foreign notation: change \bot to \foreign everywhere

[topbloke-formulae.git] / anticommit.tex

diff --git a/anticommit.tex b/anticommit.tex
index 0af8f1aba700b13905603b683a8de1d127c5b794..72469b80d69fa67aa410381faf5768262d15964e 100644 (file)
--- a/anticommit.tex
+++ b/anticommit.tex
@@ -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 $R^- \le C$.
 $\qed$
 
@@ -61,8 +61,8 @@ $D \not\isin R^-$.  Thus $D \not\isin C$.  OK.
 
 By Currently Included, $D \isin L$.
 
-By Tip Self Inpatch 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 $R^-$, $D \not\isin R^-$.
@@ -79,44 +79,59 @@ $\qed$
 
 \subsection{Unique Base}
 
-Into Base means that $C \in \pqn$, so Unique Base is not
-applicable. $\qed$
+Into Base means that $C \in \pln$, so Unique Base is not
+applicable.
 
 \subsection{Tip Contents}
 
-Again, not applicable. $\qed$
+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 $D$ s.t. $\patchof{D} = \foreign$.  $D \neq C$.
 So by Desired Contents $D \isin C \equiv D \isin L$.
 By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.