X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=article.tex;h=f73f7229ae114bc2104d3b4b8bb55b0a5b5b49e6;hb=450c63c9959be28a023c1e8a9276b0c9954fce4f;hp=609e0e3138576ccc9377a0968ce4054dd32e24eb;hpb=6768952437fc5b6b4178361ea7fc852c2e4945cd;p=topbloke-formulae.git diff --git a/article.tex b/article.tex index 609e0e3..f73f722 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 $\bot$. @@ -167,6 +168,18 @@ $\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} @@ -206,7 +219,7 @@ 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{ +\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{ \bigforall_{C \in \py} D \isin C \equiv \begin{cases} D \in \py : & D \le C \\ @@ -483,17 +496,29 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$. \begin{cases} R \in \py : & \baseof{R} \ge \baseof{L} \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\ - R \in \pn : & R \ge \baseof{L} - \land M = \baseof{L} \\ + R \in \pn : & M = \baseof{L} \\ \text{otherwise} : & \false \end{cases} }\] -\[ \eqn{ Merge Ends }{ +\[ \eqn{ Merge Acyclic }{ + L \in \pn + \implies + R \nothaspatch \p +}\] +\[ \eqn{ Removal Merge Ends }{ X \not\haspatch \p \land Y \haspatch \p \land - E \in \pendsof{X}{\py} + M \haspatch \p \implies - E \le Y + \pendsof{Y}{\py} = \pendsof{M}{\py} +}\] +\[ \eqn{ Addition Merge Ends }{ + X \not\haspatch \p \land + Y \haspatch \p \land + M \nothaspatch \p + \implies \left[ + \bigforall_{E \in \pendsof{X}{\py}} E \le Y + \right] }\] \subsection{No Replay} @@ -526,7 +551,7 @@ 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 $A \le C \equiv A \le R$. @@ -534,7 +559,7 @@ That is, $\baseof{C} = R$. $\qed$ -\subsection{Coherence and patch inclusion} +\subsection{Coherence and Patch Inclusion} Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$. This involves considering $D \in \py$. @@ -572,23 +597,119 @@ So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$. \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:} -$C \haspatch \p \equiv C \nothaspatch M$. +$C \haspatch \p \equiv M \nothaspatch \p$. \proofstarts -Merge Ends applies. +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 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. + +$\qed$ + +xxx junk after here + +%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 $L \in \py, D \not\in \py, R \in \py$:} -$D \isin Y \equiv D \le Y$. $D \not\isin X$. Recall that we -are considering $D \in \py$. +Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$. -Consider $D = C$. Thus $C \in \py, L \in \py$. -But $X \not\haspatch \p$ means xxx wip -But $X \not\haspatch \p$ means $D \not\in X$, +Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$. -so we have $L = Y, R = -X$. Thus $R \not\haspatch \p$ and by Tip Self Inpatch $R \not\in -\py$. Thus by Tip Merge $R \in \pn$ and $M = \baseof{L}$. -So by Base Acyclic, $M \nothaspatch \py$. Thus we are expecting -$C \haspatch \py$. And indeed $D \isin C$ and $D \le C$. OK. +But by Tip Merge, $\baseof{R} \ge \baseof{L}$ \end{document}