( \mathop{\hbox{\huge{$\vee$}}}_{R \in \set R} D \le R )
\lor D = C
}\]
+xxx proof tbd
\[ \eqn{Transitive Ancestors:}{
\left[ \bigforall_{ E \in \pendsof{C}{\set P} } E \le M \right] \equiv
commits, this terminates with $A'' \in \pends()$, ie $A'' \le M$
by the LHS. And $A \le A''$.
}
+
\[ \eqn{Calculation Of Ends:}{
\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
+ \begin{cases}
+ C \in \p : & \{ C \}
+ \\
+ C \not\in \p : & \displaystyle
\left\{ E \Big|
\Bigl[ \Largeexists_{A \in \set A}
E \in \pendsof{A}{\set P} \Bigr] \land
\Bigl[ \Largenexists_{B \in \set A}
E \neq B \land E \le B \Bigr]
\right\}
+ \end{cases}
}\]
-XXX proof TBD.
+xxx proof tbd
\subsection{No Replay for Merge Results}
\bigforall_{\pay{Q} \not\ni C} \pendsof{C}{\pay{Q}}
\end{gather}
+$\patchof{C}$, for each kind of Topbloke-generated commit, is stated
+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 $C \haspatch \pa{Q}$
+is in stated
+(in terms of $I \haspatch \pa{Q}$ or $I \nothaspatch \pa{Q}$
+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,
+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 $\pends$
-would have to be updated. The annotation is not needed because
-$\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$.
+would have to be updated. The annotation is not needed in that case
+because $\forall_{\py \ni C} \; \pendsof{C}{\py} = \{C\}$.
\section{Simple commit}
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 $A$:
L \haspatch \pry
}\]
-\subsection{No Replay}
+\subsection{Ordering of ${L, R^+, R^-}$:}
By Unique Tip, $R^+ \le L$. By definition of $\base$, $R^- \le R^+$
-so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$ and No Replay for
-Merge Results applies. $\qed$
+so $R^- \le L$. So $R^+ \le C$ and $R^- \le C$.
+
+(Note that the merge base $R^+ \not\le R^-$, i.e. the merge base is
+later than one of the branches to be merged.)
+
+\subsection{No Replay}
+
+No Replay for Merge Results applies. $\qed$
\subsection{Desired Contents}
Whereas if $L \haspatch \p$, $D \isin L \equiv D \le L$.
so $L \haspatch \p \implies C \haspatch \p$.
+\section{Foreign Inclusion}
+
+Consider some $D$ s.t. $\patchof{D} = \bot$. $D \neq C$.
+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 $D \isin C \equiv D \le C$. $\qed$
+
\section{Merge}
Merge commits $L$ and $R$ using merge base $M$ ($M < L, M < R$):
\subsection{No Replay}
-See No Replay for Merge Results.
+No Replay for Merge Results applies. $\qed$
\subsection{Unique Base}
\subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
-$C \haspatch \p \equiv M \nothaspatch \p$.
+$M \haspatch \p \implies C \nothaspatch \p$.
+$M \nothaspatch \p \implies C \haspatch \p$.
\proofstarts