X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=article.tex;h=b4035ddf35da3ad1c234a2b42f684784d3fd3485;hb=3064ca6a757bfdabbc11f78e579dae2f36146bf4;hp=50fc96c526c2153a95b57546a828beedfec2e297;hpb=21742f0a238a9a7d9c55666a6bd338626ea2cc79;p=topbloke-formulae.git

diff --git a/article.tex b/article.tex
index 50fc96c..b4035dd 100644
--- a/article.tex
+++ b/article.tex
@@ -34,6 +34,10 @@
 \newcommand{\py}{\pay{P}}
 \newcommand{\pn}{\pan{P}}
 
+\newcommand{\pq}{\pa{Q}}
+\newcommand{\pqy}{\pay{Q}}
+\newcommand{\pqn}{\pan{Q}}
+
 \newcommand{\pr}{\pa{R}}
 \newcommand{\pry}{\pay{R}}
 \newcommand{\prn}{\pan{R}}
@@ -313,6 +317,24 @@ xxx proof tbd
   $A \le C$ so $D \le C$.
 }
 
+\[ \eqn{Simple Foreign Inclusion:}{
+  \left[
+    C \hasparents \{ L \}
+   \land
+    \bigforall_{D} D \isin C \equiv D \isin L \lor D = C
+  \right]
+ \implies
+   \bigforall_{D \text{ s.t. } \patchof{D} = \bot}
+     D \isin C \equiv D \le C
+}\]
+\proof{
+Consider some $D$ s.t. $\patchof{D} = \bot$.
+If $D = C$, trivially true.  For $D \neq C$,
+by Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
+And by Exact Ancestors $D \le L \equiv D \le C$.
+So $D \isin C \equiv D \le C$.
+}
+
 \[ \eqn{Totally Foreign Contents:}{
   \bigforall_{C \hasparents \set A}
    \left[
@@ -340,10 +362,10 @@ We annotate each Topbloke commit $C$ with:
 \gathnext
  \baseof{C}, \text{ if } C \in \py
 \gathnext
- \bigforall_{\pa{Q}} 
-   \text{ either } C \haspatch \pa{Q} \text{ or } C \nothaspatch \pa{Q}
+ \bigforall_{\pq}
+   \text{ either } C \haspatch \pq \text{ or } C \nothaspatch \pq
 \gathnext
- \bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
+ \bigforall_{\pqy \not\ni C} \pendsof{C}{\pqy}
 \end{gather}
 
 $\patchof{C}$, for each kind of Topbloke-generated commit, is stated
@@ -352,14 +374,14 @@ in the summary in the section for that kind of commit.
 Whether $\baseof{C}$ is required, and if so what the value is, is
 stated in the proof of Unique Base for each kind of commit.
 
-$C \haspatch \pa{Q}$ or $\nothaspatch \pa{Q}$ is represented as the
-set $\{ \pa{Q} | C \haspatch \pa{Q} \}$.  Whether $C \haspatch \pa{Q}$
+$C \haspatch \pq$ or $\nothaspatch \pq$ is represented as the
+set $\{ \pq | C \haspatch \pq \}$.  Whether $C \haspatch \pq$
 is in stated
-(in terms of $I \haspatch \pa{Q}$ or $I \nothaspatch \pa{Q}$
+(in terms of $I \haspatch \pq$ or $I \nothaspatch \pq$
 for the ingredients $I$),
 in the proof of Coherence for each kind of commit.
 
-$\pendsof{C}{\pa{Q}^+}$ is computed, for all Topbloke-generated commits,
+$\pendsof{C}{\pq^+}$ is computed, for all Topbloke-generated commits,
 using the lemma Calculation of Ends, above.
 We do not annotate $\pendsof{C}{\py}$ for $C \in \py$.  Doing so would
 make it wrong to make plain commits with git because the recorded $\pends$
@@ -448,10 +470,9 @@ So:
 \[ A \nothaspatch P \implies C \nothaspatch P \]
 $\qed$
 
-\subsection{Foreign inclusion:}
+\subsection{Foreign Inclusion:}
 
-If $D = C$, trivial.  For $D \neq C$:
-$D \isin C \equiv D \isin A \equiv D \le A \equiv D \le C$.  $\qed$
+Simple Foreign Inclusion applies.  $\qed$
 
