.gitignore

[topbloke-formulae.git] / article.tex

diff --git a/article.tex b/article.tex
index 32164de0f7c0c4ab51f14c0b99a2965ddf355fd5..f40f6a2c508b4877d8fe619685ff8f810f20a63a 100644 (file)
--- a/article.tex
+++ b/article.tex
@@ -63,6 +63,9 @@
 \newcommand{\qed}{\square}
 \newcommand{\proof}[1]{{\it Proof.} #1 $\qed$}
 
+\newcommand{\gathbegin}{\begin{gather} \tag*{}}
+\newcommand{\gathnext}{\\ \tag*{}}
+
 \begin{document}
 
 \section{Notation}
@@ -215,12 +218,15 @@ by the LHS.  And $A \le A''$.
 \section{Commit annotation}
 
 We annotate each Topbloke commit $C$ with:
-\begin{gather}
-\tag*{} \patchof{C} \\
-\tag*{} \baseof{C}, \text{ if } C \in \py \\
-\tag*{} \bigforall_{\pa{Q}} 
-        \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q} \\
-\tag*{} \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
+\gathbegin
+ \patchof{C}
+\gathnext
+ \baseof{C}, \text{ if } C \in \py
+\gathnext
+ \bigforall_{\pa{Q}} 
+   \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q}
+\gathnext
+ \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
 \end{gather}
 
 We do not annotate $\pendsof{C}{\py}$ for $C \in \py$.  Doing so would
@@ -270,12 +276,42 @@ For $D = C$: $D \in \pn$ so $D \not\in \py$. OK.
 For $D \neq C$: $D \isin C \equiv D \isin A$, so by Base Acyclic for
 $A$, $D \isin C \implies D \not\in \py$. $\qed$
 
-\subsection{Coherence}
+\subsection{Coherence and patch inclusion}
+
+Need to consider $D \in \py$
 
 \subsubsection{For $A \haspatch P, D = C$:}
-\[ D \isin C \equiv \ldots \lor t \text{ so } D \haspatch C \]
-\[ D \le C \]
-OK
+
+Ancestors of $C$:
+$ D \le C $.
+
+Contents of $C$:
+$ D \isin C \equiv \ldots \lor t \text{ so } D \haspatch C $.
+
+\subsubsection{For $A \haspatch P, D \neq C$:}
+Ancestors: $ D \le C \equiv D \le A $.
+
+Contents: $ D \isin C \equiv D \isin A \lor f $
+so $ D \isin C \equiv D \isin A $.
+
+So:
+\[ A \haspatch P \implies C \haspatch P \]
+
+\subsubsection{For $A \nothaspatch P$:}
+
+Firstly, $C \not\in \py$ since if it were, $A \in \py$.  
+Thus $D \neq C$.
+
+Now by contents of $A$, $D \notin A$, so $D \notin C$.
+
+So:
+\[ A \nothaspatch P \implies C \nothaspatch P \]
+$\qed$
+
+\subsection{Foreign inclusion:}
+
+If $D = C$, trivial.  For $D \neq C$:
+$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$.  $\qed$
 
 \section{Test more symbols}