wip anticommit

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index b0a14f254ff6a6835b0269593c733b6854f21e67..01800f96132a69fb7a89ab4e9e6ab9d0a82e9519 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -32,6 +32,10 @@
 \newcommand{\py}{\pay{P}}
 \newcommand{\pn}{\pan{P}}
 
+\newcommand{\pr}{\pa{R}}
+\newcommand{\pry}{\pay{R}}
+\newcommand{\prn}{\pan{R}}
+
 %\newcommand{\hasparents}{\underaccent{1}{>}}
 %\newcommand{\hasparents}{{%
 %  \declareslashed{}{_{_1}}{0}{-0.8}{>}\slashed{>}}}
@@ -334,6 +338,29 @@ $\qed$
 If $D = C$, trivial.  For $D \neq C$:
 $D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$.  $\qed$
 
+\section{Anticommit}
+
+Given $L, R^+, R^-$ where
+$\patchof{R^+} = \pry, \patchof{R^-} = \prn$.  
+Construct $C$ which has $\pr$ removed.
+Used for removing a branch dependency.
+\gathbegin
+ C \hasparents \{ L \}
+\gathnext
+ \patchof{C} = \patchof{L}
+\gathnext
+ D \isin C \equiv
+   \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} \\
+      \text{otherwise} : & \false
+   \end{cases}
+\end{gather}
+
+xxx want to prove $D \isin C \equiv D \not\in pry \land D \isin L$.
+
 \section{Merge}
 
 Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
@@ -394,9 +421,9 @@ Need to consider only $C \in \py$, ie $L \in \py$,
 and calculate $\pendsof{C}{\pn}$.  So we will consider some
 putative ancestor $A \in \pn$ and see whether $A \le C$.
 
-$A \le C \equiv A \le L \lor A \le R \lor A = C$.
+By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$.
 But $C \in py$ and $A \in \pn$ so $A \neq C$.  
-Thus $fixme this is not really the right thing A \le L \lor A \le R$.
+Thus $A \le C \equiv A \le L \lor A \le R$.
 
 By Unique Base of L and Transitive Ancestors,
 $A \le L \equiv A \le \baseof{L}$.
@@ -408,32 +435,18 @@ $A \le R \equiv A \le \baseof{R}$.
 
 But by Tip Merge condition on $\baseof{R}$,
 $A \le \baseof{L} \implies A \le \baseof{R}$, so
-$A \le \baseof{R} \lor A \le \baseof{R} \equiv A \le \baseof{R}$.
-Thus $A \le C \equiv A \le \baseof{R}$.  Ie, $\baseof{C} =
-\baseof{R}$.
-
-UP TO HERE
+$A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
+Thus $A \le C \equiv A \le \baseof{R}$.  
+That is, $\baseof{C} = \baseof{R}$.
 
-By Tip Merge, $A \le $
+\subsubsection{For $R \in \pn$:}
 
-Let $S =
-   \begin{cases} 
-     R \in \py : & \baseof{R} \\
-     R \in \pn : & R
-   \end{cases}$.  
-Then by Tip Merge $S \ge \baseof{L}$, and $R \ge S$ so $C \ge S$.
-   
-Consider some $A \in \pn$.  If $A \le S$ then $A \le C$.
-If $A \not\le S$ then 
+By Tip Merge condition on $R$,
+$A \le \baseof{L} \implies A \le R$, so
+$A \le R \lor A \le \baseof{L} \equiv A \le R$.  
+Thus $A \le C \equiv A \le R$.  
+That is, $\baseof{C} = R$.
 
-Let $A \in \pends{C}{\pn}$.  
-Then by Calculation Of Ends $A \in \pendsof{L,\pn} \lor A \in
-\pendsof{R,\pn}$.
-
-
-
-%$\pends{C,
-
-%%\subsubsection{For $R \in \py$:}
+$\qed$
 
 \end{document}