I've distilled the Collatz conjecture to this, and I need to understand the next step which is to think about how to correctly define a $\operatorname{diam}$ operator with the desired properties. I have an idea how it is defined:
Let $S_0$ be any finite set of elements of $\Bbb Z[\frac16]$
Let $f(x)=3x+2^{\nu_2(x)}\cdot 3^{\nu_3(x)}$
Consider the infinite sequence $\overline S_0$ defined by the recurrence relation $f:S_n\to S_{n+1}$
Or in other words $S_{n+1}=\{f(x):x\in S_n\}$
The Collatz conjecture states that for all $S_0:$ $\overline S_0$ is a descending sequence of nonempty closed sets that satisfies $\operatorname{diam} S_n\to 0$ as $n\to\infty$.
The challenge is to show that $\operatorname{diam}$ exists such that:
a) this is true, and
b) $\forall s_a,s_b\in\Bbb Z[\frac16]$ such that $a\cdot\langle2\rangle\langle3\rangle\neq b\cdot\langle2\rangle\langle3\rangle: \operatorname{diam}(a,b)\neq0$
This is the amount I know about the answer:
Every number has a representation as a power series which contains a well-founded series of powers of $3$ with no terms missing, and whose coefficients are a wellordered sequence of powers of $2$. The $\operatorname{diam}(a,b)$ operator is almost certainly $\max(\text{len}(a),\text{len}(b))$ where $\text{len(a)}$ is the length of $a$'s power series. (That such a representation exists is itself equivalent to the Collatz conjecture - essentially this is the length of any number's Collatz sequence.).
The set of all possible values of the power series describes the multiplicative monoid generated by the primes greater than $3$.
The function $f$ truncates the power series.
Progress on this problem is inexorable, but painfully slow for one with limited experience so I'm inviting suggestions from those with more experience on this specific next step of the task might be approached - specifically what $\operatorname{diam}$ might look like, how one might define $\operatorname{diam}$ and show it obeys a) and b).