clarify by swapping two vars

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index d85a022ff8fb8388f380325f4ffd4ed9e4330d5d..642190f9f823431a058f3aa1b5793bc3f1751396 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -59,6 +59,8 @@
     {\hbox{\scriptsize$\forall$}}}%
 }
 
+\newcommand{\Largeexists}{\mathop{\hbox{\Large$\exists$}}}
+\newcommand{\Largenexists}{\mathop{\hbox{\Large$\nexists$}}}
 
 \newcommand{\qed}{\square}
 \newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
@@ -114,7 +116,7 @@ which are in $\set P$.
 \item[ $ \pendsof{C}{\set P} $ ]
 $ \{ E \; | \; E \in \pancsof{C}{\set P}
   \land \mathop{\not\exists}_{A \in \pancsof{C}{\set P}}
-  A \neq E \land E \le A \} $ 
+  E \neq A \land E \le A \} $ 
 i.e. all $\le$-maximal commits in $\pancsof{C}{\set P}$.
 
 \item[ $ \baseof{C} $ ]
@@ -334,15 +336,40 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 
 \subsection{Conditions}
 
-\[ \eqn{ Merges Exhaustive }{
- L \in \py => \Bigl[ R \in \py \lor R \in \pn \Bigr]
-}\]
 \[ \eqn{ Tip Merge }{
- L \in \py \land R \in \py \implies \Bigl[ \text{TBD} \Bigr]
-}\]
-\[ \eqn{ Base Merge }{
- L \in \py \land R \in \pn \implies \Bigl[ R \ge \baseof{L} \land M =
-   \baseof{L} \Bigr]
+ L \in \py \implies
+   \begin{cases}
+      R \in \py : & \baseof{R} \ge \baseof{L}
+              \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
+      R \in \pn : & R \ge \baseof{L}
+              \land M = \baseof{L} \\
+      \text{otherwise} : & \false
+   \end{cases}
 }\]
 
+\subsection{No Replay}
+
+\subsubsection{For $D=C$:} $D \isin C, D \le C$.  OK.
+
+\subsubsection{For $D \isin L \land D \isin R$:}
+$D \isin C$.  And $D \isin L \implies D \le L \implies D \le C$.  OK.
+
+\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
+$D \not\isin C$.  OK.
+
+\subsubsection{For $D \neq C \land D \not\isin L \land D \not\isin R$:}
+$D \not\isin C$.  OK.
+
+\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
+ \land D \not\isin M$:}
+$D \isin C$.  Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
+R$ so $D \le C$.  OK.
+
+\subsubsection{For $D \neq C \land (D \isin L \equiv D \not\isin R)
+ \land D \isin M$:}
+$D \not\isin C$.  Also $D \isin L \lor D \isin R$ so $D \le L \lor D \le
+R$ so $D \le C$.  OK.
+
+$\qed$
+
 \end{document}