+Then we will merge the direct contributors and the sources' ends.
+This generates more commits $\tipuc \in \pc$, but none in any other
+commit set. We maintain
+$$
+ \bigforall_{\p \isdep \pc}
+ \pancsof{\tipcc}{\p} \subset
+ \pancsof{\tipfa \p}{\p}
+$$
+\proof{
+ For $\tipcc = \tipzc$, $T$ ...WRONG WE NEED $\tipfa \p$ TO BE IN $\set E$ SOMEHOW
+}
+
+\subsection{Merge Contributors for $\pcy$}
+
+Merge $\pcn$ into $\tipc$. That is, merge with
+$L = \tipc, R = \tipfa{\pcn}, M = \baseof{\tipc}$.
+to construct $\tipu$.
+
+Merge conditions:
+
+Ingredients satisfied by construction.
+Tip Merge satisfied by construction. Merge Acyclic follows
+from Perfect Contents and $\hasdep$ being acyclic.
+
+Removal Merge Ends: For $\p = \pc$, $M \nothaspatch \p$; OK.
+For $\p \neq \pc$, by Tip Contents,
+$M \haspatch \p \equiv L \haspatch \p$, so we need only
+worry about $X = R, Y = L$; ie $L \haspatch \p$,
+$M = \baseof{L} \haspatch \p$.
+By Tip Contents for $L$, $D \le L \equiv D \le M$. OK.~~$\qed$
+
+WIP UP TO HERE
+
+Addition Merge Ends: If $\py \isdep \pcn$, we have already
+done the execution phase for $\pcn$ and $\py$. By
+Perfect Contents for $\pcn$, $\tipfa \pcn \haspatch \p$ i.e.
+$R \haspatch \p$. So we only need to worry about $Y = R = \tipfa \pcn$.
+By Tip Dependencies $\tipfa \pcn \ge \tipfa \py$.
+And by Tip Sources $\tipfa \py \ge $
+
+want to prove $E \le \tipfc$ where $E \in \pendsof{\tipcc}{\py}$
+
+$\pancsof{\tipcc}{\py} = $
+
+
+computed $\tipfa \py$, and by Perfect Contents for $\py$
+