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refactor for coherence cases - simple
author
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Wed, 21 Mar 2012 21:34:24 +0000
(21:34 +0000)
committer
Ian Jackson
<ijackson@chiark.greenend.org.uk>
Wed, 21 Mar 2012 21:34:24 +0000
(21:34 +0000)
simple.tex
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diff --git
a/simple.tex
b/simple.tex
index dfcf907fbb9ea6d43ddbd1ade98f8eaea57f6c38..ae6fc5551fd797dedeeeced6423cf5708b0d0f05 100644
(file)
--- a/
simple.tex
+++ b/
simple.tex
@@
-49,7
+49,21
@@
$\qed$
\subsection{Coherence and Patch Inclusion}
\subsection{Coherence and Patch Inclusion}
-Need to consider $D \in \py$
+$$
+\begin{cases}
+ L \haspatch \p : & C \haspatch \p \\
+ L \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+\proofstarts
+~
+
+Firstly, if $L \haspatch \p$, $\exists_{F \in \py} F \le L$
+and this $F$ is also $\le C$
+so $C \zhaspatch \p \implies C \haspatch \p$.
+We will prove $\zhaspatch$
+
+We need to consider $D \in \py$.
\subsubsection{For $L \haspatch \p, D = C$:}
\subsubsection{For $L \haspatch \p, D = C$:}
@@
-57,18
+71,14
@@
Ancestors of $C$:
$ D \le C $.
Contents of $C$:
$ D \le C $.
Contents of $C$:
-$ D \isin C \equiv \ldots \lor \true \text{ so } D \haspatch C $.
+$ D \isin C \equiv \ldots \lor \true$. So $ D \zhaspatch C $.
+OK.
\subsubsection{For $L \haspatch \p, D \neq C$:}
Ancestors: $ D \le C \equiv D \le L $.
Contents: $ D \isin C \equiv D \isin L \lor f $
so $ D \isin C \equiv D \isin L $, i.e. $ C \zhaspatch P $.
\subsubsection{For $L \haspatch \p, D \neq C$:}
Ancestors: $ D \le C \equiv D \le L $.
Contents: $ D \isin C \equiv D \isin L \lor f $
so $ D \isin C \equiv D \isin L $, i.e. $ C \zhaspatch P $.
-By $\haspatch$ for $L$, $\exists_{F \in \py} F \le L$
-and this $F$ is also $\le C$. So $\haspatch$.
-
-So:
-\[ L \haspatch \p \implies C \haspatch \p \]
\subsubsection{For $L \nothaspatch \p$:}
\subsubsection{For $L \nothaspatch \p$:}
@@
-76,9
+86,8
@@
Firstly, $C \not\in \py$ since if it were, $L \in \py$.
Thus $D \neq C$.
Now by contents of $L$, $D \notin L$, so $D \notin C$.
Thus $D \neq C$.
Now by contents of $L$, $D \notin L$, so $D \notin C$.
+OK.
-So:
-\[ L \nothaspatch \p \implies C \nothaspatch \p \]
$\qed$
\subsection{Foreign Inclusion:}
$\qed$
\subsection{Foreign Inclusion:}