merge remove redundant element in condition

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 1735ffc75a9c72b194b2d0a7bcb09ef86c633048..5459b88726b9e2264e74bd0261cd67bc54ae34b3 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -168,6 +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}
@@ -484,8 +496,7 @@ We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
    \begin{cases}
       R \in \py : & \baseof{R} \ge \baseof{L}
               \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
-      R \in \pn : & R \ge \baseof{L}
-              \land M = \baseof{L} \\
+      R \in \pn : & M = \baseof{L} \\
       \text{otherwise} : & \false
    \end{cases}
 }\]
@@ -540,7 +551,7 @@ That is, $\baseof{C} = \baseof{R}$.
 
 \subsubsection{For $R \in \pn$:}
 
-By Tip Merge condition on $R$,
+By Tip Merge condition on $R$ and since $M \le R$,
 $A \le \baseof{L} \implies A \le R$, so
 $A \le R \lor A \le \baseof{L} \equiv A \le R$.  
 Thus $A \le C \equiv A \le R$.  
@@ -636,8 +647,8 @@ R$.  And $D \neq C$.  So $D \not\isin C$.  $\qed$
 
 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.
+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 $R \not\in \py$
@@ -655,11 +666,13 @@ $C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
 
 $D \neq C$.  By Tip Contents of $L$,
 $D \isin L \equiv D \isin \baseof{L}$, and by Tip Merge condition,
-$D \isin L \equiv D \isin M$.  xxx up to here
-
+$D \isin L \equiv D \isin M$.  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 $D \not\in \py, R \in \py$:}
 
+xxx up to here
 
 %D \in \py$:}