merge tip contents done

diff --git a/article.tex b/article.tex
index faf41526029fb4b0c489e4b8fedfe072d7f62dad..f73f7229ae114bc2104d3b4b8bb55b0a5b5b49e6 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -673,7 +673,24 @@ Thus $D \isin C \equiv D \isin \baseof{C}$.  OK.
 
 \subsubsection{For $D \not\in \py, R \in \py$:}
 
-xxx up to here
+$D \neq C$.
+
+By Tip Contents
+$D \isin L \equiv D \isin \baseof{L}$ and
+$D \isin R \equiv D \isin \baseof{R}$.
+
+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 $\merge$,
+$D \isin C \equiv D \isin R$.  But from Unique Base,
+$\baseof{C} = R$ so $D \isin C \equiv D \isin \baseof{C}$.  OK.
+
+$\qed$
+
+xxx junk after here
 
 %D \in \py$:}