wip merge tip contents

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 39a12875eb76a7266a477d76063a98ac6e5457de..3da2db93f9e42a580cbe201f3889de5d0037cda5 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -46,6 +46,7 @@
 
 \renewcommand{\implies}{\Rightarrow}
 \renewcommand{\equiv}{\Leftrightarrow}
+\renewcommand{\nequiv}{\nLeftrightarrow}
 \renewcommand{\land}{\wedge}
 \renewcommand{\lor}{\vee}
 
@@ -167,6 +168,18 @@ $\displaystyle D \isin C \equiv
   \end{cases}
 $ 
 
+Some (overlapping) alternative formulations:
+
+$\displaystyle D \isin C \equiv
+  \begin{cases}
+    D \isin L \equiv D \isin R     : & D = C \lor D \isin L \\
+    D \isin L \equiv D \isin R     : & D = C \lor D \isin R \\
+    D \isin L \nequiv D \isin R  : & D = C \lor D \not\isin M \\
+    D \isin M \equiv D \isin L     : & D = C \lor D \isin R \\
+    D \isin M \equiv D \isin R     : & D = C \lor D \isin L \\
+  \end{cases}
+$
+
 \end{basedescript}
 \newpage
 \section{Invariants}
@@ -633,29 +646,34 @@ R$.  And $D \neq C$.  So $D \not\isin C$.  $\qed$
 
 \subsection{Tip Contents}
 
-xxx up to here
-
 We need worry only about $C \in \py$.  
 And $\patchof{C} = \patchof{L}$
-so $L \in \py$ so $L \haspatch \p$.  We will use the coherence and
-patch inclusion of $C$ as just proved.
-
-Firstly we prove $C \haspatch \p$: If $R \in \py$, this is true by
-coherence/inclusion $C \haspatch \p$.  If $R \not\in \py$ then
-by Tip Merge 
+so $L \in \py$ so $L \haspatch \p$.  We will use the unique base,
+and coherence and patch inclusion, of $C$ as just proved.
 
+Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
+\p$ and by coherence/inclusion $C \haspatch \p$ .  If $R \not\in \py$
+then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
+of $\nothaspatch$, $M \nothaspatch \p$.  So by coherence/inclusion $C
+\haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
 
 We will consider some $D$ and prove the Exclusive Tip Contents form.
 
+\subsubsection{For $D \in \py$:}
+$C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
+\le C$.  OK.
 
-So by definition of
-$\haspatch$, $D \isin C \equiv D \le C$.  OK.
+\subsubsection{For $D \not\in \py, R \not\in \py$:}
 
-\subsubsection{For $L \in \py, D \in \py, $:}
-$R \haspatch \p$ so 
+$D \neq C$.  By Tip Contents of $L$,
+$D \isin L \equiv D \isin \baseof{L}$, and by Tip Merge condition,
+$D \isin L \equiv D \isin M$.  So by definition of $\merge$, $D \isin
+C \equiv D \isin R$.  And $R = \baseof{C}$ by Unique Base of $C$.
+Thus $D \isin C \equiv D \isin \baseof{C}$.  OK.
 
-\subsubsection{For $L \in \py, D \not\in \py, R \in \py$:}
+\subsubsection{For $D \not\in \py, R \in \py$:}
 
+xxx up to here
 
 %D \in \py$:}