Resolution of the Collatz Conjecture: A Rigorous Analysis of Collatz Sequences and their Unique Cycle

1. Introduction

Important Note:

This preprint presents a work-in-progress approach to resolving the Collatz Conjecture. The proof outlined here is currently undergoing rigorous verification and is subject to further refinement. We strongly encourage the mathematical community to review, comment on, and critically analyze the presented arguments.

Key points:

• This is a preliminary version and not a final, peer-reviewed publication.
• The proof may contain errors or oversights that require correction.
• Constructive feedback and scrutiny from experts in number theory, dynamical systems, and related fields are highly welcomed.
• The author(s) are open to collaboration to strengthen and verify the proof.

Please direct any comments, suggestions, or identified issues to:

eduardo.diedrich@outlook.com.ar

We appreciate your interest and potential contributions to this ongoing research effort.

Let ${\mathbb{N}}^{+}$ denote the set of positive integers.

Definition 1 

(Collatz Function). The Collatz function $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ is defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathrm{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathrm{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

Definition 2 

(Inverse Collatz Function). The inverse Collatz function $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ is defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Definition 3 

(Collatz Sequence). For any $n\in {\mathbb{N}}^{+}$, the Collatz sequence starting at n is the sequence ${\left(a_k\right)}_{k\ge 0}$ defined by:

$$\begin{array}{cc}\hfill a_0& =n\hfill \\ \hfill a_{k+1}& =C\left(a_k\right)\mathrm{for}k\ge 0\hfill \end{array}$$

Conjecture 1 

(Collatz Conjecture). For all $n\in {\mathbb{N}}^{+}$, there exists $k\in \mathbb{N}$ such that $C^k\left(n\right)=1$, where $C^k$ denotes k successive applications of C.

The Collatz conjecture, also known as the $3n+1$ problem, has been one of the most famous unsolved problems in mathematics. Proposed by Lothar Collatz in 1937, it concerns a sequence defined as follows: start with any positive integer n. If n is even, divide it by 2. If n is odd, multiply it by 3 and add 1. Repeat this process with the resulting number. The conjecture states that no matter what number you start with, you will always eventually reach 1.

Despite its simple formulation, the Collatz conjecture resisted proof for over 80 years, challenging mathematicians and computer scientists alike. Its importance lies not only in its intrinsic mathematical interest but also in its connections to number theory, dynamical systems, and algorithmic complexity.

This paper presents a rigorous approach to analyzing and resolving the Collatz conjecture. Our method focuses on establishing fundamental properties of Collatz sequences through careful mathematical analysis and proof. The key innovations lie in:

• Comprehensive treatment of sequence properties
• Analysis of the inverse Collatz function
• Logical progression towards a complete resolution of the conjecture

Our approach offers several advantages:

• It provides a rigorous analysis of the structural properties of Collatz sequences.
• It establishes key theorems that characterize the behavior of all Collatz sequences.
• It presents a logical framework that culminates in a complete resolution of the conjecture.
• It utilizes the properties of the inverse Collatz function to gain new insights into the problem.

This paper provides a complete proof of the Collatz conjecture by rigorously establishing a series of properties and theorems that, taken together, demonstrate that all Collatz sequences eventually reach the cycle $\{1,4,2\}$. Given the significance and long-standing nature of this problem, we emphasize the need for thorough peer review and verification by the mathematical community.

The rest of this paper is organized as follows:

• Section 2 introduces the key concepts and definitions.
• The next sections present the main theorems and their proofs, including the Bounded Subsequence Property, the uniqueness of cycles, and the boundedness of all Collatz sequences.
• Section 6 presents the culminating theorem that resolves the Collatz conjecture.
• Section 7 discusses the implications of our results and potential future research directions.

2. Background and Comparative Results

2.1. Historical Context and Related Work

The Collatz Conjecture, proposed by Lothar Collatz in 1937, has been a central problem in number theory and discrete dynamical systems for over 80 years. Numerous approaches have been attempted to prove the conjecture, with varying degrees of success. This section provides an overview of key related works and compares them to our approach.

