X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?p=topbloke-formulae.git;a=blobdiff_plain;f=merge.tex;h=e386ada6794715cb9d814880135ab7bf1827f30b;hp=f5038213bc5ce7b97b067cc00fcaa63d5ac0bd76;hb=26a840c1b223aa5b9d9fd1dc9e7c1922c87cf5f5;hpb=670d1b8cb1403203123ffd2d6c510f82bb7a335c

diff --git a/merge.tex b/merge.tex
index f503821..e386ada 100644
--- a/merge.tex
+++ b/merge.tex
@@ -115,7 +115,8 @@ This involves considering $D \in \py$.
 
 \subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
 $D \not\isin L \land D \not\isin R$.  $C \not\in \py$ (otherwise $L
-\in \py$ ie $L \haspatch \p$ by Tip Self Inpatch for $L$).  So $D \neq C$.
+\in \py$ ie $\neg[ L \nothaspatch \p ]$ by Tip Self Inpatch for $L$).
+So $D \neq C$.
 Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
 
 \subsubsection{For $L \haspatch \p, R \haspatch \p$:}
@@ -158,9 +159,9 @@ 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 $C \in \py, L \in \py$.  By Tip Contents
-for $L$, $L \isin L$ so $\neg [ L \nothaspatch \p ]$.
-Therefore we must have $L=Y$, $R=X$.
+Consider $D = C$:  Thus $C \in \py, L \in \py$.
+By Tip Self Inpatch, $\neg[ L \nothaspatch \p ]$ so $L \neq R$,
+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.
@@ -233,14 +234,16 @@ 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.
 
 $\qed$