X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=article.tex;h=0688c7ad02524d0d5c0d399682f3ac46946d1e1f;hb=26c18196858d61ec8bd051837aa9ccb08f68ccaf;hp=1fa67c2330d692aed86011628931177731e8a99a;hpb=2af3a4453d3b0ad5dcd7b5014486ff927b0cdf52;p=topbloke-formulae.git diff --git a/article.tex b/article.tex index 1fa67c2..0688c7a 100644 --- a/article.tex +++ b/article.tex @@ -46,6 +46,7 @@ \renewcommand{\implies}{\Rightarrow} \renewcommand{\equiv}{\Leftrightarrow} +\renewcommand{\nequiv}{\nLeftrightarrow} \renewcommand{\land}{\wedge} \renewcommand{\lor}{\vee} @@ -65,8 +66,8 @@ \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}{% @@ -121,7 +122,7 @@ A patch $\p$ consists of two sets of commits $\pn$ and $\py$, which 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 $\bot$. @@ -151,10 +152,13 @@ $\displaystyle \bigforall_{D \in \py} D \isin C \equiv D \le C $. $\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: @@ -195,6 +199,23 @@ We maintain these each time we construct a new commit. \\ \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 ) @@ -206,7 +227,7 @@ Ie, the two limbs of the RHS of Tip Contents are mutually exclusive. 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 \\ @@ -263,7 +284,7 @@ XXX proof TBD. \subsection{No Replay for Merge Results} -If we are constructing $C$, given +If we are constructing $C$, with, \gathbegin \mergeof{C}{L}{M}{R} \gathnext @@ -319,6 +340,8 @@ A simple single-parent forward commit $C$ as made by git-commit. \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. @@ -406,15 +429,15 @@ Used for removing a branch dependency. \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} @@ -458,7 +481,12 @@ $\qed$ \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 @@ -474,6 +502,7 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): \gathnext \mergeof{C}{L}{M}{R} \end{gather} +We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$. \subsection{Conditions} @@ -482,11 +511,41 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): \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} }\] +\[ \eqn{ Merge Acyclic }{ + L \in \pn + \implies + R \nothaspatch \p +}\] +\[ \eqn{ Removal Merge Ends }{ + X \not\haspatch \p \land + Y \haspatch \p \land + M \haspatch \p + \implies + \pendsof{Y}{\py} = \pendsof{M}{\py} +}\] +\[ \eqn{ Addition Merge Ends }{ + X \not\haspatch \p \land + Y \haspatch \p \land + M \nothaspatch \p + \implies \left[ + \bigforall_{E \in \pendsof{X}{\py}} E \le Y + \right] +}\] + +\subsection{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} @@ -518,7 +577,7 @@ That is, $\baseof{C} = \baseof{R}$. \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 $A \le C \equiv A \le R$. @@ -526,23 +585,21 @@ That is, $\baseof{C} = R$. $\qed$ -\subsection{Coherence and patch inclusion} +\subsection{Coherence and Patch Inclusion} -Need to determine $C \haspatch P$ based on $L,M,R \haspatch P$. +Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$. This involves considering $D \in \py$. -We will use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$. - -\subsubsection{For $L \nothaspatch P, R \nothaspatch P$:} +\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. $C \nothaspatch P$. +\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 $L \haspatch P, R \haspatch P$:} +\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 $C \haspatch P$. +Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$. For $D \neq C$: $D \le C \equiv D \le L \lor D \le R \equiv D \isin L \lor D \isin R$. @@ -550,18 +607,144 @@ For $D \neq C$: $D \le C \equiv D \le L \lor D \le R 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 $C \haspatch P$. +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 $C \haspatch P$. +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 $C \haspatch P$. +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. + +$\qed$ -So indeed $L \haspatch P \land R \haspatch P \implies C \haspatch P$. +\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 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 $R \not\in \py$ +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 an arbitrary commit $D$ +and prove the Exclusive Tip Contents form. + +\subsubsection{For $D \in \py$:} +$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D +\le C$. 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, +$D \isin L \equiv D \isin M$. 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 +$D \isin R \equiv D \isin \baseof{R}$. + +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 $L, M, R$: +$D \isin L \equiv D \le L$; +$D \isin M \equiv D \le M$; +$D \isin R \equiv D \le R$. + +\subsubsection{For $D = C$:} + +$D \isin C$ and $D \le C$. OK. + +\subsubsection{For $D \neq C, D \isin M$:} + +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. + +\subsubsection{For $D \neq C, D \not\isin M, D \isin X$:} + +By $\merge$, $D \isin C$. +And $D \isin X$ means $D \le X$ so $D \le C$. +OK. + +\subsubsection{For $D \neq C, D \not\isin M, D \not\isin L, D \not\isin R$:} + +By $\merge$, $D \not\isin C$. +And $D \not\le L, D \not\le R$ so $D \not\le C$. +OK + +$\qed$ \end{document}