move merge X and Y def up

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index b71bb326c09fcf3eb8387c075b3f7ded903087b6..a0adacbdee9b8d02bb1f2421f8158b1a43b505b2 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -10,6 +10,8 @@
 
 \renewcommand{\ge}{\geqslant}
 \renewcommand{\le}{\leqslant}
+\newcommand{\nge}{\ngeqslant}
+\newcommand{\nle}{\nleqslant}
 
 \newcommand{\has}{\sqsupseteq}
 \newcommand{\isin}{\sqsubseteq}
@@ -53,7 +55,8 @@
 \newcommand{\pancsof}[2]{\pancs ( #1 , #2 ) }
 \newcommand{\pendsof}[2]{\pends ( #1 , #2 ) }
 
-\newcommand{\merge}[4]{{\mathcal M}(#1,#2,#3,#4)}
+\newcommand{\merge}{{\mathcal M}}
+\newcommand{\mergeof}[4]{\merge(#1,#2,#3,#4)}
 %\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
 
 \newcommand{\patch}{{\mathcal P}}
@@ -78,7 +81,8 @@
 \newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}
 
 \newcommand{\qed}{\square}
-\newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
+\newcommand{\proofstarts}{{\it Proof:}}
+\newcommand{\proof}[1]{\proofstarts #1 $\qed$}
 
 \newcommand{\gathbegin}{\begin{gather} \tag*{}}
 \newcommand{\gathnext}{\\ \tag*{}}
@@ -152,7 +156,7 @@ patch is applied to a non-Topbloke branch and then bubbles back to
 the Topbloke patch itself, we hope that git's merge algorithm will
 DTRT or that the user will no longer care about the Topbloke patch.
 
-\item[ $\displaystyle \merge{C}{L}{M}{R} $ ]
+\item[ $\displaystyle \mergeof{C}{L}{M}{R} $ ]
 The contents of a git merge result:
 
 $\displaystyle D \isin C \equiv
@@ -259,15 +263,15 @@ XXX proof TBD.
 
 \subsection{No Replay for Merge Results}
 
-If we are constructing $C$, given
+If we are constructing $C$, with,
 \gathbegin
-  \merge{C}{L}{M}{R}
+  \mergeof{C}{L}{M}{R}
 \gathnext
   L \le C
 \gathnext
   R \le C
 \end{gather}
-No Replay is preserved.  {\it Proof:}
+No Replay is preserved.  \proofstarts
 
 \subsubsection{For $D=C$:} $D \isin C, D \le C$.  OK.
 
@@ -397,7 +401,7 @@ Used for removing a branch dependency.
 \gathnext
  \patchof{C} = \patchof{L}
 \gathnext
- \merge{C}{L}{R^+}{R^-}
+ \mergeof{C}{L}{R^+}{R^-}
 \end{gather}
 
 \subsection{Conditions}
@@ -420,19 +424,37 @@ Merge Results applies. $\qed$
 
 \subsection{Desired Contents}
 
-\[ $D \isin C \equiv [ D \not\in \pry \land D \isin L$ ] \lor D = C \]
-{\it Proof.}
+\[ D \isin C \equiv [ D \notin \pry \land D \isin L ] \lor D = C \]
+\proofstarts
 
 \subsubsection{For $D = C$:}
 
 Trivially $D \isin C$.  OK.
 
-\subsubsection{For $D \not\le C$:}
+\subsubsection{For $D \neq C, D \not\le L$:}
 
+By No Replay $D \not\isin L$.  Also $D \not\le R^-$ hence
+$D \not\isin R^-$.  Thus $D \not\isin C$.  OK.
 
+\subsubsection{For $D \neq C, D \le L, D \in \pry$:}
 
-\subsubsection{For $D \in R^+$:}
-By Currently Included, 
+By Currently Included, $D \isin L$.
+
+By Tip Self Inpatch, $D \isin R^+ \equiv D \le R^+$, but by
+by Unique Tip, $D \le R^+ \equiv D \le L$.  
+So $D \isin R^+$.
+
+By Base Acyclic, $D \not\isin R^-$.
+
+Apply $\merge$: $D \not\isin C$.  OK.
+
+\subsubsection{For $D \neq C, D \le L, D \notin \pry$:}
+
+By Tip Contents for $R^+$, $D \isin R^+ \equiv D \isin R^-$.
+
+Apply $\merge$: $D \isin C \equiv D \isin L$.  OK.
+
+$\qed$
 
 \subsection{Unique Base}
 
@@ -440,6 +462,8 @@ Need to consider only $C \in \py$, ie $L \in \py$.
 
 xxx tbd
 
+xxx need to finish anticommit
+
 \section{Merge}
 
 Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
@@ -448,8 +472,9 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 \gathnext
  \patchof{C} = \patchof{L}
 \gathnext
- \merge{C}{L}{M}{R}
+ \mergeof{C}{L}{M}{R}
 \end{gather}
+We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
 
 \subsection{Conditions}
 
@@ -502,4 +527,40 @@ That is, $\baseof{C} = R$.
 
 $\qed$
 
+\subsection{Coherence and patch inclusion}
+
+Need to determine $C \haspatch P$ based on $L,M,R \haspatch P$.
+This involves considering $D \in \py$.  
+
+\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 Self Inpatch).  So $D \neq C$.
+Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch P$.
+
+\subsubsection{For $L \haspatch P, R \haspatch P$:}
+$D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
+(Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.)
+
+Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch P$.
+
+For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
+ \equiv D \isin L \lor D \isin R$.  
+(Likewise $D \le C \equiv D \le X \lor D \le Y$.)
+
+Consider $D \neq C, D \isin X \land D \isin Y$:
+By $\merge$, $D \isin C$.  Also $D \le X$ 
+so $D \le C$.  OK for $C \haspatch P$.
+
+Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
+By $\merge$, $D \not\isin C$.  
+And $D \not\le X \land D \not\le Y$ so $D \not\le C$.  
+OK for $C \haspatch P$.
+
+Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
+$D \not\le X$ so $D \not\le M$ so $D \not\isin M$.  
+Thus by $\merge$, $D \isin C$.  And $D \le Y$ so $D \le C$.
+OK for $C \haspatch P$.
+
+So indeed $L \haspatch P \land R \haspatch P \implies C \haspatch P$.
+
 \end{document}