After misinterpreting the basis of Craig Feinstein's proof of the unprovability of the Collatz conjecture and writing him about it, he kindly led me to the point. His proof merely requires a string describing a Collatz trajectory which does not always close (never quite closes) the door to the addition of still more path elements to the trajectory.

My response to my incorrect view uses a viewpoint incorporating the entire predecessor tree, not just within l.d.a.s. It employs an {e,s,b} notation in the version of the tree like that I've called the abstract tree. This predecessor tree (as does the l.d.a. predecessor tree) provides exactly the unfinished path construction feature that Feinstein needs. At every stage in the development of the abstract tree, a node represents an infinite set of elements in some {c[d]} which must include members in {0[3]}, {1[3]}, and {2[3]} with equal density. Those elements in {0[3]} are leaf elements. The paths to them are complete. (The power of 2 multiples of all those leaf elements, which would be represented by appended 0-bits in the parity bit string or by appended e's in the {esb} notation, are indicated by arrows pointing behind the pane of the tree of odd integers, but otherwise omitted and ignored. Since all those power of 2 multiples trivially reduce to their parental odd integer, we have concerned ourselves solely with odd integers in this work.)

The paths reaching those {1[3]} and {2[3]} sets are not
complete. Another symbol must be appended to the string to specify
whether the next step is a *b*- or an *s*-step,
respectively. That goes on at every level of development of the
tree. So Feinstein could as well cast his proof in terms of
my {e,s,b} notation, which one might find more
convincing because the strings would then represent legitimate
realizable Collatz trajectories rather than some arbitrary random bit
string. So, in so far it goes, his proof is
convincing. The development of the abstract predecessor tree
does, in fact, never terminate. Feinstein's claim is
that that proves the unprovability of the Collatz conjecture.

But I've been aware of that infinite development of the abstract tree and the paths it represents. I've covered many of its aspects in other pages on this site. The abstract tree of l.d.a.s consists of an infinite tree of infinite sets. You not only never get to stop adding generations of nodes in the abstract tree, you also never reach the last instantiation of the set of odd integers constituting each node in the abstract tree.

In my web page on "counting", I go through a number of different infinite summations of the densities of the integers in various infinite sets in one or two dimensions (and, in other pages, a bunch of finite summations) to gain/give an impression of how (rapidly) the paths reach leaf nodes in the l.d.a.s. (But similar treatments applied to the tree as a whole rather than just plucked out l.d.a.s would give very similar results.) It turns out that in the infinite sums, all the l.d.a.s (substitute paths if the sums are applied to the entire tree instead of just its component l.d.a.s) do reach leaf nodes. So while there is always further to go, the infinite summations show that every path does eventually terminate. The finite summations show that it doesn't take so very long to achieve a fraction of the densities of the integers included in the l.d.a.s very very close to 1.

The density of the even integers in the infinite series of powers of 2 multiples of the odd numbers in the predecessor tree have also been summed, and prove to have the identical overall density that the odd numbers have.

Other pages of my web site make the same point about complete
coverage of the integers, but approach the issue
from the point of view of actively placing integers in the
abstract tree intead of waiting for them to appear during tree
generation.
If you were to build the predecessor tree to some large depth, and give
me a list of integers not yet included I can algorithmically determine
just where each of those integers fits into the undeveloped
portions of the abstract tree. One page
describes the algorithm. There are lots
of such
integers -- at a depth of 20, there are 2^{19} terminal
nodes, each of
which has a {1[3]} and a {2[3]} unfinished set, so just counting the first
instance of each set at level 20, there are 2^{20} such
integers. Of course, their combined density is very small
indeed.

My argument is that the only operations in that algorithm are simple arithmetic operations, clearly applicable to integers of any magnitude. Using this algorithm every odd integer (and the not-yet-discovered l.d.a. which contains it) can be placed in the abstract tree. Their powers of 2 multiples (even numers) tag along. Whence every integer exists in the abstract tree. In the best tradition of paradoxes, in spite of the lack of closure of the paths, every odd positive integer does appear somewhere in the predecessor tree.

I count this as just
one more instance of a
paradox.
Considering the unlimited addition of new paths makes it appear that no
conclusion be reached, yet considering the density of the integers
included at each stage makes it clear that a conclusion is being reached
asymptotically. The toad **can** get to the top of his hole, and
the overtaking runner **can** draw even.

Admittedly, I don't know how to map particular instances of l.d.a.s in the abstract tree into specific locations in the binary (or general) tree. You might therefore insist that the umpteenth instance of some x-millionth path is in fact the root of a counter-example predecessor tree, and I can't counter that on these grounds. However, the Collatz predecessor graph does have a single root node, which serves to answer that possibility.

But, topping it all off, there's the graph theoretical argument that nothing but a binary tree can be constructed out of the unique nodes which represent every odd integer. That's where the the last step in the proof ultimately lies, and that resolves the paradox in favor of belief in the summation of the densities of the integers, whence the existence (and its promulgation) of a proof of the Collatz conjecture.

Recall that the contrapositive goes like this: If P then S implies that if NOT S then NOT P.

Thanks, again, to Craig Feinstein for his additional hint that an axiom provides the precarious part of his proof.

Let's say that Feinstein has shown that employing the axiom that any infinite process(P) leads to unprovability(S). By the contrapositive, to NOT have unprovability (NOT S) we must NOT employ the axiom about use of infinite processes (NOT P). Or, stated positively, to have any hope of proof of a proposition involving infinite processes we must admit infinite processes in the proof.

It seems quite clear and perfectly obvious when you think about it.

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