2.1.1. Terras’s Probabilistic Approach (1976)

Terras, R. ("A stopping time problem on the positive integers." Acta Arithmetica, vol. 30, no. 3, 1976, pp. 241-252) explored a probabilistic approach, demonstrating that almost all Collatz sequences reach a value smaller than their initial value. Terras’s work shares similarities with our analysis of convergence properties.

2.1.2. Lagarias’s Comprehensive Analysis (1985)

Lagarias, J. C. ("The 3x+1 problem and its generalizations." American Mathematical Monthly, vol. 92, no. 1, 1985, pp. 3-23) conducted extensive work on the Collatz Conjecture and its generalizations. His analysis of the Collatz function’s properties, particularly regarding the absence of non-trivial cycles, aligns with our findings in the G-graph structure.

2.1.3. Tao’s Almost-All Result (2019)

Tao, T. ("Almost all orbits of the Collatz map attain almost bounded values." arXiv preprint arXiv:1909.03562, 2019) provided a significant breakthrough by proving that the Collatz conjecture holds for "almost all" starting values, in a probabilistic sense. While our approach is deterministic, Tao’s work complements our findings by providing strong probabilistic evidence for the conjecture’s validity.

3. Preliminaries

3.1. Basic Definitions

Definition 4 

(Well-Ordering Principle). For any non-empty set S of natural numbers, there exists a least element in S. Formally:

$$\forall S\subseteq \mathbb{N},(S\ne \varnothing )\to (\exists m\in S)(\forall n\in S)(m\le n)$$

Where:

• S is a set of natural numbers
• $\mathbb{N}$ is the set of all natural numbers
• m and n are natural numbers
• ≤ is the less than or equal to relation on natural numbers

Remark 1. 

This principle is equivalent to the following statement:

$$\forall P\left(x\right),[\exists n\in \mathbb{N},P(n\left)\right]\to [\exists m\in \mathbb{N},(P\left(m\right)\wedge (\forall k\in \mathbb{N},k<m\to \neg P(k\left)\right)\left)\right]$$

Where $P\left(x\right)$ is any predicate on natural numbers.

Theorem 2 

(Pigeonhole Principle). Let A be a finite set and B be a non-empty finite set. Let $f:A\to B$ be a function and $n=\left|A\right|$, $m=\left|B\right|$. If $n>m$, then there exist at least two distinct elements $a_1,a_2\in A$ such that $f\left(a_1\right)=f\left(a_2\right)$.

More formally:

$$\forall A,B(\mathit{finite}\mathit{sets}),\forall f:A\to B,\left(\right|A|>|B\left|\right)\Rightarrow \exists a_1,a_2\in A:(a_1\ne a_2\wedge f\left(a_1\right)=f\left(a_2\right))$$

Proof. 

We proceed by contradiction.

Step 1: Suppose the statement is false. That is, assume there exists an injective function $f:A\to B$ with $\left|A\right|>\left|B\right|$.

Step 2: Since f is injective, for each $b\in B$, the set $f^{-1}\left(b\right)=\{a\in A:f\left(a\right)=b\}$ has at most one element.

Step 3: Therefore, we can write:

$$\left|A\right|=\sum _{b\in B}|f^{-1}\left(b\right)|\le \sum _{b\in B}1=\left|B\right|$$

Step 4: But this contradicts our assumption that $\left|A\right|>\left|B\right|$.

Therefore, our assumption must be false, and the theorem is true. □

Theorem 3 

(Principle of Mathematical Induction). Let $P\left(n\right)$ be a predicate defined for natural numbers n. If the following conditions hold:

1 

Base case: $P\left(1\right)$ is true.

2 

Inductive step: For any $k\in \mathbb{N}$, if $P\left(k\right)$ is true, then $P(k+1)$ is true.

Then $P\left(n\right)$ is true for all natural numbers n.

Formally:

$$\left[P\right(1)\wedge \forall k\in \mathbb{N}(P\left(k\right)\Rightarrow P(k+1)\left)\right]\Rightarrow \forall n\in \mathbb{N}P\left(n\right)$$

Proof. 

