By Foreign Contents of $L$, $\patchof{M} = \bot$ as well.
So by Foreign Contents for any $A \in \{L,M,R\}$,
$\forall_{\p, D \in \py} D \not\le A$
-so by No Replay for A $D \not\isin A$.
+so by No Replay for $A$, $D \not\isin A$.
Thus $\pendsof{A}{\py} = \{ \}$ and the RHS of both Merge Ends
conditions are satisifed.
(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 $L \neq R$,
+By Tip Self Inpatch, $\neg[ L \nothaspatch \p ]$ 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 $\merge$, $D \isin C$,