wip merge complex - C = D

[topbloke-formulae.git] / article.tex

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@@ -572,14 +572,19 @@ So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
 
 \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
 
-$C \haspatch \p \equiv C \nothaspatch M$.
+$C \haspatch \p \equiv M \nothaspatch \p$.
 
 \proofstarts
 
-Merge Ends applies.
-
+Merge Ends applies.  Recall that we are considering $D \in \py$.
 $D \isin Y \equiv D \le Y$.  $D \not\isin X$.
-
-Consider $D = C$.  
+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 $L \haspatch \p$, 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.
 
 \end{document}