Recall that we are considering $D \in \py$.
$D \isin Y \equiv D \le Y$. $D \not\isin X$.
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 $\nothaspatch$).
+various cases that
+if $M \haspatch \p$, $D \not\isin C$,
+whereas if $M \nothaspatch \p$, $D \isin C \equiv \land D \le C$.
Consider $D = C$: Thus $C \in \py, L \in \py$.
By Tip Own Contents, $\neg[ L \nothaspatch \p ]$ so $L \neq X$,