X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=7a48a25cf994a282b5c1d74d019af0f45aafb7d2;hp=fd45642df697518317032ed9c9956c62baf7c1dd;hb=dbc2fa88cece12d33ca5788ad5d359f77676a802;hpb=5d0609c1fcad5f04f3dbac80335577173d038ce4

diff --git a/merge.tex b/merge.tex
index fd45642..7a48a25 100644
--- a/merge.tex
+++ b/merge.tex
@@ -59,10 +59,11 @@ 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 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.
+By Foreign Contents of $L$, $\patchof{M} = \bot$ as well.
+So by Foreign Contents for any $A \in \{L,M,R\}$,
+$\forall_{\p, D \in \py} D \not\le A$
+so $\pendsof{A}{\py} = \{ \}$ and the RHS of both Merge Ends
+conditions are satisifed.
 
 So a plain git merge of non-Topbloke branches meets the conditions and
 is therefore consistent with our model.
@@ -110,19 +111,29 @@ $\qed$
 
 \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 $L \haspatch \p$ by Tip Self Inpatch for $L$).  So $D \neq C$.
+\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 $D \neq C$: $D \le C \equiv D \le L \lor D \le R
  \equiv D \isin L \lor D \isin R$.
@@ -130,40 +141,40 @@ 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 $D \isin C \equiv M \nothaspatch \p \land D \le C$
-(which suffices by definition of $\haspatch$ and $\nothaspatch$).
-
-Consider $D = C$:  Thus $C \in \py, L \in \py$.  By Tip Contents
-for $L$, $L \isin L$ so $\neg [ L \nothaspatch \p ]$.
-Therefore we must have $L=Y$, $R=X$.
+various cases that
+if $M \haspatch \p$, $D \not\isin C$,
+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 $L \neq X$,
+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.
+and $D \le C$.  OK.
 
 Consider $D \neq C, M \nothaspatch \p, D \isin Y$:
 $D \le Y$ so $D \le C$.