merge tip contentss: clarify arbitrariness of D

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index cbd56840b63d5579073f30786bcac5ed76fa1773..faf41526029fb4b0c489e4b8fedfe072d7f62dad 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}
 
@@ -65,8 +66,8 @@
 \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 +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}
@@ -206,7 +219,7 @@ 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{
+\[ \eqntag{{\it Corollary - equivalent to Tip Contents}}{
   \bigforall_{C \in \py} D \isin C \equiv
   \begin{cases}
     D \in \py : & D \le C \\
@@ -483,11 +496,15 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
    \begin{cases}
       R \in \py : & \baseof{R} \ge \baseof{L}
               \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
-      R \in \pn : & R \ge \baseof{L}
-              \land M = \baseof{L} \\
+      R \in \pn : & M = \baseof{L} \\
       \text{otherwise} : & \false
    \end{cases}
 }\]
+\[ \eqn{ Merge Acyclic }{
+    L \in \pn
+   \implies
+    R \nothaspatch \p
+}\]
 \[ \eqn{ Removal Merge Ends }{
     X \not\haspatch \p \land
     Y \haspatch \p \land
@@ -534,7 +551,7 @@ That is, $\baseof{C} = \baseof{R}$.
 
 \subsubsection{For $R \in \pn$:}
 
-By Tip Merge condition on $R$,
+By Tip Merge condition on $R$ and since $M \le R$,
 $A \le \baseof{L} \implies A \le R$, so
 $A \le R \lor A \le \baseof{L} \equiv A \le R$.  
 Thus $A \le C \equiv A \le R$.  
@@ -542,7 +559,7 @@ That is, $\baseof{C} = R$.
 
 $\qed$
 
-\subsection{Coherence and patch inclusion}
+\subsection{Coherence and Patch Inclusion}
 
 Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
 This involves considering $D \in \py$.  
@@ -616,4 +633,66 @@ 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
+of $C$, and its Coherence and Patch Inclusion, 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 an arbitrary commit $D$
+and prove the Exclusive Tip Contents form.
+
+\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$:}
+
+
+
+xxx the coherence is not that useful ?
+
+$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}