wip traversal

[topbloke-formulae.git] / anticommit.tex

diff --git a/anticommit.tex b/anticommit.tex
index eab42047d6d7cb0163a5b134b3b8ce9bd75f336a..49931279917b36872d16a2f179788458ee4a7d17 100644 (file)
--- a/anticommit.tex
+++ b/anticommit.tex
@@ -8,7 +8,7 @@ Used for removing a branch dependency.
 \gathnext
  \patchof{C} = \patchof{L}
 \gathnext
- \mergeof{C}{L}{R^+}{R^-}
+ \commitmergeof{C}{L}{R^+}{R^-}
 \end{gather}
 
 \subsection{Conditions}
@@ -19,7 +19,7 @@ R^+ \in \pry \land R^- = \baseof{R^+}
 \[ \eqn{ Into Base }{
  L \in \pln
 }\]
-\[ \eqn{ Unique Tip }{
+\[ \eqn{ Correct Tip }{
  \pendsof{L}{\pry} = \{ R^+ \}
 }\]
 \[ \eqn{ Currently Included }{
@@ -28,7 +28,7 @@ R^+ \in \pry \land R^- = \baseof{R^+}
 
 \subsection{Ordering of Ingredients:}
 
-By Unique Tip, $R^+ \le L$.  By definition of $\base$, $R^- \le R^+$
+By Correct Tip, $R^+ \le L$.  By definition of $\base$, $R^- \le R^+$
 so $R^- \le L$.  So $R^+ \le C$ and $R^- \le C$.
 $\qed$
 
@@ -37,7 +37,7 @@ is a descendant, not an ancestor, of the 2nd parent.)
 
 \subsection{No Replay}
 
-By $\merge$,
+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.  $\qed$
@@ -62,18 +62,18 @@ $D \not\isin R^-$.  Thus $D \not\isin C$.  OK.
 By Currently Included, $D \isin L$.
 
 By Tip Own Contents for $R^+$, $D \isin R^+ \equiv D \le R^+$, but
-by Unique Tip, $D \le R^+ \equiv D \le L$.
+by Correct Tip, $D \le R^+ \equiv D \le L$.
 So $D \isin R^+$.
 
 By Base Acyclic for $R^-$, $D \not\isin R^-$.
 
-Apply $\merge$: $D \not\isin C$.  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 $\merge$: $D \isin C \equiv D \isin L$.  OK.
+Apply \commitmergename: $D \isin C \equiv D \isin L$.  OK.
 
 $\qed$
 
@@ -131,7 +131,7 @@ Single Parent Unique Tips applies.  $\qed$
 
 \subsection{Foreign Inclusion}
 
-Consider some $D$ s.t. $\patchof{D} = \bot$.  $D \neq C$.
+Consider some $D \in \foreign$.  $D \neq C$.
 So by Desired Contents $D \isin C \equiv D \isin L$.
 By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.