The densities of the integers in the nodes add to 1, indicating that enough room is there to contain all the odd integers. The nodes in the abstract predecessor tree are all disjoint, and the integers within any given node are disjoint. By the pigeon-hole principle, all odd positive integers must be present in the abstract predecessor tree. Whence the edges in every left descent assemblage are directed as well. We are not able to map any given l.d.a. into its position in the binary predecessor tree, but we can say how any given l.d.a. will behave in the binary predecessor tree when the Collatz trajectory is calculated.
Extensions join their parental l.d.a.
The trajectory will move up a left descent to its header extension and then upward and leftward to the integer parental to the whole set of extensions of which this l.d.a.'s parent is a member. That integer may again be identified as to its l.d.a., transplanted into the binary predecessor tree, and run through its Collatz trajectory, with the identification of another integer in its turn. Continuing iterations must eventually reach the root. This argument depends on two key points. If these arguments are supportable it would seem that the proof outlined will be successful.