X-Git-Url: http://www.chiark.greenend.org.uk/ucgi/~ian/git?a=blobdiff_plain;f=article.tex;h=0428df707a2978eae6c336b40447f29b5b820385;hb=1814499fa14d3774567205fa57a95b3adbfe4fb5;hp=70d6d88200ffa76188a4e3fb553043e4094288a2;hpb=bfa0996c930a2507501eec79b4a89333d35d47aa;p=topbloke-formulae.git

diff --git a/article.tex b/article.tex
index 70d6d88..0428df7 100644
--- a/article.tex
+++ b/article.tex
@@ -369,7 +369,10 @@ Topbloke strips the metadata when exporting.
 Trivial.
 
 \subsection{Unique Base}
-If $A, C \in \py$ then $\baseof{C} = \baseof{A}$. $\qed$
+If $A, C \in \py$ then by Calculation of Ends for
+$C, \py, C \not\in \py$:
+$\pendsof{C}{\pn} = \pendsof{A}{\pn}$ so
+$\baseof{C} = \baseof{A}$. $\qed$
 
 \subsection{Tip Contents}
 We need to consider only $A, C \in \py$.  From Tip Contents for $A$:
@@ -396,7 +399,9 @@ Need to consider only $A, C \in \pn$.
 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
-$A$, $D \isin C \implies D \not\in \py$. $\qed$
+$A$, $D \isin C \implies D \not\in \py$.
+
+$\qed$
 
 \subsection{Coherence and patch inclusion}
 
@@ -435,10 +440,55 @@ $\qed$
 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{Create Base}
+
+Given $L$, create a Topbloke base branch initial commit $B$.
+\gathbegin
+ B \hasparents \{ L \}
+\gathnext
+ \patchof{B} = \pa{B}
+\gathnext
+ D \isin B \equiv D \isin L \lor D = B
+\end{gather}
+
+\subsection{Conditions}
+
+\[ \eqn{ Ingredients }{
+ \patchof{L} = \pa{L} \lor \patchof{L} = \bot
+}\]
+\[ \eqn{ Non-recursion }{
+ L \not\in \pa{B}
+}\]
+
+\subsection{No Replay}
+
+If $\patchof{L} = \pa{L}$, trivial by Base Acyclic for $L$.
+
+If $\patchof{L} = \bot$, xxx
+
+Trivial from Base Acyclic for $L$.  $\qed$
+
+\subsection{Unique Base}
+
+Not applicable. $\qed$
+
+\subsection{Tip Contents}
+
+Not applicable. $\qed$
+
+\subsection{Base Acyclic}
+
+xxx
+
+xxx unfinished
+
+\section{Create Tip}
+
+xxx tbd\
+
 \section{Anticommit}
 
-Given $L, R^+, R^-$ where
-$R^+ \in \pry, R^- = \baseof{R^+}$.  
+Given $L$ and $\pr$ as represented by $R^+, R^-$.
 Construct $C$ which has $\pr$ removed.
 Used for removing a branch dependency.
 \gathbegin
@@ -451,6 +501,9 @@ Used for removing a branch dependency.
 
 \subsection{Conditions}
 
+\[ \eqn{ Ingredients }{
+R^+ \in \pry \land R^- = \baseof{R^+}
+}\]
 \[ \eqn{ Into Base }{
  L \in \pn
 }\]
@@ -465,9 +518,10 @@ Used for removing a branch dependency.
 
 By Unique Tip, $R^+ \le L$.  By definition of $\base$, $R^- \le R^+$
 so $R^- \le L$.  So $R^+ \le C$ and $R^- \le C$.
+$\qed$
 
-(Note that the merge base $R^+ \not\le R^-$, i.e. the merge base is
-later than one of the branches to be merged.)
+(Note that $R^+ \not\le R^-$, i.e. the merge base
+is a descendant, not an ancestor, of the 2nd parent.)
 
 \subsection{No Replay}
 
@@ -542,6 +596,8 @@ So $L \nothaspatch \p \implies C \nothaspatch \p$.
 Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
 so $L \haspatch \p \implies C \haspatch \p$.
 
+$\qed$
+
 \section{Foreign Inclusion}
 
 Consider some $D$ s.t. $\patchof{D} = \bot$.  $D \neq C$.
@@ -549,11 +605,13 @@ So by Desired Contents $D \isin C \equiv D \isin L$.
 By Foreign Inclusion of $D$ in $L$, $D \isin L \equiv D \le L$.
 
 And $D \le C \equiv D \le L$.
-Thus $D \isin C \equiv D \le C$.  $\qed$
+Thus $D \isin C \equiv D \le C$.
+
+$\qed$
 
 \section{Merge}
 
-Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
+Merge commits $L$ and $R$ using merge base $M$:
 \gathbegin
  C \hasparents \{ L, R \}
 \gathnext
@@ -564,7 +622,9 @@ Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
 We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
 
 \subsection{Conditions}
-
+\[ \eqn{ Ingredients }{
+ M \le L, M \le R
+}\]
 \[ \eqn{ Tip Merge }{
  L \in \py \implies
    \begin{cases}
@@ -727,7 +787,9 @@ $C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
 Consider some $D \in \py$.
 
 By Base Acyclic of $L$, $D \not\isin L$.  By the above, $D \not\isin
-R$.  And $D \neq C$.  So $D \not\isin C$.  $\qed$
+R$.  And $D \neq C$.  So $D \not\isin C$.
+
+$\qed$
 
 \subsection{Tip Contents}