X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=7ec582e666ef94e6b55ba1e682e1866a1048cb03;hp=71e09608bcb95d8147f06340e81461a70437cdd3;hb=c39f3f1f5a5df2a4d628f335f7aaad44c0f68913;hpb=f7c24d718e10f4673853f9495b57701a24d25e94 diff --git a/article.tex b/article.tex index 71e0960..7ec582e 100644 --- a/article.tex +++ b/article.tex @@ -8,6 +8,13 @@ \usepackage{mdwlist} %\usepackage{accents} +\usepackage{fancyhdr} +\pagestyle{fancy} +\lhead[\rightmark]{} + +\let\stdsection\section +\renewcommand\section{\newpage\stdsection} + \renewcommand{\ge}{\geqslant} \renewcommand{\le}{\leqslant} \newcommand{\nge}{\ngeqslant} @@ -34,6 +41,10 @@ \newcommand{\py}{\pay{P}} \newcommand{\pn}{\pan{P}} +\newcommand{\pq}{\pa{Q}} +\newcommand{\pqy}{\pay{Q}} +\newcommand{\pqn}{\pan{Q}} + \newcommand{\pr}{\pa{R}} \newcommand{\pry}{\pay{R}} \newcommand{\prn}{\pan{R}} @@ -105,7 +116,7 @@ $\set X$. \item[ $ C \ge D $ ] $C$ is a descendant of $D$ in the git commit -graph. This is a partial order, namely the transitive closure of +graph. This is a partial order, namely the transitive closure of $ D \in \set X $ where $ C \hasparents \set X $. \item[ $ C \has D $ ] @@ -122,21 +133,21 @@ 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 $\pn$. -None of these sets overlap. Hence: +All of these sets are disjoint. Hence: \item[ $ \patchof{ C } $ ] -Either $\p$ s.t. $ C \in \p $, or $\bot$. +Either $\p$ s.t. $ C \in \p $, or $\bot$. A function from commits to patches' sets $\p$. \item[ $ \pancsof{C}{\set P} $ ] -$ \{ A \; | \; A \le C \land A \in \set P \} $ +$ \{ A \; | \; A \le C \land A \in \set P \} $ i.e. all the ancestors of $C$ which are in $\set P$. \item[ $ \pendsof{C}{\set P} $ ] $ \{ E \; | \; E \in \pancsof{C}{\set P} \land \mathop{\not\exists}_{A \in \pancsof{C}{\set P}} - E \neq A \land E \le A \} $ + E \neq A \land E \le A \} $ i.e. all $\le$-maximal commits in $\pancsof{C}{\set P}$. \item[ $ \baseof{C} $ ] @@ -150,12 +161,15 @@ $\displaystyle \bigforall_{D \in \py} D \isin C \equiv D \le C $. \item[ $ C \nothaspatch \p $ ] $\displaystyle \bigforall_{D \in \py} D \not\isin C $. -~ Informally, $C$ has none of the contents of $\p$. +~ Informally, $C$ has none of the contents of $\p$. -Non-Topbloke commits are $\nothaspatch \p$ for all $\p$; if a Topbloke +Commits on Non-Topbloke branches 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: @@ -166,18 +180,6 @@ $\displaystyle D \isin C \equiv (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\ \text{otherwise} : & D \not\isin M \end{cases} -$ - -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} @@ -205,14 +207,52 @@ We maintain these each time we construct a new commit. \\ \[\eqn{Foreign Inclusion:}{ \bigforall_{D \text{ s.t. } \patchof{D} = \bot} D \isin C \equiv D \leq C }\] +\[\eqn{Foreign Contents:}{ + \bigforall_{C \text{ s.t. } \patchof{C} = \bot} + D \le C \implies \patchof{D} = \bot +}\] \section{Some lemmas} -\[ \eqn{Exclusive Tip Contents:}{ - \bigforall_{C \in \py} +\subsection{Alternative (overlapping) formulations of $\mergeof{C}{L}{M}{R}$} +$$ + D \isin C \equiv + \begin{cases} + D \isin L \equiv D \isin R : & D = C \lor D \isin L \\ + D \isin L \nequiv D \isin R : & D = C \lor D \not\isin M \\ + D \isin L \equiv D \isin M : & D = C \lor D \isin R \\ + D \isin L \nequiv D \isin M : & D = C \lor D \isin L \\ + \text{as above with L and R exchanged} + \end{cases} +$$ +\proof{ ~ Truth table (ordered by original definition): \\ + \begin{tabular}{cccc|c|cc} + $D = C$ & + $\isin L$ & + $\isin M$ & + $\isin R$ & $\isin C$ & + $L$ vs. $R$ & $L$ vs. $M$ + \\\hline + y & ? & ? & ? & y & ? & ? \\ + n & y & y & y & y & $\equiv$ & $\equiv$ \\ + n & y & n & y & y & $\equiv$ & $\nequiv$ \\ + n & n & y & n & n & $\equiv$ & $\nequiv$ \\ + n & n & n & n & n & $\equiv$ & $\equiv$ \\ + n & y & y & n & n & $\nequiv$ & $\equiv$ \\ + n & n & y & y & n & $\nequiv$ & $\nequiv$ \\ + n & y & n & n & y & $\nequiv$ & $\nequiv$ \\ + n & n & n & y & y & $\nequiv$ & $\equiv$ \\ + \end{tabular} \\ + And original definition is symmetrical in $L$ and $R$. +} + +\subsection{Exclusive Tip Contents} +Given Base Acyclic for $C$, +$$ + \bigforall_{C \in \py} \neg \Bigl[ D \isin \baseof{C} \land ( D \in \py \land D \le C ) \Bigr] -}\] +$$ Ie, the two limbs of the RHS of Tip Contents are mutually exclusive. \proof{ @@ -227,9 +267,11 @@ So by Base Acyclic $D \isin B \implies D \notin \py$. \end{cases} }\] -\[ \eqn{Tip Self Inpatch:}{ +\subsection{Tip Self Inpatch} +Given Exclusive Tip Contents and Base Acyclic for $C$, +$$ \bigforall_{C \in \py} C \haspatch \p -}\] +$$ Ie, tip commits contain their own patch. \proof{ @@ -238,22 +280,27 @@ $ \bigforall_{C \in \py}\bigforall_{D \in \py} D \isin C \equiv D \le C $ } -\[ \eqn{Exact Ancestors:}{ +\subsection{Exact Ancestors} +$$ \bigforall_{ C \hasparents \set{R} } + \left[ D \le C \equiv ( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R ) \lor D = C -}\] + \right] +$$ +\proof{ ~ Trivial.} -\[ \eqn{Transitive Ancestors:}{ +\subsection{Transitive Ancestors} +$$ \left[ \bigforall_{ E \in \pendsof{C}{\set P} } E \le M \right] \equiv \left[ \bigforall_{ A \in \pancsof{C}{\set P} } A \le M \right] -}\] +$$ \proof{ The implication from right to left is trivial because $ \pends() \subset \pancs() $. -For the implication from left to right: +For the implication from left to right: by the definition of $\mathcal E$, for every such $A$, either $A \in \pends()$ which implies $A \le M$ by the LHS directly, @@ -262,48 +309,97 @@ in which case we repeat for $A'$. Since there are finitely many commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$ by the LHS. And $A \le A''$. } -\[ \eqn{Calculation Of Ends:}{ + +\subsection{Calculation of Ends} +$$ \bigforall_{C \hasparents \set A} \pendsof{C}{\set P} = + \begin{cases} + C \in \p : & \{ C \} + \\ + C \not\in \p : & \displaystyle \left\{ E \Big| - \Bigl[ \Largeexists_{A \in \set A} + \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] + \Bigl[ \Largenexists_{B \in \set A, F \in \pendsof{B}{\p}} + E \neq F \land E \le F \Bigr] \right\} -}\] -XXX proof TBD. - -\subsection{No Replay for Merge Results} - -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 $D \isin L \land D \isin R$:} -$D \isin C$. And $D \isin L \implies D \le L \implies D \le C$. OK. - -\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:} -$D \not\isin C$. OK. + \end{cases} +$$ +\proof{ +Trivial for $C \in \set P$. For $C \not\in \set P$, +$\pancsof{C}{\set P} = \bigcup_{A \in \set A} \pancsof{A}{\set P}$. +So $\pendsof{C}{\set P} \subset \bigcup_{E in \set E} \pendsof{E}{\set P}$. +Consider some $E \in \pendsof{A}{\set P}$. If $\exists_{B,F}$ as +specified, then either $F$ is going to be in our result and +disqualifies $E$, or there is some other $F'$ (or, eventually, +an $F''$) which disqualifies $F$. +Otherwise, $E$ meets all the conditions for $\pends$. +} -\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R) - \land D \not\isin M$:} -$D \isin C$. Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le -R$ so $D \le C$. OK. +\subsection{Ingredients Prevent Replay} +$$ + \left[ + {C \hasparents \set A} \land + \\ + \bigforall_{D} + \left( + D \isin C \implies + D = C \lor + \Largeexists_{A \in \set A} D \isin A + \right) + \right] \implies \left[ \bigforall_{D} + D \isin C \implies D \le C + \right] +$$ +\proof{ + Trivial for $D = C$. Consider some $D \neq C$, $D \isin C$. + By the preconditions, there is some $A$ s.t. $D \in \set A$ + and $D \isin A$. By No Replay for $A$, $D \le A$. And + $A \le C$ so $D \le C$. +} -\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R) - \land D \isin M$:} -$D \not\isin C$. OK. +\subsection{Simple Foreign Inclusion} +$$ + \left[ + C \hasparents \{ L \} + \land + \bigforall_{D} D \isin C \equiv D \isin L \lor D = C + \right] + \implies + \left[ + \bigforall_{D \text{ s.t. } \patchof{D} = \bot} + D \isin C \equiv D \le C + \right] +$$ +\proof{ +Consider some $D$ s.t. $\patchof{D} = \bot$. +If $D = C$, trivially true. For $D \neq C$, +by Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$. +And by Exact Ancestors $D \le L \equiv D \le C$. +So $D \isin C \equiv D \le C$. +} -$\qed$ +\subsection{Totally Foreign Contents} +$$ + \left[ + C \hasparents \set A \land + \patchof{C} = \bot \land + \bigforall_{A \in \set A} \patchof{A} = \bot + \right] + \implies + \left[ + \bigforall_{D} + D \le C + \implies + \patchof{D} = \bot + \right] +$$ +\proof{ +Consider some $D \le C$. If $D = C$, $\patchof{D} = \bot$ trivially. +If $D \neq C$ then $D \le A$ where $A \in \set A$. By Foreign +Contents of $A$, $\patchof{D} = \bot$. +} \section{Commit annotation} @@ -313,45 +409,65 @@ We annotate each Topbloke commit $C$ with: \gathnext \baseof{C}, \text{ if } C \in \py \gathnext - \bigforall_{\pa{Q}} - \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q} + \bigforall_{\pq} + \text{ either } C \haspatch \pq \text{ or } C \nothaspatch \pq \gathnext - \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}} + \bigforall_{\pqy \not\ni C} \pendsof{C}{\pqy} \end{gather} +$\patchof{C}$, for each kind of Topbloke-generated commit, is stated +in the summary in the section for that kind of commit. + +Whether $\baseof{C}$ is required, and if so what the value is, is +stated in the proof of Unique Base for each kind of commit. + +$C \haspatch \pq$ or $\nothaspatch \pq$ is represented as the +set $\{ \pq | C \haspatch \pq \}$. Whether $C \haspatch \pq$ +is in stated +(in terms of $I \haspatch \pq$ or $I \nothaspatch \pq$ +for the ingredients $I$) +in the proof of Coherence for each kind of commit. + +$\pendsof{C}{\pq^+}$ is computed, for all Topbloke-generated commits, +using the lemma Calculation of Ends, above. We do not annotate $\pendsof{C}{\py}$ for $C \in \py$. Doing so would make it wrong to make plain commits with git because the recorded $\pends$ -would have to be updated. The annotation is not needed because -$\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$. +would have to be updated. The annotation is not needed in that case +because $\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$. \section{Simple commit} A simple single-parent forward commit $C$ as made by git-commit. \begin{gather} -\tag*{} C \hasparents \{ A \} \\ -\tag*{} \patchof{C} = \patchof{A} \\ -\tag*{} D \isin C \equiv D \isin A \lor D = C +\tag*{} C \hasparents \{ L \} \\ +\tag*{} \patchof{C} = \patchof{L} \\ +\tag*{} D \isin C \equiv D \isin L \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. + +Ingredients Prevent Replay applies. $\qed$ \subsection{Unique Base} -If $A, C \in \py$ then $\baseof{C} = \baseof{A}$. $\qed$ +If $L, C \in \py$ then by Calculation of Ends, +$\pendsof{C}{\pn} = \pendsof{L}{\pn}$ so +$\baseof{C} = \baseof{L}$. $\qed$ \subsection{Tip Contents} -We need to consider only $A, C \in \py$. From Tip Contents for $A$: -\[ D \isin A \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A ) \] +We need to consider only $L, C \in \py$. From Tip Contents for $L$: +\[ D \isin L \equiv D \isin \baseof{L} \lor ( D \in \py \land D \le L ) \] Substitute into the contents of $C$: -\[ D \isin C \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A ) +\[ D \isin C \equiv D \isin \baseof{L} \lor ( D \in \py \land D \le L ) \lor D = C \] -Since $D = C \implies D \in \py$, -and substituting in $\baseof{C}$, this gives: +Since $D = C \implies D \in \py$, +and substituting in $\baseof{C}$, from Unique Base above, this gives: \[ D \isin C \equiv D \isin \baseof{C} \lor - (D \in \py \land D \le A) \lor + (D \in \py \land D \le L) \lor (D = C \land D \in \py) \] \[ \equiv D \isin \baseof{C} \lor - [ D \in \py \land ( D \le A \lor D = C ) ] \] + [ D \in \py \land ( D \le L \lor D = C ) ] \] So by Exact Ancestors: \[ D \isin C \equiv D \isin \baseof{C} \lor ( D \in \py \land D \le C ) \] @@ -359,18 +475,20 @@ $\qed$ \subsection{Base Acyclic} -Need to consider only $A, C \in \pn$. +Need to consider only $L, C \in \pn$. For $D = C$: $D \in \pn$ so $D \not\in \py$. OK. -For $D \neq C$: $D \isin C \equiv D \isin A$, so by Base Acyclic for -$A$, $D \isin C \implies D \not\in \py$. $\qed$ +For $D \neq C$: $D \isin C \equiv D \isin L$, so by Base Acyclic for +$L$, $D \isin C \implies D \not\in \py$. + +$\qed$ \subsection{Coherence and patch inclusion} Need to consider $D \in \py$ -\subsubsection{For $A \haspatch P, D = C$:} +\subsubsection{For $L \haspatch P, D = C$:} Ancestors of $C$: $ D \le C $. @@ -378,36 +496,172 @@ $ D \le C $. Contents of $C$: $ D \isin C \equiv \ldots \lor \true \text{ so } D \haspatch C $. -\subsubsection{For $A \haspatch P, D \neq C$:} -Ancestors: $ D \le C \equiv D \le A $. +\subsubsection{For $L \haspatch P, D \neq C$:} +Ancestors: $ D \le C \equiv D \le L $. -Contents: $ D \isin C \equiv D \isin A \lor f $ -so $ D \isin C \equiv D \isin A $. +Contents: $ D \isin C \equiv D \isin L \lor f $ +so $ D \isin C \equiv D \isin L $. So: -\[ A \haspatch P \implies C \haspatch P \] +\[ L \haspatch P \implies C \haspatch P \] -\subsubsection{For $A \nothaspatch P$:} +\subsubsection{For $L \nothaspatch P$:} -Firstly, $C \not\in \py$ since if it were, $A \in \py$. +Firstly, $C \not\in \py$ since if it were, $L \in \py$. Thus $D \neq C$. -Now by contents of $A$, $D \notin A$, so $D \notin C$. +Now by contents of $L$, $D \notin L$, so $D \notin C$. So: -\[ A \nothaspatch P \implies C \nothaspatch P \] +\[ L \nothaspatch P \implies C \nothaspatch P \] +$\qed$ + +\subsection{Foreign Inclusion:} + +Simple Foreign Inclusion applies. $\qed$ + +\subsection{Foreign Contents:} + +Only relevant if $\patchof{C} = \bot$, and in that case Totally +Foreign Contents applies. $\qed$ + +\section{Create Base} + +Given a starting point $L$ and a proposed patch $\pq$, +create a Topbloke base branch initial commit $B$. +\gathbegin + B \hasparents \{ L \} +\gathnext + \patchof{B} = \pqn +\gathnext + D \isin B \equiv D \isin L \lor D = B +\end{gather} + +\subsection{Conditions} + +\[ \eqn{ Create Acyclic }{ + \pendsof{L}{\pqy} = \{ \} +}\] + +\subsection{No Replay} + +Ingredients Prevent Replay applies. $\qed$ + +\subsection{Unique Base} + +Not applicable. + +\subsection{Tip Contents} + +Not applicable. + +\subsection{Base Acyclic} + +Consider some $D \isin B$. If $D = B$, $D \in \pqn$. +If $D \neq B$, $D \isin L$, so by No Replay $D \le L$ +and by Create Acyclic +$D \not\in \pqy$. $\qed$ + +\subsection{Coherence and Patch Inclusion} + +Consider some $D \in \py$. +$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$ +and $D \le B \equiv D \le L$. + +Thus $L \haspatch \p \implies B \haspatch P$ +and $L \nothaspatch \p \implies B \nothaspatch P$. + +$\qed$. + +\subsection{Foreign Inclusion} + +Simple Foreign Inclusion applies. $\qed$ + +\subsection{Foreign Contents} + +Not applicable. + +\section{Create Tip} + +Given a Topbloke base $B$ for a patch $\pq$, +create a tip branch initial commit B. +\gathbegin + C \hasparents \{ B \} +\gathnext + \patchof{B} = \pqy +\gathnext + D \isin C \equiv D \isin B \lor D = C +\end{gather} + +\subsection{Conditions} + +\[ \eqn{ Ingredients }{ + \patchof{B} = \pqn +}\] +\[ \eqn{ No Sneak }{ + \pendsof{B}{\pqy} = \{ \} +}\] + +\subsection{No Replay} + +Ingredients Prevent Replay applies. $\qed$ + +\subsection{Unique Base} + +Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$. $\qed$ + +\subsection{Tip Contents} + +Consider some arbitrary commit $D$. If $D = C$, trivially satisfied. + +If $D \neq C$, $D \isin C \equiv D \isin B$, +which by Unique Base, above, $ \equiv D \isin \baseof{B}$. +By Base Acyclic of $B$, $D \isin B \implies D \not\in \pqy$. + + +$\qed$ + +\subsection{Base Acyclic} + +Not applicable. + +\subsection{Coherence and Patch Inclusion} + +$$ +\begin{cases} + \p = \pq \lor B \haspatch \p : & C \haspatch \p \\ + \p \neq \pq \land B \nothaspatch \p : & C \nothaspatch \p +\end{cases} +$$ + +\proofstarts +~ Consider some $D \in \py$. + +\subsubsection{For $\p = \pq$:} + +By Base Acyclic, $D \not\isin B$. So $D \isin C \equiv D = C$. +By No Sneak, $D \not\le B$ so $D \le C \equiv D = C$. Thus $C \haspatch \pq$. + +\subsubsection{For $\p \neq \pq$:} + +$D \neq C$. So $D \isin C \equiv D \isin B$, +and $D \le C \equiv D \le B$. + $\qed$ -\subsection{Foreign inclusion:} +\subsection{Foreign Inclusion} + +Simple Foreign Inclusion applies. $\qed$ -If $D = C$, trivial. For $D \neq C$: -$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$ +\subsection{Foreign Contents} -\section{Anticommit} +Not applicable. -Given $L, R^+, R^-$ where -$R^+ \in \pry, R^- = \baseof{R^+}$. -Construct $C$ which has $\pr$ removed. +\section{Dependency Removal} + +Given $L$ which contains $\pr$ as represented by $R^+, R^-$. +Construct $C$ which has $\pr$ removed by applying a single +commit which is the anticommit of $\pr$. Used for removing a branch dependency. \gathbegin C \hasparents \{ L \} @@ -419,21 +673,34 @@ Used for removing a branch dependency. \subsection{Conditions} +\[ \eqn{ Ingredients }{ +R^+ \in \pry \land R^- = \baseof{R^+} +}\] +\[ \eqn{ Into Base }{ + L \in \pqn +}\] \[ \eqn{ Unique Tip }{ \pendsof{L}{\pry} = \{ R^+ \} }\] \[ \eqn{ Currently Included }{ L \haspatch \pry }\] -\[ \eqn{ Not Self }{ - L \not\in \{ R^+ \} -}\] -\subsection{No Replay} +\subsection{Ordering of Ingredients:} 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. $\qed$ +so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$. +$\qed$ + +(Note that $R^+ \not\le R^-$, i.e. the merge base +is a descendant, not an ancestor, of the 2nd parent.) + +\subsection{No Replay} + +By $\merge$, +$D \isin C \implies D \isin L \lor D \isin R^- \lor D = C$. +So, by Ordering of Ingredients, +Ingredients Prevent Replay applies. $\qed$ \subsection{Desired Contents} @@ -446,18 +713,19 @@ 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 +By No Replay for $L$, $D \not\isin L$. +Also, by Ordering of Ingredients, $D \not\le R^-$ 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, $D \le R^+ \equiv D \le L$. +By Tip Self Inpatch for $R^+$, $D \isin R^+ \equiv D \le R^+$, but by +by Unique Tip, $D \le R^+ \equiv D \le L$. So $D \isin R^+$. -By Base Acyclic, $D \not\isin R^-$. +By Base Acyclic for $R^-$, $D \not\isin R^-$. Apply $\merge$: $D \not\isin C$. OK. @@ -471,15 +739,179 @@ $\qed$ \subsection{Unique Base} -Need to consider only $C \in \py$, ie $L \in \py$. +Into Base means that $C \in \pqn$, so Unique Base is not +applicable. $\qed$ + +\subsection{Tip Contents} + +Again, not applicable. $\qed$ + +\subsection{Base Acyclic} + +By Into Base and Base Acyclic for $L$, $D \isin L \implies D \not\in \pqy$. +And by Into Base $C \not\in \pqy$. +Now from Desired Contents, above, $D \isin C +\implies D \isin L \lor D = C$, which thus +$\implies D \not\in \pqy$. $\qed$. + +\subsection{Coherence and Patch Inclusion} + +Need to consider some $D \in \py$. By Into Base, $D \neq C$. + +\subsubsection{For $\p = \pr$:} +By Desired Contents, above, $D \not\isin C$. +So $C \nothaspatch \pr$. + +\subsubsection{For $\p \neq \pr$:} +By Desired Contents, $D \isin C \equiv D \isin L$ +(since $D \in \py$ so $D \not\in \pry$). + +If $L \nothaspatch \p$, $D \not\isin L$ so $D \not\isin C$. +So $L \nothaspatch \p \implies C \nothaspatch \p$. + +Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$. +so $L \haspatch \p \implies C \haspatch \p$. + +$\qed$ + +\subsection{Foreign Inclusion} + +Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq C$. +So by Desired Contents $D \isin C \equiv D \isin L$. +By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$. + +And $D \le C \equiv D \le L$. +Thus $D \isin C \equiv D \le C$. + +$\qed$ + +\subsection{Foreign Contents} + +Not applicable. + +\section{Dependency Insertion} + +Given $L$ construct $C$ which additionally +contains $\pr$ as represented by $R^+$ and $R^-$. +This may even be used for reintroducing a previous-removed branch +dependency. +\gathbegin + C \hasparents \{ L, R^+ \} +\gathnext + \patchof{C} = \patchof{L} +\gathnext + \mergeof{C}{L}{R^-}{R^+} +\end{gather} + +\subsection{Conditions} + +\[ \eqn{ Ingredients }{ + R^- = \baseof{R^+} +}\] +\[ \eqn{ Into Base }{ + L \in \pqn +}\] +\[ \eqn{ Currently Excluded }{ + L \nothaspatch \pr +}\] +\[ \eqn{ Inserted's Ends }{ + E \in \pendsof{L}{\pry} \implies E \le R^+ +}\] +\[ \eqn{ Others' Ends }{ + \bigforall_{\p \patchisin \L} + E \in \pendsof{R^+}{\py} \implies E \le L +}\] +\[ \eqn{ Insertion Acyclic }{ + R^+ \nothaspatch \pq +}\] + +\subsection{No Replay} + +By $\merge$, +$D \isin C \implies D \isin L \lor D \isin R^+ \lor D = C$. +So Ingredients Prevent Replay applies. $\qed$ + +\subsection{Unique Base} + +Not applicable. + +\subsection{Tip Contents} + +Not applicable. + +\subsection{Base Acyclic} + +Consider some $D \isin C$. We will show that $D \not\in \pqy$. +By $\merge$, $D \isin L \lor D \isin R^+ \lor D = C$. + +For $D \isin L$, Base Acyclic for L suffices. For $D \isin R^+$, +Insertion Acyclic suffices. For $D = C$, trivial. $\qed$. + +\subsection{Coherence and Patch Inclusion} + +$$ +\begin{cases} + \p = \pr \lor L \haspatch \p : & C \haspatch \p \\ + \p \neq \pr \land L \nothaspatch \p : & C \nothaspatch \p +\end{cases} +$$ + +\proofstarts +~ Consider some $D \in \py$. +$D \neq C$ so $D \le C \equiv D \le L \lor D \le R^+$. + +\subsubsection{For $\p = \pr$:} + +$D \not\isin L$ by Currently Excluded. +$D \not\isin R^-$ by Base Acyclic. +So by $\merge$, $D \isin C \equiv D \isin R^+$, +which by Tip Self Inpatch of $R^+$, $\equiv D \le R^+$. + +And by definition of $\pancs$, +$D \le L \equiv D \in \pancsof{L}{R^+}$. +Applying Transitive Ancestors to Inserted's Ends gives +$A \in \pancsof{L}{R^+} \implies A \le R^+$. +So $D \le L \implies D \le R^+$. +Thus $D \le C \equiv D \le R^+$. + +So $D \isin C \equiv D \le C$, i.e. $C \haspatch \pr$. +OK. + +\subsubsection{For $\p \neq \pr$:} + +By Exclusive Tip Contents for $R^+$ ($D \not\in \pry$ case) +$D \isin R^+ \equiv D \isin R^-$. +So by $\merge$, $D \isin C \equiv D \isin L$. + +If $L \nothaspatch \p$, $D \not\isin L$ so $C \nothaspatch \p$. OK. + +If $L \haspatch \p$, Others' Ends applies; by Transitive +Ancestors, $A \in \pancsof{R^+}{\py} \implies A \le L$. +So $D \le R^+ \implies D \le L$, +since $D \le R^+ \equiv D \in \pancsof{R^+}{\py}$. +Thus $D \le C \equiv D \le L$. +And by $\haspatch$, $D \le L \equiv D \isin L$ so +$D \isin C \equiv D \le C$. Thus $C \haspatch \p$. +OK. + +$\qed$ + +\subsection{Foreign Inclusion} + +Consider some $D$ s.t. $\patchof{D} = \bot$. + +By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$. +So by $\merge$, $D \isin C \equiv D \isin L$. + +xxx up to here, need new condition + +$D \neq C$. -xxx tbd -xxx need to finish anticommit \section{Merge} -Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): +Merge commits $L$ and $R$ using merge base $M$: \gathbegin C \hasparents \{ L, R \} \gathnext @@ -490,14 +922,15 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$): We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$. \subsection{Conditions} - +\[ \eqn{ Ingredients }{ + M \le L, M \le R +}\] \[ \eqn{ Tip Merge }{ L \in \py \implies \begin{cases} R \in \py : & \baseof{R} \ge \baseof{L} \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\ - R \in \pn : & R \ge \baseof{L} - \land M = \baseof{L} \\ + R \in \pn : & M = \baseof{L} \\ \text{otherwise} : & \false \end{cases} }\] @@ -521,10 +954,31 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$. \bigforall_{E \in \pendsof{X}{\py}} E \le Y \right] }\] +\[ \eqn{ Foreign Merges }{ + \patchof{L} = \bot \equiv \patchof{R} = \bot +}\] + +\subsection{Non-Topbloke merges} + +We require both $\patchof{L} = \bot$ and $\patchof{R} = \bot$ +(Foreign Merges, above). +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. + +So a plain git merge of non-Topbloke branches meets the conditions and +is therefore consistent with our model. \subsection{No Replay} -See No Replay for Merge Results. +By definition of $\merge$, +$D \isin C \implies D \isin L \lor D \isin R \lor D = C$. +So, by Ingredients, +Ingredients Prevent Replay applies. $\qed$ \subsection{Unique Base} @@ -533,7 +987,7 @@ and calculate $\pendsof{C}{\pn}$. So we will consider some putative ancestor $A \in \pn$ and see whether $A \le C$. By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$. -But $C \in py$ and $A \in \pn$ so $A \neq C$. +But $C \in py$ and $A \in \pn$ so $A \neq C$. Thus $A \le C \equiv A \le L \lor A \le R$. By Unique Base of L and Transitive Ancestors, @@ -547,27 +1001,27 @@ $A \le R \equiv A \le \baseof{R}$. But by Tip Merge condition on $\baseof{R}$, $A \le \baseof{L} \implies A \le \baseof{R}$, so $A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$. -Thus $A \le C \equiv A \le \baseof{R}$. +Thus $A \le C \equiv A \le \baseof{R}$. 