X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=f5038213bc5ce7b97b067cc00fcaa63d5ac0bd76;hp=f153811f3e7864b0ebad85fb3db274f9adc96099;hb=670d1b8cb1403203123ffd2d6c510f82bb7a335c;hpb=cdbdcb99ffda2c58f9f41e444cbc385e11cd0ab8 diff --git a/merge.tex b/merge.tex index f153811..f503821 100644 --- a/merge.tex +++ b/merge.tex @@ -59,7 +59,9 @@ 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 +And by Foreign Contents for (wlog) Y, $\forall_{\p, D \in \py} D \not\le Y$ +so then by No Replay $D \not\isin Y$ +so $\neg [ Y \haspatch \p ]$ so neither Merge Ends condition applies. So a plain git merge of non-Topbloke branches meets the conditions and @@ -79,7 +81,7 @@ 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, @@ -98,7 +100,7 @@ That is, $\baseof{C} = \baseof{R}$. \subsubsection{For $R \in \pn$:} -By Tip Merge condition on $R$ and since $M \le R$, +By Tip Merge condition 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 $A \le C \equiv A \le R$. @@ -156,27 +158,29 @@ 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 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 $D = C$: Thus $C \in \py, L \in \py$. By Tip Contents +for $L$, $L \isin L$ so $\neg [ L \nothaspatch \p ]$. +Therefore we must have $L=Y$, $R=X$. +By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so +by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$, +and $D \le C$, consistent with $C \haspatch \p$. OK. -Consider $D \neq C, M \nothaspatch P, D \isin Y$: +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 $D \neq C, M \nothaspatch P, D \not\isin Y$: +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 $D \neq C, M \haspatch P, D \isin Y$: +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 $D \neq C, M \haspatch P, D \not\isin Y$: +Consider $D \neq C, M \haspatch \p, D \not\isin Y$: By $\merge$, $D \not\isin C$. OK. $\qed$ @@ -216,8 +220,8 @@ $C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D \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 $\merge$, $D \isin +$D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition, +$D \isin L \equiv D \isin M$. So by $\merge$, $D \isin C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$. Thus $D \isin C \equiv D \isin \baseof{C}$. OK.