X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=pseudomerge.tex;h=9ea20452abed952f324ac3a32e9e6ade913de286;hb=28ed5bb07153a37a5e48c9bfa7498eb48b946abf;hp=1e9422c447b6f143837553c40a9afefe2cfeeed4;hpb=05d6e35d69c91ddf88e8d5a5febea740bbfac22f;p=topbloke-formulae.git

diff --git a/pseudomerge.tex b/pseudomerge.tex
index 1e9422c..9ea2045 100644
--- a/pseudomerge.tex
+++ b/pseudomerge.tex
@@ -15,7 +15,16 @@ but whose contents are exactly those of $L$.
 \subsection{Conditions}
 
 \[ \eqn{ Base Only }{
- \patchof{L} \neq \py
+ L \in \pn
+}\]
+
+\[ \eqn{ Ingredients }{
+ \bigforall_{R \in \set R}
+    R \in \pn
+     \lor
+    R \in \foreign
+     \lor
+    R \in \pqy
 }\]
 
 \[ \eqn{ Unique Tips }{
@@ -25,25 +34,22 @@ but whose contents are exactly those of $L$.
 }\]
 
 \[ \eqn{ Foreign Unaffected }{
- \bigforall_{ D \text{ s.t. } \patchof{D} = \bot }
-  \left[ \bigexists_{A \in \set A} D \le A \right]
-   \implies
-  D \le L
+ \pendsof{C}{\foreign} = \pendsof{L}{\foreign}
 }\]
- TODO THAT IS IMPOSSIBLE TO CALCULATE
 
 \subsection{Lemma: Foreign Identical}
 
-$\patchof{D} = \bot \implies \big[ D \le C \equiv D \le L \big]$.
+$\isforeign{D} \implies \big[ D \le C \equiv D \le L \big]$.
 
 \proof{
-If $D \le L$, trivially $D \le C$; so conversely
-$D \not\le C \implies D \not\le L$.  
-Whereas if $D \le C$, either $D \le L$ or
-$\exists{A \in \set A} D \le A$ (since $D \neq C$),
-in which case by Foreign Unaffected $D \le L$.
+Trivial by Foreign Unaffected and the definition of $\pends$
 }
 
+It might seem that bare git commits might also be psuedo-merges ---
+e.g., merges made directly with {\tt git merge -s ours}.  However, by
+our definition of $\has$, these are considered simply as normal merges
+(\autoref{commit-merge}).
+
 \subsection{No Replay}
 
 Ingredients Prevent Replay applies:
@@ -87,3 +93,34 @@ True by Foreign Identical, and Foreign Inclusion of $L$.
 \subsection{Foreign Contents}
 
 Not applicable.
+
+\subsection{Bases' Children}
+
+We need to consider this for $D=L$ and also for $D=R$ ($R \in \set
+R$).
+
+For $D=L$, if $L \in \pn$ then $C \in \pn$, OK; whereas if
+$L \not \in \pn$ Bases' Children is inapplicable.
+
+For $D=R$, 
+xxx up to here?
+
+If $L \in \py, R \in \py$: not applicable for either $D=L$ or $D=R$.
+
+If $L \in \py, R \in \pn$: not applicable for $L$, OK for $R$.
+
+Other possibilities for $L \in \py$ are excluded by Tip Merge.
+
+If $L \in \pn, R \in \pn$: satisfied for both $L$ and $R$.
+
+If $L \in \pn, R \in \foreign$: satisfied for $L$, not applicable for
+$R$.
+
+If $L \in \pn, R \in \pqy$: satisfied for $L$, not applicable for
+$R$.
+
+Other possibilities for $L \in \pn$ are excluded by Base Merge.
+
+If $L \in \foreign$: not applicable for $L$; nor for $R$, by Foreign Merges.
+
+