X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=anticommit.tex;h=49931279917b36872d16a2f179788458ee4a7d17;hp=90304e08e12857df301487c46ca34be34989cd0c;hb=8a69195a6c694ebddaf1d383ca0301041b05c2b4;hpb=35c23b7d2f99196a118cf034c601b42dc31e8815

diff --git a/anticommit.tex b/anticommit.tex
index 90304e0..4993127 100644
--- 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}
@@ -17,9 +17,9 @@ Used for removing a branch dependency.
 R^+ \in \pry \land R^- = \baseof{R^+}
 }\]
 \[ \eqn{ Into Base }{
- L \in \pqn
+ L \in \pln
 }\]
-\[ \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.  $\qed$
@@ -61,25 +61,25 @@ $D \not\isin R^-$.  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
-by Unique Tip, $D \le R^+ \equiv D \le L$.
+By Tip Own Contents for $R^+$, $D \isin R^+ \equiv D \le R^+$, but
+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.
 
 $\qed$
 
 \subsection{Unique Base}
 
-Into Base means that $C \in \pqn$, so Unique Base is not
+Into Base means that $C \in \pln$, so Unique Base is not
 applicable.
 
 \subsection{Tip Contents}
@@ -88,35 +88,50 @@ Again, not applicable.
 
 \subsection{Base Acyclic}
 
-By Into Base and Base Acyclic for $L$, $D \isin L \implies D \not\in \pqy$.
-And by Into Base $C \not\in \pqy$.
+By Into Base and Base Acyclic for $L$, $D \isin L \implies D \not\in \ply$.
+And by Into Base $C \not\in \ply$.
 Now from Desired Contents, above, $D \isin C
 \implies D \isin L \lor D = C$, which thus
-$\implies D \not\in \pqy$.  $\qed$.
+$\implies D \not\in \ply$.  $\qed$.
 
 \subsection{Coherence and Patch Inclusion}
 
-Need to consider some $D \in \py$.  By Into Base, $D \neq C$.
+$$
+\begin{cases}
+  \p = \pr                            : & C \nothaspatch \p  \\
+  \p \neq \pr \land L \nothaspatch \p : & C \nothaspatch \p  \\
+  \p \neq \pr \land L \haspatch    \p : & C \haspatch    \p
+\end{cases}
+$$
+\proofstarts
+~ Need to consider some $D \in \py$.  By Into Base, $D \neq C$.
 
 \subsubsection{For $\p = \pr$:}
 By Desired Contents, above, $D \not\isin C$.
-So $C \nothaspatch \pr$.
+OK.
 
 \subsubsection{For $\p \neq \pr$:}
 By Desired Contents, $D \isin C \equiv D \isin L$
 (since $D \in \py$ so $D \not\in \pry$).
 
 If $L \nothaspatch \p$, $D \not\isin L$ so $D \not\isin C$.
-So $L \nothaspatch \p \implies C \nothaspatch \p$.
+OK.
 
-Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
-so $L \haspatch \p \implies C \haspatch \p$.
+Whereas, if $L \haspatch \p$, $D \isin L \equiv D \le L$,
+so $C \zhaspatch \p$;
+and $\exists_{F \in \py} F \le L$ and this $F \le C$.
+Thus $C \haspatch \p$.
+OK.
 
 $\qed$
 
+\subsection{Unique Tips:}
+
+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 $L$, $D \isin L \equiv D \le L$.