\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 $\neg[ L \nothaspatch \p ]$ by Tip Self Inpatch for $L$).
+\in \py$ ie $\neg[ L \nothaspatch \p ]$ by Tip Own Contents for $L$).
So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
(which suffices by definition of $\haspatch$ and $\nothaspatch$).
Consider $D = C$: Thus $C \in \py, L \in \py$.
-By Tip Self Inpatch, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,
+By Tip Own Contents, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,
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 $\merge$, $D \isin C$,
~ian / topbloke-formulae.git / blobdiff