\usepackage{mdwlist}
%\usepackage{accents}
+\usepackage{fancyhdr}
+\pagestyle{fancy}
+\lhead[\rightmark]{}
+
\renewcommand{\ge}{\geqslant}
\renewcommand{\le}{\leqslant}
\newcommand{\nge}{\ngeqslant}
\right\}
\end{cases}
}\]
-xxx proof tbd
+\proof{
+Trivial for $C \in \set P$. For $C \not\in \set P$,
+$\pancsof{C}{\set P} = \bigcup_{A \in \set A} \pancsof{A}{\set P}$.
+So $\pendsof{C}{\set P} \subset \bigcup_{E in \set E} \pendsof{E}{\set P}$.
+Consider some $E \in \pendsof{A}{\set P}$. If $\exists_{B,F}$ as
+specified, then either $F$ is going to be in our result and
+disqualifies $E$, or there is some other $F'$ (or, eventually,
+an $F''$) which disqualifies $F$.
+Otherwise, $E$ meets all the conditions for $\pends$.
+}
\[ \eqn{Ingredients Prevent Replay:}{
\left[
A simple single-parent forward commit $C$ as made by git-commit.
\begin{gather}
-\tag*{} C \hasparents \{ A \} \\
-\tag*{} \patchof{C} = \patchof{A} \\
-\tag*{} D \isin C \equiv D \isin A \lor D = C
+\tag*{} C \hasparents \{ L \} \\
+\tag*{} \patchof{C} = \patchof{L} \\
+\tag*{} D \isin C \equiv D \isin L \lor D = C
\end{gather}
This also covers Topbloke-generated commits on plain git branches:
Topbloke strips the metadata when exporting.
Ingredients Prevent Replay applies. $\qed$
\subsection{Unique Base}
-If $A, C \in \py$ then by Calculation of Ends for
+If $L, C \in \py$ then by Calculation of Ends for
$C, \py, C \not\in \py$:
-$\pendsof{C}{\pn} = \pendsof{A}{\pn}$ so
-$\baseof{C} = \baseof{A}$. $\qed$
+$\pendsof{C}{\pn} = \pendsof{L}{\pn}$ so
+$\baseof{C} = \baseof{L}$. $\qed$
\subsection{Tip Contents}
-We need to consider only $A, C \in \py$. From Tip Contents for $A$:
-\[ D \isin A \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A ) \]
+We need to consider only $L, C \in \py$. From Tip Contents for $L$:
+\[ D \isin L \equiv D \isin \baseof{L} \lor ( D \in \py \land D \le L ) \]
Substitute into the contents of $C$:
-\[ D \isin C \equiv D \isin \baseof{A} \lor ( D \in \py \land D \le A )
+\[ D \isin C \equiv D \isin \baseof{L} \lor ( D \in \py \land D \le L )
\lor D = C \]
Since $D = C \implies D \in \py$,
and substituting in $\baseof{C}$, this gives:
\[ D \isin C \equiv D \isin \baseof{C} \lor
- (D \in \py \land D \le A) \lor
+ (D \in \py \land D \le L) \lor
(D = C \land D \in \py) \]
\[ \equiv D \isin \baseof{C} \lor
- [ D \in \py \land ( D \le A \lor D = C ) ] \]
+ [ D \in \py \land ( D \le L \lor D = C ) ] \]
So by Exact Ancestors:
\[ D \isin C \equiv D \isin \baseof{C} \lor ( D \in \py \land D \le C
) \]
\subsection{Base Acyclic}
-Need to consider only $A, C \in \pn$.
+Need to consider only $L, C \in \pn$.
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
-$A$, $D \isin C \implies D \not\in \py$.
+For $D \neq C$: $D \isin C \equiv D \isin L$, so by Base Acyclic for
+$L$, $D \isin C \implies D \not\in \py$.
$\qed$
Need to consider $D \in \py$
-\subsubsection{For $A \haspatch P, D = C$:}
+\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 $.
-\subsubsection{For $A \haspatch P, D \neq C$:}
-Ancestors: $ D \le C \equiv D \le A $.
+\subsubsection{For $L \haspatch P, D \neq C$:}
+Ancestors: $ D \le C \equiv D \le L $.
-Contents: $ D \isin C \equiv D \isin A \lor f $
-so $ D \isin C \equiv D \isin A $.
+Contents: $ D \isin C \equiv D \isin L \lor f $
+so $ D \isin C \equiv D \isin L $.
So:
-\[ A \haspatch P \implies C \haspatch P \]
+\[ L \haspatch P \implies C \haspatch P \]
-\subsubsection{For $A \nothaspatch P$:}
+\subsubsection{For $L \nothaspatch P$:}
-Firstly, $C \not\in \py$ since if it were, $A \in \py$.
+Firstly, $C \not\in \py$ since if it were, $L \in \py$.
Thus $D \neq C$.
-Now by contents of $A$, $D \notin A$, so $D \notin C$.
+Now by contents of $L$, $D \notin L$, so $D \notin C$.
So:
-\[ A \nothaspatch P \implies C \nothaspatch P \]
+\[ L \nothaspatch P \implies C \nothaspatch P \]
$\qed$
\subsection{Foreign Inclusion:}