\bigforall_{C \hasparents \set A}
\pendsof{C}{\set P} =
\begin{cases}
- C \in \p : & \{ C \}
+ C \in \set P : & \{ C \}
\\
- C \not\in \p : & \displaystyle
+ C \not\in \set 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, F \in \pendsof{B}{\p}}
+ \Bigl[ \Largenexists_{B \in \set A, F \in \pendsof{B}{\set P}}
E \neq F \land E \le F \Bigr]
\right\}
\end{cases}
\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 $\pendsof{C}{\set P} \subset \bigcup_{E in \set E} \pendsof{E}{\set P}$.
+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 $F$.
+an $F''$) which disqualifies $F$ and $E$.
Otherwise, $E$ meets all the conditions for $\pends$.
}
+\subsection{Single Parent Unique Tips}
+
+Unique Tips is satisfied for single-parent commits. Formally,
+given a conformant commit $A$,
+$$
+ \Big[
+ C \hasparents \{ A \}
+ \Big] \implies \left[
+ \bigforall_{P \patchisin C} \pendsof{C}{\py} = \{ T \}
+ \right]
+$$
+\proof{
+ Trivial for $C \in \py$.
+ For $C \not\in \py$, $\pancsof{C}{\py} = \pancsof{A}{\py}$,
+ so Unique Tips of $A$ suffices.
+}
+
\subsection{Ingredients Prevent Replay}
Given conformant commits $A \in \set A$,
$$