$B \not\in \py$ so $D \neq B$. So $D \isin B \equiv D \isin L$
and $D \le B \equiv D \le L$.
-Thus $L \haspatch \p \implies B \haspatch P$
-and $L \nothaspatch \p \implies B \nothaspatch P$.
+Thus $L \haspatch \p \equiv B \haspatch P$
+and $L \nothaspatch \p \equiv B \nothaspatch P$.
$\qed$