X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=465fcba55b4ac0962c881302e4b632a412c59014;hp=1ca4af73574a7343d367fed94ebc945d105dc7a3;hb=eacdc5333e25e51df71366d5a55ae8a6b0c5eb86;hpb=410132e37a1f6e95cbfa9792d63b50b9390e59e4

diff --git a/merge.tex b/merge.tex
index 1ca4af7..465fcba 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}
 \[ \eqn{ Ingredients }{
@@ -46,6 +47,13 @@ $L \in \pn$, $R \in \pry$, $M = \baseof{R}$.
     \bigforall_{E \in \pendsof{X}{\py}} E \le Y
    \right]
 }\]
+\[ \eqn{ Suitable Tip }{
+    \bigexists_T
+      \pendsof{J}{\py} = \{ T \}
+     \land
+      \forall_{E \in \pendsof{K}{\py}} T \ge E
+    , \text{where} \{J,K\} = \{L,R\}
+}\]
 \[ \eqn{ Foreign Merges }{
     \patchof{L} = \bot \implies \patchof{R} = \bot
 }\]
@@ -111,14 +119,23 @@ $\qed$
 
 \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 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 $D \isin R \equiv D \le R$.
@@ -150,25 +167,21 @@ 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 $D \le C$.  OK.
 
@@ -253,6 +266,17 @@ 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 Tip is a suitable $T$ for
+Unique Tips.
+
+$\qed$
+
 \subsection{Foreign Inclusion}
 
 Consider some $D$ s.t. $\patchof{D} = \bot$.