X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=fd45642df697518317032ed9c9956c62baf7c1dd;hp=f6a3e9304b5cc404dd8121d08b39cc95d642b3c8;hb=57f83cd8bc6bcf23e45739c00c83c9a8672ae701;hpb=f11a862c007560f325ac57e618aef52e3c03d61c

diff --git a/merge.tex b/merge.tex
index f6a3e93..fd45642 100644
--- a/merge.tex
+++ b/merge.tex
@@ -158,10 +158,12 @@ We will show for each of
 various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
 (which suffices by definition of $\haspatch$ and $\nothaspatch$).
 
-Consider $D = C$:  Thus $C \in \py, L \in \py$, and by Tip
-Self Inpatch for $L$, $L \haspatch \p$, so $L=Y, R=X$.  By Tip Merge,
-$M=\baseof{L}$.  So by Base Acyclic $D \not\isin M$, i.e.
-$M \nothaspatch \p$.  And indeed $D \isin C$ and $D \le C$.  OK.
+Consider $D = C$:  Thus $C \in \py, L \in \py$.  By Tip Contents
+for $L$, $L \isin L$ so $\neg [ L \nothaspatch \p ]$.
+Therefore we must have $L=Y$, $R=X$.
+By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
+by Base Acyclic $M \nothaspatch \p$.  By $\merge$, $D \isin C$,
+and $D \le C$, consistent with $C \haspatch \p$.  OK.
 
 Consider $D \neq C, M \nothaspatch \p, D \isin Y$:
 $D \le Y$ so $D \le C$.
@@ -218,8 +220,8 @@ $C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
 \subsubsection{For $D \not\in \py, R \not\in \py$:}
 
 $D \neq C$.  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 $\merge$, $D \isin
+$D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition,
+$D \isin L \equiv D \isin M$.  So by $\merge$, $D \isin
 C \equiv D \isin R$.  And $R = \baseof{C}$ by Unique Base of $C$.
 Thus $D \isin C \equiv D \isin \baseof{C}$.  OK.
 
@@ -231,14 +233,16 @@ By Tip Contents
 $D \isin L \equiv D \isin \baseof{L}$ and
 $D \isin R \equiv D \isin \baseof{R}$.
 
+Apply Tip Merge condition.
 If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
 Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
 $\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
 $D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
 
-So $D \isin M \equiv D \isin L$ and by $\merge$,
+So $D \isin M \equiv D \isin L$ so by $\merge$,
 $D \isin C \equiv D \isin R$.  But from Unique Base,
-$\baseof{C} = R$ so $D \isin C \equiv D \isin \baseof{C}$.  OK.
+$\baseof{C} = \baseof{R}$.
+Therefore $D \isin C \equiv D \isin \baseof{C}$.  OK.
 
 $\qed$