Note that all definitions used in this proof that are not defined here are defined in the Definitions section.
Consider \(g\left(x\right)\), a Parity Cycle function that takes an odd integer input, performs the Collatz Conjecture according to some Parity Cycle, and returns an odd integer that is also in the domain of \(g\). Let the Parity Cycle corresponding to \(g\left(x\right)\) be:
$$1 \text{o}\text{d}\text{d},{p}_{1} \text{e}\text{v}\text{e}\text{n}\text{s},1 \text{o}\text{d}\text{d},{p}_{2} \text{e}\text{v}\text{e}\text{n}\text{s},1 \text{o}\text{d}\text{d},\cdots ,1 \text{o}\text{d}\text{d},{p}_{b} \text{e}\text{v}\text{e}\text{n}\text{s}.$$
By the Collatz Conjecture function, each “odd” corresponds to multiplying by 3 and adding 1 to \(x\). Similarly, each “even” corresponds to dividing \(x\) by 2.
Thus, \(g\left(x\right)\) can be computed in terms of the \({p}_{i}\)’s:
$$g\left(x\right)=\frac{3\left(\frac{3\left(\frac{3\left(\frac{\left(3x+1\right)}{{2}^{{p}_{1}}}\right)+1}{{2}^{{p}_{2}}}\right)+1}{⋮}\right)+1}{{2}^{{p}_{b}}}$$
$$=\frac{{3}^{b}x+{\sum }_{i=1}^{b}{3}^{b-i}{2}^{{\sum }_{j=0}^{i-1}{p}_{j}}}{{2}^{{\sum }_{i=1}^{b}{p}_{i}}}$$
$$=\frac{rx+q}{{2}^{s}}.$$
Note that by the definition of \(q\), \(q\equiv 1\hspace{0.25em}(mod\hspace{0.25em}2)\).
By the definition of a Parity Cycle, \({g}^{a}\left(x\right)\) for any positive integer \(a\) is in the domain of \(g\). So, similar to how we expanded \(g\left(x\right)\), we can also expand \({g}^{a}\left(x\right)\) using the concise form:
$${g}^{a}\left(x\right)=\frac{r\left(\frac{r\left(\frac{r\left(\frac{\left(rx+q\right)}{{2}^{s}}\right)+q}{{2}^{s}}\right)+q}{⋮}\right)+q}{{2}^{s}}$$
$$=\frac{{r}^{a}x+{\sum }_{i=1}^{a}q{r}^{a-i}{2}^{\left(i-1\right)s}}{{2}^{as}}$$
$$=\frac{{r}^{a}x+q\left({\sum }_{i=1}^{a}{r}^{a-i}{2}^{\left(i-1\right)s}\right)}{{2}^{as}}.$$
The sum in the numerator encased by parentheses is a geometric sequence with common ratio \({2}^{s}/r\), so this form can be simplified by taking the sum of a geometric sequence:
$${r}^{a-1}+{r}^{a-2}{2}^{s}+{r}^{a-3}{2}^{2s}+\dots +{2}^{\left(a-1\right)s}$$
$$=\frac{{2}^{as}-{r}^{a}}{{2}^{s}-r}.$$
Substituting this for the geometric series in the numerator of \({g}^{a}\left(x\right)\), we get:
$${g}^{a}\left(x\right)=\frac{{r}^{a}x+q\left(\frac{{2}^{as}-{r}^{a}}{{2}^{s}-r}\right)}{{2}^{as}}$$
$${g}^{a}\left(x\right)={\left(\frac{r}{{2}^{s}}\right)}^{a}\left(x-\frac{q}{{2}^{s}-r}\right)+\frac{q}{{2}^{s}-r}.$$
Note that once \(x\) is determined, \(r\), \(q\), and \(s\) are uniquely defined constants. Also note again that \({g}^{a}\left(x\right)\) is an odd integer, so \({g}^{a}\left(x\right)\equiv 1\hspace{0.25em}(mod\hspace{0.25em}2)\).
$${g}^{a}\left(x\right)={\left(\frac{r}{{2}^{s}}\right)}^{a}\left(x-\frac{q}{{2}^{s}-r}\right)+\frac{q}{{2}^{s}-r}\equiv 1\hspace{0.25em}(mod\hspace{0.25em}2).$$
Multiplying both sides of this congruence statement by the odd number \({2}^{s}-r\), we get
$${\left(\frac{r}{{2}^{s}}\right)}^{a}\left(x\left({2}^{s}-r\right)-q\right)+q\equiv {2}^{s}-r\equiv 1\hspace{0.25em}(mod\hspace{0.25em}2)$$
since \(r\equiv 1\hspace{0.25em}(mod\hspace{0.25em}2)\). We know \(q\equiv 1\hspace{0.25em}(mod\hspace{0.25em}2)\) by definition, so when we subtract \(q\) on both sides, we get:
$${\left(\frac{r}{{2}^{s}}\right)}^{a}\left(x\left({2}^{s}-r\right)-q\right)\equiv 1-q\equiv 0\hspace{0.25em}(mod\hspace{0.25em}2).$$
Thus, we see that
$${\left(\frac{r}{{2}^{s}}\right)}^{a}\left(x\left({2}^{s}-r\right)-q\right)$$
is an even integer. By the definition of \(g\), \(a\) can be arbitrarily large and every other variable is constant, including \(x\). From here, we have a few cases.
a) Case 1
b) Our first case is when
$$x\left({2}^{s}-r\right)-q\ne 0.$$
In this case, since \(r\) and \({2}^{s}\) are relatively prime by definition,
$${2}^{as} | x\left({2}^{s}-r\right)-q$$
for all \(a\). However, \(x\left({2}^{s}-r\right)-q\) is constant, and \({2}^{as}\) is unbounded since \(a\) is unbounded. Therefore, the quotient of the two cannot be an integer and this case is impossible.
c) Case 2
d) Our second case is when
$$x\left({2}^{s}-r\right)-q=0.$$
Rearranging this, we find:
$$x=\frac{q}{{2}^{s}-r}.$$
Therefore,
$${g}^{a}\left(x\right)={\left(\frac{r}{{2}^{s}}\right)}^{a}\left(x-\frac{q}{{2}^{s}-r}\right)+\frac{q}{{2}^{s}-r}$$
$$={\left(\frac{r}{{2}^{s}}\right)}^{a}\left(0\right)+\frac{q}{{2}^{s}-r}$$
$$=0+\frac{q}{{2}^{s}-r}$$
for all \(a\). Thus, this case corresponds to Looping, not Divergence. Specifically, not only do the parities of the numbers repeat after every iteration of the Parity Cycle corresponding to \(g\), but the numbers themselves repeat.
Before examining this any further, we see that it is not possible for a number to generate an infinite Diverging Collatz Reduction Sequence for which the corresponding infinitely long Parity Sequence is periodic with finite period. Therefore, a number can only Diverge if its infinitely long Parity Sequence is aperiodic, proving our finding on Divergence.
Now, we return to Case 2 with respect to Looping. We see that a number can lead to Looping if and only if
$$x=\frac{q}{{2}^{s}-r}.$$
Further, since \(x>0\) and \(q>0\), we must have \({2}^{s}-r>0\), or
if \(x\) leads to Looping.
Hence, we have also proved our finding on Looping, and the proof is complete.\(▫\)