$D \isin L \equiv D \isin \baseof{L}$ and
$D \isin R \equiv D \isin \baseof{R}$.
+Apply Tip Merge condition.
If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
$\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
$D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
-So $D \isin M \equiv D \isin L$ and by $\merge$,
+So $D \isin M \equiv D \isin L$ so by $\merge$,
$D \isin C \equiv D \isin R$. But from Unique Base,
-$\baseof{C} = R$ so $D \isin C \equiv D \isin \baseof{C}$. OK.
+$\baseof{C} = \baseof{R}$.
+Therefore $D \isin C \equiv D \isin \baseof{C}$. OK.
$\qed$