X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=article.tex;h=b26a120098b953606abe0fdee77d83abe9543735;hp=3efe2f28afecbb0a804a5fd83938b33c70819cd8;hb=e855f75b357a56be5f66cd8adb8deb08b10c55a9;hpb=c0cde0bfdd200f540e0ca3c8c4c78a8888b77b67 diff --git a/article.tex b/article.tex index 3efe2f2..b26a120 100644 --- a/article.tex +++ b/article.tex @@ -109,7 +109,7 @@ $\set X$. \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 $ ] @@ -129,18 +129,18 @@ is to be taken as applying to both $\py$ and $\pn$. None of these sets overlap. 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} $ ] @@ -154,13 +154,13 @@ $\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 $\p$. 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 relevant Topbloke branches, we hope that +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. @@ -173,7 +173,7 @@ $\displaystyle D \isin C \equiv (D \not\isin L \land D \not\isin R) \land D \neq C : & \false \\ \text{otherwise} : & D \not\isin M \end{cases} -$ +$ \end{basedescript} \newpage @@ -218,14 +218,29 @@ We maintain these each time we construct a new commit. \\ \text{as above with L and R exchanged} \end{cases} }\] -\proof{ - Truth table xxx - - Original definition is symmetrical in $L$ and $R$. +\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 $R$. } \[ \eqn{Exclusive Tip Contents:}{ - \bigforall_{C \in \py} + \bigforall_{C \in \py} \neg \Bigl[ D \isin \baseof{C} \land ( D \in \py \land D \le C ) \Bigr] }\] @@ -260,7 +275,7 @@ $ \bigforall_{C \in \py}\bigforall_{D \in \py} ( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R ) \lor D = C }\] -xxx proof tbd +\proof{ ~ Trivial.} \[ \eqn{Transitive Ancestors:}{ \left[ \bigforall_{ E \in \pendsof{C}{\set P} } E \le M \right] \equiv @@ -270,7 +285,7 @@ xxx proof tbd \proof{ The implication from right to left is trivial because $ \pends() \subset \pancs() $. -For the implication from left to right: +For the implication from left to right: by the definition of $\mathcal E$, for every such $A$, either $A \in \pends()$ which implies $A \le M$ by the LHS directly, @@ -288,14 +303,23 @@ by the LHS. And $A \le A''$. \\ C \not\in \p : & \displaystyle \left\{ E \Big| - \Bigl[ \Largeexists_{A \in \set A} + \Bigl[ \Largeexists_{A \in \set A} E \in \pendsof{A}{\set P} \Bigr] \land - \Bigl[ \Largenexists_{B \in \set A} - E \neq B \land E \le B \Bigr] + \Bigl[ \Largenexists_{B \in \set A, F \in \pendsof{B}{\p}} + E \neq F \land E \le F \Bigr] \right\} \end{cases} }\] -xxx proof tbd +\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 $\pends$. +} \[ \eqn{Ingredients Prevent Replay:}{ \left[ @@ -317,6 +341,24 @@ xxx proof tbd $A \le C$ so $D \le C$. } +\[ \eqn{Simple Foreign Inclusion:}{ + \left[ + C \hasparents \{ L \} + \land + \bigforall_{D} D \isin C \equiv D \isin L \lor D = C + \right] + \implies + \bigforall_{D \text{ s.t. } \patchof{D} = \bot} + D \isin C \equiv D \le C +}\] +\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 $D \isin C \equiv D \le C$. +} + \[ \eqn{Totally Foreign Contents:}{ \bigforall_{C \hasparents \set A} \left[ @@ -397,7 +439,7 @@ We need to consider only $A, C \in \py$. From Tip Contents for $A$: Substitute into the contents of $C$: \[ D \isin C \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A ) \lor D = C \] -Since $D = C \implies D \in \py$, +Since $D = C \implies D \in \py$, and substituting in $\baseof{C}$, this gives: \[ D \isin C \equiv D \isin \baseof{C} \lor (D \in \py \land D \le A) \lor @@ -411,7 +453,7 @@ $\qed$ \subsection{Base Acyclic} -Need to consider only $A, C \in \pn$. +Need to consider only $A, C \in \pn$. For $D = C$: $D \in \pn$ so $D \not\in \py$. OK. @@ -443,7 +485,7 @@ So: \subsubsection{For $A \nothaspatch P$:} -Firstly, $C \not\in \py$ since if it were, $A \in \py$. +Firstly, $C \not\in \py$ since if it were, $A \in \py$. Thus $D \neq C$. Now by contents of $A$, $D \notin A$, so $D \notin C$. @@ -452,10 +494,9 @@ So: \[ A \nothaspatch P \implies C \nothaspatch P \] $\qed$ -\subsection{Foreign inclusion:} +\subsection{Foreign Inclusion:} -If $D = C$, trivial. For $D \neq C$: -$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$. $\qed$ +Simple Foreign Inclusion applies. $\qed$ \subsection{Foreign Contents:} @@ -513,11 +554,7 @@ $\qed$. \subsection{Foreign Inclusion} -Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq B$ -so $D \isin B \equiv D \isin L$. -By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$. -And by Exact Ancestors $D \le L \equiv D \le B$. -So $D \isin B \equiv D \le B$. $\qed$ +Simple Foreign Inclusion applies. $\qed$ \subsection{Foreign Contents} @@ -549,7 +586,7 @@ Ingredients Prevent Replay applies. $\qed$ \subsection{Unique Base} -Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$. +Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$. $\qed$ \subsection{Tip Contents} @@ -567,14 +604,35 @@ Not applicable. \subsection{Coherence and Patch Inclusion} -Consider some $D \in \py$. +$$ +\begin{cases} + \p = \pq \lor B \haspatch \p : & C \haspatch \p \\ + \p \neq \pq \land B \nothaspatch \p : & C \nothaspatch \p +\end{cases} +$$ + +\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, $D \le B \equiv D = C$. Thus $C \haspatch \pq$. -xxx up to here +\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} + +Simple Foreign Inclusion applies. $\qed$ + +\subsection{Foreign Contents} + +Not applicable. \section{Anticommit} @@ -639,7 +697,7 @@ $D \not\isin R^-$. Thus $D \not\isin C$. OK. 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 Unique Tip, $D \le R^+ \equiv D \le L$. So $D \isin R^+$. By Base Acyclic, $D \not\isin R^-$. @@ -784,7 +842,7 @@ 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, @@ -798,15 +856,15 @@ $A \le R \equiv A \le \baseof{R}$. 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 $R$ 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, $\baseof{C} = R$. $\qed$ @@ -814,7 +872,7 @@ $\qed$ \subsection{Coherence and Patch Inclusion} Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$. -This involves considering $D \in \py$. +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 @@ -828,20 +886,20 @@ $D \isin L \equiv D \le L$ 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 $C \haspatch \p$. @@ -854,7 +912,7 @@ $M \nothaspatch \p \implies C \haspatch \p$. \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 @@ -867,12 +925,12 @@ $M=\baseof{L}$. 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. @@ -900,7 +958,7 @@ $\qed$ \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 Unique Base of $C$, and its Coherence and Patch Inclusion, as just proved.