wip merge tip contents

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index aef968a66d09117b2e396a4419f12ed55f948f95..3da2db93f9e42a580cbe201f3889de5d0037cda5 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -46,6 +46,7 @@
 
 \renewcommand{\implies}{\Rightarrow}
 \renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
 \renewcommand{\land}{\wedge}
 \renewcommand{\lor}{\vee}
 
@@ -121,7 +122,7 @@ 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 $\bot$.  
@@ -167,6 +168,18 @@ $\displaystyle D \isin C \equiv
   \end{cases}
 $ 
 
+Some (overlapping) alternative formulations:
+
+$\displaystyle D \isin C \equiv
+  \begin{cases}
+    D \isin L \equiv D \isin R     : & D = C \lor D \isin L \\
+    D \isin L \equiv D \isin R     : & D = C \lor D \isin R \\
+    D \isin L \nequiv D \isin R  : & D = C \lor D \not\isin M \\
+    D \isin M \equiv D \isin L     : & D = C \lor D \isin R \\
+    D \isin M \equiv D \isin R     : & D = C \lor D \isin L \\
+  \end{cases}
+$
+
 \end{basedescript}
 \newpage
 \section{Invariants}
@@ -488,6 +501,11 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
       \text{otherwise} : & \false
    \end{cases}
 }\]
+\[ \eqn{ Merge Acyclic }{
+    L \in \pn
+   \implies
+    R \nothaspatch \p
+}\]
 \[ \eqn{ Removal Merge Ends }{
     X \not\haspatch \p \land
     Y \haspatch \p \land
@@ -616,14 +634,54 @@ 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}
 
+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 unique base,
+and coherence and patch inclusion, of $C$ as just proved.
+
+Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+\p$ and by coherence/inclusion $C \haspatch \p$ .  If $R \not\in \py$
+then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
+of $\nothaspatch$, $M \nothaspatch \p$.  So by coherence/inclusion $C
+\haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
+
 We will consider some $D$ and prove the Exclusive Tip Contents form.
-We use the Coherence of $C$ as just proved.
+
+\subsubsection{For $D \in \py$:}
+$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
+\le C$.  OK.
+
+\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 $\merge$, $D \isin
+C \equiv D \isin R$.  And $R = \baseof{C}$ by Unique Base of $C$.
+Thus $D \isin C \equiv D \isin \baseof{C}$.  OK.
+
+\subsubsection{For $D \not\in \py, R \in \py$:}
+
+xxx up to here
+
+%D \in \py$:}
+
+
 
 xxx the coherence is not that useful ?
 
-\subsubsection{For $L \in \py, D \in \py$:}
+$L \haspatch \p$ by 
 
 xxx need to recheck this