notation: strip word "merge" from \setmergeof etc.; use new definition of \commitmerg...

[topbloke-formulae.git] / simple.tex

diff --git a/simple.tex b/simple.tex
index 28dc6e53bd5f15a9b14d27dc323d67340958df02..297103adf419b066cb5269773de20409c14c0135 100644 (file)
--- a/simple.tex
+++ b/simple.tex
@@ -47,44 +47,59 @@ $L$, $D \isin C \implies D \not\in \py$.
 
 $\qed$
 
-\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
 
-\subsubsection{For $L \haspatch P, D = C$:}
+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 $L \haspatch \p, D = C$:}
 
 Ancestors 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 $L \haspatch P, D \neq C$:}
+\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 $.
-
-So:
-\[ L \haspatch P \implies C \haspatch P \]
+so $ D \isin C \equiv D \isin L $, i.e. $ C \zhaspatch P $.
+OK.
 
-\subsubsection{For $L \nothaspatch P$:}
+\subsubsection{For $L \nothaspatch \p$:}
 
 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 $D \notin C$.
+Now by $\nothaspatch$, $D \not\isin L$ so $D \not\isin C$.
+OK.
 
-So:
-\[ L \nothaspatch P \implies C \nothaspatch P \]
 $\qed$
 
+\subsection{Unique Tips:}
+
+Single Parent Unique Tips applies.  $\qed$
+
 \subsection{Foreign Inclusion:}
 
 Simple Foreign Inclusion applies.  $\qed$
 
 \subsection{Foreign Contents:}
 
-Only relevant if $\patchof{C} = \bot$, and in that case Totally
+Only relevant if $\isforeign{C}$, and in that case Totally
 Foreign Contents applies. $\qed$