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 $L \neq X$,
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 $\merge$, $D \isin C$,
-and $D \le C$, consistent with $C \haspatch \p$. OK.
+and $D \le C$. OK.
Consider $D \neq C, M \nothaspatch \p, D \isin Y$:
$D \le Y$ so $D \le C$.
~ian / topbloke-formulae.git / blobdiff