X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=6ff5d81e51d67193858caf678fa0a1185a60c7fd;hp=5dc99ee548760ebeb80f908a8d1dd959416bd928;hb=588087cf4ec7d7c90cb21ac1b3790b86283ba992;hpb=23862a524934c97cc30445f0250c943df589c729 diff --git a/merge.tex b/merge.tex index 5dc99ee..6ff5d81 100644 --- a/merge.tex +++ b/merge.tex @@ -10,8 +10,9 @@ Merge commits $L$ and $R$ using merge base $M$: \end{gather} We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$. -This can also be used for dependency re-insertion, by setting -$L \in \pn$, $R \in \pry$, $M = \baseof{R}$. +This can also be used for dependency re-insertion, by setting $L \in +\pn$, $R \in \pry$, $M = \baseof{R}$, provided that the Conditions are +satisfied; in particular, provided that $L \ge \baseof{R}$. \subsection{Conditions} \[ \eqn{ Ingredients }{ @@ -46,6 +47,14 @@ $L \in \pn$, $R \in \pry$, $M = \baseof{R}$. \bigforall_{E \in \pendsof{X}{\py}} E \le Y \right] }\] +\[ \eqn{ Suitable Tips }{ + \bigforall_{\p \neq \patchof{L}, \; C \haspatch \p} + \bigexists_T + \pendsof{J}{\py} = \{ T \} + \land + \forall_{E \in \pendsof{K}{\py}} T \ge E + , \text{where} \{J,K\} = \{L,R\} +}\] \[ \eqn{ Foreign Merges }{ \patchof{L} = \bot \implies \patchof{R} = \bot }\] @@ -111,20 +120,29 @@ $\qed$ \subsection{Coherence and Patch Inclusion} -Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$. -This involves considering $D \in \py$. +$$ +\begin{cases} + L \nothaspatch \p \land R \nothaspatch \p : & C \nothaspatch \p \\ + L \haspatch \p \land R \haspatch \p : & C \haspatch \p \\ + \text{otherwise} \land M \haspatch \p : & C \nothaspatch \p \\ + \text{otherwise} \land M \nothaspatch \p : & C \haspatch \p +\end{cases} +$$ +\proofstarts +~ Consider $D \in \py$. \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 $\neg[ L \nothaspatch \p ]$ by Tip Own Contents for $L$). +\in \py$ ie $L \haspatch \p$ by Tip Own Contents for $L$). So $D \neq C$. Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$. +OK. \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 $C \zhaspatch \p$. For $D \neq C$: $D \le C \equiv D \le L \lor D \le R \equiv D \isin L \lor D \isin R$. @@ -132,41 +150,41 @@ 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 $C \zhaspatch \p$. 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 $C \zhaspatch \p$. 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 $C \zhaspatch \p$. -So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$. +So, in all cases, $C \zhaspatch \p$. +And by $L \haspatch \p$, $\exists_{F \in \py} F \le L$ +and this $F \le C$ so indeed $C \haspatch \p$. \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:} -$M \haspatch \p \implies C \nothaspatch \p$. -$M \nothaspatch \p \implies C \haspatch \p$. - -\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 if $M \haspatch \p$, $D \not\isin C$, -whereas if $M \nothaspatch \p$, $D \isin C \equiv \land D \le C$. +whereas if $M \nothaspatch \p$, $D \isin C \equiv D \le C$. +And by $Y \haspatch \p$, $\exists_{F \in \py} F \le Y$ and this +$F \le C$ so this suffices. Consider $D = C$: Thus $C \in \py, L \in \py$. -By Tip Own Contents, $\neg[ L \nothaspatch \p ]$ so $L \neq X$, +By Tip Own Contents, $L \haspatch \p$ so $L \neq X$, therefore we must have $L=Y$, $R=X$. -By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so +Conversely $R \not\in \py$ +so by Tip Merge $M = \baseof{L}$. Thus $M \in \pn$ so by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$, -and $D \le C$, consistent with $C \haspatch \p$. OK. +and $D \le C$. OK. Consider $D \neq C, M \nothaspatch \p, D \isin Y$: $D \le Y$ so $D \le C$. @@ -249,6 +267,17 @@ Therefore $D \isin C \equiv D \isin \baseof{C}$. OK. $\qed$ +\subsection{Unique Tips} + +For $L \in \py$, trivially $\pendsof{C}{\py} = C$ so $T = C$ is +suitable. + +For $L \not\in \py$, $\pancsof{C}{\py} = \pancsof{L}{\py} \cup +\pancsof{R}{\py}$. So $T$ from Suitable Tips is a suitable $T$ for +Unique Tips. + +$\qed$ + \subsection{Foreign Inclusion} Consider some $D$ s.t. $\patchof{D} = \bot$.