+Let us focus on the upper part. There are three possibilities of what can happen
+when we visit a~vertex:
+
+\itemize\ibull
+
+\:We delete it: Every vertex deleted has to have been created at some time in the past.
+New vertices are created only during inserts and melds (when joining two trees) and
+we have already shown that these operations have constant amortized complexity. So the
+same must hold for deletions.
+
+\:We recurse twice and concatenate the lists: The lists are disassembled only when
+they reach the root of the tree, otherwise they are only concatenated. We can easily
+model the situation by a~binary tree forest similar to the meld forest. There are~$n$
+leaves and every internal vertex has outdegree two, so the total number of concatenations
+is at most~$n$. Each of them can be performed in constant time as the list is doubly linked.
+
+\:We recurse only once: This occurs only if the rank is even and the gap between the
+rank of this vertex and its sons is small. It therefore cannot happen twice in a~row,
+so it is clearly dominated by the other possibilities.
+
+\endlist
+\>The total cost of all operations on the upper part is therefore $\O(n)$.
+\qed
+
+It remains to analyse the rest of the \<DeleteMin> operation.
+
+\lemma
+Every \<DeleteMin> takes $\O(1)$ time amortized.
+
+\proof