in \py$.
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.
+so $L \in \py$ so $L \haspatch \p$. We will use the Unique Base
+of $C$, and its Coherence and Patch Inclusion, 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 $R \not\in \py$
+\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
+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.
+We will consider an arbitrary commit $D$
+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
$D \neq C$. 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
-
+$D \isin L \equiv D \isin M$. So by definition of $\merge$, $D \isin
+C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
+Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
\subsubsection{For $D \not\in \py, R \in \py$:}
+$D \neq C$.
+
+By Tip Contents
+$D \isin L \equiv D \isin \baseof{L}$ and
+$D \isin R \equiv D \isin \baseof{R}$.
-%D \in \py$:}
+If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
+Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
+$\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
+$D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
+So $D \isin M \equiv D \isin L$ and by $\merge$,
+$D \isin C \equiv D \isin R$. But from Unique Base,
+$\baseof{C} = R$ so $D \isin C \equiv D \isin \baseof{C}$. OK.
+$\qed$
+
+\subsection{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+By Foreign Inclusion of $L, M, R$:
+$D \isin L \equiv D \le L$;
+$D \isin M \equiv D \le M$;
+$D \isin R \equiv D \le R$.
-xxx the coherence is not that useful ?
+\subsubsection{For $D = C$:}
+
+$D \isin C$ and $D \le C$. OK.
-$L \haspatch \p$ by
+\subsubsection{For $D \neq C, D \isin M$:}
-xxx need to recheck this
+Thus $D \le M$ so $D \le L$ and $D \le R$ so $D \isin L$ and $D \isin
+R$. So by $\merge$, $D \isin C$. And $D \le C$. OK.
-$C \in \py$ $C \haspatch P$ so $D \isin C \equiv D \le C$. OK.
+\subsubsection{For $D \neq C, D \not\isin M, D \isin X$:}
-\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+By $\merge$, $D \isin C$.
+And $D \isin X$ means $D \le X$ so $D \le C$.
+OK.
-Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$.
+\subsubsection{For $D \neq C, D \not\isin M, D \not\isin L, D \not\isin R$:}
-Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$.
+By $\merge$, $D \not\isin C$.
+And $D \not\le L, D \not\le R$ so $D \not\le C$.
+OK
-But by Tip Merge, $\baseof{R} \ge \baseof{L}$
+$\qed$
\end{document}
~ian / topbloke-formulae.git / blobdiff