\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 $L \haspatch \p$ by Tip Self Inpatch for $L$). So $D \neq C$.
+\in \py$ ie $\neg[ L \nothaspatch \p ]$ by Tip Self Inpatch for $L$).
+So $D \neq C$.
Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
\subsubsection{For $L \haspatch \p, R \haspatch \p$:}
various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
(which suffices by definition of $\haspatch$ and $\nothaspatch$).
-Consider $D = C$: Thus $C \in \py, L \in \py$. By Tip Contents
-for $L$, $L \isin L$ so $\neg [ L \nothaspatch \p ]$.
-Therefore we must have $L=Y$, $R=X$.
+Consider $D = C$: Thus $C \in \py, L \in \py$.
+By Tip Self Inpatch, $\neg[ L \nothaspatch \p ]$ so $L \neq R$,
+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.
$D \isin L \equiv D \isin \baseof{L}$ and
$D \isin R \equiv D \isin \baseof{R}$.
+Apply Tip Merge condition.
If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
$\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
$D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
-So $D \isin M \equiv D \isin L$ and by $\merge$,
+So $D \isin M \equiv D \isin L$ so by $\merge$,
$D \isin C \equiv D \isin R$. But from Unique Base,
-$\baseof{C} = R$ so $D \isin C \equiv D \isin \baseof{C}$. OK.
+$\baseof{C} = \baseof{R}$.
+Therefore $D \isin C \equiv D \isin \baseof{C}$. OK.
$\qed$
~ian / topbloke-formulae.git / blobdiff