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 $A \le C \equiv A \le R$. +$A \le R \lor A \le \baseof{L} \equiv A \le R$. +Thus $A \le C \equiv A \le R$. 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 $D \in \py$. +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 $D \neq C$. +\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch for $L$). So $D \neq C$. Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$. \subsubsection{For $L \haspatch \p, R \haspatch \p$:} @@ -577,20 +1031,20 @@ $D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$. Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$. For $D \neq C$: $D \le C \equiv D \le L \lor D \le R - \equiv D \isin L \lor D \isin R$. + \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 $D \le X$ +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 $D \not\le C$. +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 $D \not\isin M$. +$D \not\le X$ 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 $C \haspatch \p$. @@ -598,11 +1052,12 @@ So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$. \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:} -$C \haspatch \p \equiv M \nothaspatch \p$. +$M \haspatch \p \implies C \nothaspatch \p$. +$M \nothaspatch \p \implies C \haspatch \p$. \proofstarts -One of the Merge Ends conditions 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 @@ -610,17 +1065,17 @@ 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, +Self Inpatch for $L$, $L \haspatch \p$, 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 $D \le C$. +$D \le Y$ 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 +$D \in \pancsof{X}{\py}$, 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. @@ -642,22 +1097,25 @@ $C \in \pn$ 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 $D \not\isin C$. $\qed$ +R$. And $D \neq C$. So $D \not\isin C$. + +$\qed$ \subsection{Tip Contents} -We need worry only about $C \in \py$. +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 @@ -667,30 +1125,64 @@ $C \haspatch \p$ 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 $D \not\in \py, R \in \py$:} +$D \neq C$. -%D \in \py$:} +By Tip Contents +$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. -xxx the coherence is not that useful ? +$\qed$ + +\subsection{Foreign Inclusion} -$L \haspatch \p$ by +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 need to recheck this +\subsubsection{For $D = C$:} -$C \in \py$ $C \haspatch P$ so $D \isin C \equiv D \le C$. OK. +$D \isin C$ and $D \le C$. OK. -\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:} +\subsubsection{For $D \neq C, D \isin M$:} -Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$. +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 + +$\qed$ -Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$. +\subsection{Foreign Contents} -But by Tip Merge, $\baseof{R} \ge \baseof{L}$ +Only relevant if $\patchof{L} = \bot$, in which case +$\patchof{C} = \bot$ and by Foreign Merges $\patchof{R} = \bot$, +so Totally Foreign Contents applies. $\qed$ \end{document}