\Big[
C \hasparents \{ A \}
\Big] \implies \left[
- \bigforall_{P \patchisin C} \pendsof{C}{\p} = \{ T \}
+ \bigforall_{P \patchisin C} \pendsof{C}{\py} = \{ T \}
\right]
$$
\proof{
- Trivial for $C \in \p$.
- For $C \not\in \p$, $\pancsof{C}{\p} = \pancsof{A}{\p}$,
+ Trivial for $C \in \py$.
+ For $C \not\in \py$, $\pancsof{C}{\py} = \pancsof{A}{\py}$,
so Unique Tips of $A$ suffices.
}