X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=article.tex;h=39a12875eb76a7266a477d76063a98ac6e5457de;hb=dc267ea3dd7ebf73ee7b8c000c483e69207153c5;hp=dd4538988cdb51cc42c6dee6ab12edaf486126db;hpb=d2c96fc2700752ce2ad9e29032b6fd7581194f32;p=topbloke-formulae.git

diff --git a/article.tex b/article.tex
index dd45389..39a1287 100644
--- a/article.tex
+++ b/article.tex
@@ -65,8 +65,8 @@
 \newcommand{\patchof}[1]{\patch ( #1 ) }
 \newcommand{\baseof}[1]{\base ( #1 ) }
 
+\newcommand{\eqntag}[2]{ #2 \tag*{\mbox{#1}} }
 \newcommand{\eqn}[2]{ #2 \tag*{\mbox{\bf #1}} }
-\newcommand{\corrolary}[1]{ #1 \tag*{\mbox{\it Corrolary.}} }
 
 %\newcommand{\bigforall}{\mathop{\hbox{\huge$\forall$}}}
 \newcommand{\bigforall}{%
@@ -121,7 +121,7 @@ A patch $\p$ consists of two sets of commits $\pn$ and $\py$, which
 are respectively the base and tip git branches.  $\p$ may be used
 where the context requires a set, in which case the statement
 is to be taken as applying to both $\py$ and $\pn$.
-All these sets are distinct.  Hence:
+None of these sets overlap.  Hence:
 
 \item[ $ \patchof{ C } $ ]
 Either $\p$ s.t. $ C \in \p $, or $\bot$.  
@@ -206,7 +206,7 @@ Ie, the two limbs of the RHS of Tip Contents are mutually exclusive.
 Let $B = \baseof{C}$ in $D \isin \baseof{C}$.  Now $B \in \pn$.
 So by Base Acyclic $D \isin B \implies D \notin \py$.
 }
-\[ \corrolary{
+\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
   \bigforall_{C \in \py} D \isin C \equiv
   \begin{cases}
     D \in \py : & D \le C \\
@@ -488,16 +488,25 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
       \text{otherwise} : & \false
    \end{cases}
 }\]
-\[ \eqn{ Merge Ends }{
+\[ \eqn{ Merge Acyclic }{
+    L \in \pn
+   \implies
+    R \nothaspatch \p
+}\]
+\[ \eqn{ Removal Merge Ends }{
     X \not\haspatch \p \land
-    Y \haspatch \p
+    Y \haspatch \p \land
+    M \haspatch \p
   \implies
-  \begin{cases}
-    M \haspatch \p :    & \displaystyle
-                          \bigforall_{E \in \pendsof{Y}{\py}} E \le M \\
-    M \nothaspatch \p : & \displaystyle
-                          \bigforall_{E \in \pendsof{X}{\py}} E \le Y
-  \end{cases}
+    \pendsof{Y}{\py} = \pendsof{M}{\py}
+}\]
+\[ \eqn{ Addition Merge Ends }{
+    X \not\haspatch \p \land
+    Y \haspatch \p \land
+    M \nothaspatch \p
+   \implies \left[
+    \bigforall_{E \in \pendsof{X}{\py}} E \le Y
+   \right]
 }\]
 
 \subsection{No Replay}
@@ -580,7 +589,8 @@ $C \haspatch \p \equiv M \nothaspatch \p$.
 
 \proofstarts
 
-Merge Ends applies.  Recall that we are considering $D \in \py$.
+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 $D \isin C \equiv M \nothaspatch \p \land D \le C$
@@ -597,16 +607,74 @@ $D \not\isin M$ so by $\merge$, $D \isin C$.  OK.
 
 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 Merge Ends and 
+$D \in \pancsof{X}{\py}$, 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 $D \neq C, M \haspatch P, D \isin Y$:
-$D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Merge Ends
-and Transitive Ancestors $D \le M$.
+$D \le Y$ 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 $D \neq C, M \haspatch P, D \not\isin Y$:
 By $\merge$, $D \not\isin C$.  OK.
 
+$\qed$
+
+\subsection{Base Acyclic}
+
+This applies when $C \in \pn$.
+$C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
+
+Consider some $D \in \py$.
+
+By Base Acyclic of $L$, $D \not\isin L$.  By the above, $D \not\isin
+R$.  And $D \neq C$.  So $D \not\isin C$.  $\qed$
+
+\subsection{Tip Contents}
+
+xxx up to here
+
+We need worry only about $C \in \py$.  
+And $\patchof{C} = \patchof{L}$
+so $L \in \py$ so $L \haspatch \p$.  We will use the coherence and
+patch inclusion of $C$ as just proved.
+
+Firstly we prove $C \haspatch \p$: If $R \in \py$, this is true by
+coherence/inclusion $C \haspatch \p$.  If $R \not\in \py$ then
+by Tip Merge 
+
+
+We will consider some $D$ and prove the Exclusive Tip Contents form.
+
+
+So by definition of
+$\haspatch$, $D \isin C \equiv D \le C$.  OK.
+
+\subsubsection{For $L \in \py, D \in \py, $:}
+$R \haspatch \p$ so 
+
+\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+
+
+%D \in \py$:}
+
+
+
+xxx the coherence is not that useful ?
+
+$L \haspatch \p$ by 
+
+xxx need to recheck this
+
+$C \in \py$ $C \haspatch P$ so $D \isin C \equiv D \le C$.  OK.
+
+\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+
+Tip Contents for $L$: $D \isin L \equiv D \isin \baseof{L}$.
+
+Tip Contents for $R$: $D \isin R \equiv D \isin \baseof{R}$.
+
+But by Tip Merge, $\baseof{R} \ge \baseof{L}$
+
 \end{document}