 \subsection{Foreign Contents:}
 
@@ -464,7 +485,7 @@ Given $L$, create a Topbloke base branch initial commit $B$.
 \gathbegin
  B \hasparents \{ L \}
 \gathnext
- \patchof{B} = \pan{Q}
+ \patchof{B} = \pqn
 \gathnext
  D \isin B \equiv D \isin L \lor D = B
 \end{gather}
@@ -474,8 +495,8 @@ Given $L$, create a Topbloke base branch initial commit $B$.
 \[ \eqn{ Ingredients }{
  \patchof{L} = \pa{L} \lor \patchof{L} = \bot
 }\]
-\[ \eqn{ Non-recursion }{
- L \not\in \pa{Q}
+\[ \eqn{ Create Acyclic }{
+ L \not\haspatch \pq
 }\]
 
 \subsection{No Replay}
@@ -492,19 +513,15 @@ Not applicable.
 
 \subsection{Base Acyclic}
 
-Consider some $D \isin B$.  If $D = B$, $D \in \pn$, OK.
-
-If $D \neq B$, $D \isin L$.  By No Replay of $D$ in $L$, $D \le L$.
-Thus by Foreign Contents of $L$, $\patchof{D} = \bot$.  OK.
-
-xxx this is wrong
-
-$\qed$
+Consider some $D \isin B$.  If $D = B$, $D \in \pqn$.
+If $D \neq B$, $D \isin L$, and by Create Acyclic
+$D \not\in \pqy$.  $\qed$
 
 \subsection{Coherence and Patch Inclusion}
 
 Consider some $D \in \p$.
-$B \not\in \py$ so $D \neq B$.  So $D \isin B \equiv D \isin L$.
+$B \not\in \py$ so $D \neq B$.  So $D \isin B \equiv D \isin L$
+and $D \le B \equiv D \le L$.
 
 Thus $L \haspatch \p \implies B \haspatch P$
 and $L \nothaspatch \p \implies B \nothaspatch P$.
@@ -513,11 +530,7 @@ $\qed$.
 
 \subsection{Foreign Inclusion}
 
-Consider some $D$ s.t. $\patchof{D} = \bot$.  $D \neq B$
-so $D \isin B \equiv D \isin L$.
-By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
-And by Exact Ancestors $D \le L \equiv D \le B$.
-So $D \isin B \equiv D \le B$.  $\qed$
+Simple Foreign Inclusion applies. $\qed$
 
 \subsection{Foreign Contents}
 
@@ -525,7 +538,71 @@ Not applicable.
 
 \section{Create Tip}
 
-xxx tbd
+Given a Topbloke base $B$, create a tip branch initial commit B.
+\gathbegin
+ C \hasparents \{ B \}
+\gathnext
+ \patchof{B} = \pqy
+\gathnext
+ D \isin C \equiv D \isin B \lor D = C
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{B} = \pqn
+}\]
+\[ \eqn{ No Sneak }{
+ \pendsof{B}{\pqy} = \{ \}
+}\]
+
+\subsection{No Replay}
+
+Ingredients Prevent Replay applies.  $\qed$
+
+\subsection{Unique Base}
+
+Trivially, $\pendsof{C}{\pqn} = \{B\}$ so $\baseof{C} = B$.  $\qed$
+
+\subsection{Tip Contents}
+
+Consider some arbitrary commit $D$.  If $D = C$, trivially satisfied.
+
+If $D \neq C$, $D \isin C \equiv D \isin B$.
+By Base Acyclic of $B$, $D \isin B \implies D \not\in \pqy$.
+So $D \isin C \equiv D \isin \baseof{B}$.
+
+$\qed$
+
+\subsection{Base Acyclic}
+
+Not applicable.
+
+\subsection{Coherence and Patch Inclusion}
+
+$$
+\begin{cases}
+  \p = \pq    \lor B \haspatch \p : & C \haspatch \p \\
+  \p \neq \pq \land B \nothaspatch \p : & C \nothaspatch \p
+\end{cases}
+$$
+
+\proofstarts
+~ Consider some $D \in \py$.
+
+\subsubsection{For $\p = \pq$:}
+
+By Base Acyclic, $D \not\isin B$.  So $D \isin C \equiv D = C$.
+By No Sneak, $D \le B \equiv D = C$.  Thus $C \haspatch \pq$.
+
+\subsubsection{For $\p \neq \pq$:}
+
+$D \neq C$.  So $D \isin C \equiv D \isin B$,
+and $D \le C \equiv D \le B$.
+
+$\qed$
+
+xxx up to here
 
 \section{Anticommit}