internal notation: invent \commitmergename and use it everywhere

perl -i~ -pe 's/\$\\commitmerge\b\$/\\commitmergename/' *.tex

plus the actual definition

diff --git a/anticommit.tex b/anticommit.tex
index ee23c7f40b141f6a655aca1be2882c291ddea522..49931279917b36872d16a2f179788458ee4a7d17 100644 (file)
--- a/anticommit.tex
+++ b/anticommit.tex
@@ -37,7 +37,7 @@ is a descendant, not an ancestor, of the 2nd parent.)
 
 \subsection{No Replay}
 
-By $\commitmerge$,
+By \commitmergename,
 $D \isin C \implies D \isin L \lor D \isin R^- \lor D = C$.
 So, by Ordering of Ingredients,
 Ingredients Prevent Replay applies.  $\qed$
@@ -67,13 +67,13 @@ So $D \isin R^+$.
 
 By Base Acyclic for $R^-$, $D \not\isin R^-$.
 
-Apply $\commitmerge$: $D \not\isin C$.  OK.
+Apply \commitmergename: $D \not\isin C$.  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 $\commitmerge$: $D \isin C \equiv D \isin L$.  OK.
+Apply \commitmergename: $D \isin C \equiv D \isin L$.  OK.
 
 $\qed$
 

diff --git a/article.tex b/article.tex
index 7bff6920bd91db6129b61de48da10bb779183756..2253d43bf91abe7088d77607f70bddcd0ac957d6 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -87,6 +87,7 @@
 \newcommand{\commitmerge}{{\mathcal M}}
 \newcommand{\commitmergeof}[4]{\commitmerge(#1,#2,#3,#4)}
 %\newcommand{\merge}[4]{{#2 {{\frac{ #1 }{ #3 } #4}}}}
+\newcommand{\commitmergename}{${\mathcal M}$}
 
 \newcommand{\patch}{{\mathcal P}}
 \newcommand{\base}{{\mathcal B}}

diff --git a/merge.tex b/merge.tex
index 9022249b8456dd6b0e815a32233352a46bc6f40a..9daaa00450ae562db1d3af581b856e08de0e4af0 100644 (file)
--- a/merge.tex
+++ b/merge.tex
@@ -79,7 +79,7 @@ is therefore consistent with our model.
 
 \subsection{No Replay}
 
-By definition of $\commitmerge$,
+By definition of \commitmergename,
 $D \isin C \implies D \isin L \lor D \isin R \lor D = C$.
 So, by Ingredients,
 Ingredients Prevent Replay applies.  $\qed$
@@ -135,7 +135,7 @@ $$
 $D \not\isin L \land D \not\isin R$.  $C \not\in \py$ (otherwise $L
 \in \py$ ie $L \haspatch \p$ by Tip Own Contents for $L$).
 So $D \neq C$.
-Applying $\commitmerge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
+Applying \commitmergename gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
 OK.
 
 \subsubsection{For $L \haspatch \p, R \haspatch \p$:}
@@ -149,17 +149,17 @@ For $D \neq C$: $D \le C \equiv D \le L \lor D \le 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 $\commitmerge$, $D \isin C$.  Also $D \le X$
+By \commitmergename, $D \isin C$.  Also $D \le X$
 so $D \le C$.  OK for $C \zhaspatch \p$.
 
 Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
-By $\commitmerge$, $D \not\isin C$.
+By \commitmergename, $D \not\isin C$.
 And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
 OK for $C \zhaspatch \p$.
 
 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 $\commitmerge$, $D \isin C$.  And $D \le Y$ so $D \le C$.
+Thus by \commitmergename, $D \isin C$.  And $D \le Y$ so $D \le C$.
 OK for $C \zhaspatch \p$.
 
 So, in all cases, $C \zhaspatch \p$.
@@ -183,26 +183,26 @@ By Tip Own Contents, $L \haspatch \p$ so $L \neq X$,
 therefore we must have $L=Y$, $R=X$.
 Conversely $R \not\in \py$
 so by Tip Merge $M = \baseof{L}$.  Thus $M \in \pn$ so
-by Base Acyclic $M \nothaspatch \p$.  By $\commitmerge$, $D \isin C$,
+by Base Acyclic $M \nothaspatch \p$.  By \commitmergename, $D \isin C$,
 and $D \le C$.  OK.
 
 Consider $D \neq C, M \nothaspatch \p, D \isin Y$:
 $D \le Y$ so $D \le C$.
-$D \not\isin M$ so by $\commitmerge$, $D \isin C$.  OK.
+$D \not\isin M$ so by \commitmergename, $D \isin C$.  OK.
 
 Consider $D \neq C, M \nothaspatch \p, D \not\isin Y$:
 $D \not\le Y$.  If $D \le X$ then
 $D \in \pancsof{X}{\py}$, so by Addition Merge Ends and
 Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
-Thus $D \not\le C$.  By $\commitmerge$, $D \not\isin C$.  OK.
+Thus $D \not\le C$.  By \commitmergename, $D \not\isin C$.  OK.
 
 Consider $D \neq C, M \haspatch \p, D \isin Y$:
 $D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
 and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
-Thus $D \isin M$.  By $\commitmerge$, $D \not\isin C$.  OK.
+Thus $D \isin M$.  By \commitmergename, $D \not\isin C$.  OK.
 
 Consider $D \neq C, M \haspatch \p, D \not\isin Y$:
-By $\commitmerge$, $D \not\isin C$.  OK.
+By \commitmergename, $D \not\isin C$.  OK.
 
 $\qed$
 
@@ -242,7 +242,7 @@ $C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
 
 $D \neq C$.  By Tip Contents of $L$,
 $D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition,
-$D \isin L \equiv D \isin M$.  So by $\commitmerge$, $D \isin
+$D \isin L \equiv D \isin M$.  So by \commitmergename, $D \isin
 C \equiv D \isin R$.  And $R = \baseof{C}$ by Unique Base of $C$.
 Thus $D \isin C \equiv D \isin \baseof{C}$.  OK.
 
@@ -260,7 +260,7 @@ Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
 $\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
 $D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
 
-So $D \isin M \equiv D \isin L$ so by $\commitmerge$,
+So $D \isin M \equiv D \isin L$ so by \commitmergename,
 $D \isin C \equiv D \isin R$.  But from Unique Base,
 $\baseof{C} = \baseof{R}$.
 Therefore $D \isin C \equiv D \isin \baseof{C}$.  OK.
@@ -293,17 +293,17 @@ $D \isin C$ and $D \le C$.  OK.
 \subsubsection{For $D \neq C, D \isin M$:}
 
 Thus $D \le M$ so $D \le L$ and $D \le R$ so $D \isin L$ and $D \isin
-R$.  So by $\commitmerge$, $D \isin C$.  And $D \le C$.  OK.
+R$.  So by \commitmergename, $D \isin C$.  And $D \le C$.  OK.
 
 \subsubsection{For $D \neq C, D \not\isin M, D \isin X$:}
 
-By $\commitmerge$, $D \isin C$.
+By \commitmergename, $D \isin C$.
 And $D \isin X$ means $D \le X$ so $D \le C$.
 OK.
 
 \subsubsection{For $D \neq C, D \not\isin M, D \not\isin L, D \not\isin R$:}
 
-By $\commitmerge$, $D \not\isin C$.
+By \commitmergename, $D \not\isin C$.
 And $D \not\le L, D \not\le R$ so $D \not\le C$.
 OK