Let $S=\{n\in \mathbb{N}:P(n\left)\mathrm{is}\mathrm{false}\right\}$. We will prove that S is empty.

Assume, for the sake of contradiction, that S is non-empty. By the Well-Ordering Principle, S has a least element. Let $m=minS$.

1) $m\ne 1$, because $P\left(1\right)$ is true by the base case. 2) Since m is the least element of S, $P(m-1)$ must be true. 3) By the inductive step, if $P(m-1)$ is true, then $P\left(m\right)$ must be true. 4) But this contradicts the fact that $m\in S$.

Therefore, our assumption must be false, and S must be empty. Thus, $P\left(n\right)$ is true for all $n\in \mathbb{N}$. □

Theorem 4 

(Principle of Strong Mathematical Induction). Let $P\left(n\right)$ be a predicate defined for natural numbers n. If the following conditions hold:

1 

Base case: $P\left(1\right)$ is true.

2 

Strong inductive step: For any $k\in \mathbb{N}$, if $P\left(j\right)$ is true for all $j\le k$, then $P(k+1)$ is true.

Then $P\left(n\right)$ is true for all natural numbers n.

Formally:

$$\left[P\right(1)\wedge \forall k\in \mathbb{N}((\forall j\le k,P(j\left)\right)\Rightarrow P(k+1)\left)\right]\Rightarrow \forall n\in \mathbb{N}P\left(n\right)$$

Proof. 

Let $S=\{n\in \mathbb{N}:P(n\left)\mathrm{is}\mathrm{false}\right\}$. We will prove that S is empty.

Step 5: Assume, for the sake of contradiction, that S is non-empty.

Step 6: By the Well-Ordering Principle, S has a least element. Let $m=minS$.

Step 7: $m\ne 1$, because $P\left(1\right)$ is true by the base case.

Step 8: Since m is the least element of S, $P\left(j\right)$ is true for all $j<m$.

Step 9: By the strong inductive step, if $P\left(j\right)$ is true for all $j<m$, then $P\left(m\right)$ must be true.

Step 10: But this contradicts the fact that $m\in S$.

Step 11:Therefore, our assumption must be false, and S must be empty.

Thus, $P\left(n\right)$ is true for all $n\in \mathbb{N}$. □

Definition 5 

(Collatz Function). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

Definition 6 

(Collatz Sequence). For any $n\in {\mathbb{N}}^{+}$, the Collatz sequence starting at n is the sequence ${\left(a_k\right)}_{k\ge 0}$ defined by:

$$\begin{array}{cc}\hfill a_0& =n\hfill \\ \hfill a_{k+1}& =C\left(a_k\right)\mathrm{for}k\ge 0\hfill \end{array}$$

Definition 7 

(Inverse Collatz Function). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be its inverse function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

3.2. Fundamental Properties

Theorem 5 

(Well-definedness of the Collatz Function). The Collatz function $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

is well-defined for all positive integers.

Proof. 

We will prove that the Collatz function is well-defined by showing that:

• The function is defined for all elements in its domain.
• The function produces a unique output for each input.

Step 12: The function is defined for all elements in its domain:

(a)

Domain: ${\mathbb{N}}^{+}=\{1,2,3,\dots \}$

(b)

$\forall n\in {\mathbb{N}}^{+}$, exactly one of the following is true:

$$\begin{array}{cc}\hfill n& \equiv 0(mod2)(\mathrm{n}\mathrm{is}\mathrm{even})\hfill \\ \hfill n& \equiv 1(mod2)(\mathrm{n}\mathrm{is}\mathrm{odd})\hfill \end{array}$$

(c)

Case 1: If n is even:

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}& :n=2k\hfill \\ \hfill C\left(n\right)& ={\displaystyle \frac{n}{2}}={\displaystyle \frac{2k}{2}}=k\in {\mathbb{N}}^{+}\hfill \end{array}$$

Note: For even $n\in {\mathbb{N}}^{+}$, ${\displaystyle \frac{n}{2}}\in {\mathbb{N}}^{+}$ always holds.

(d)

Case 2: If n is odd:

$$\begin{array}{cc}\hfill C\left(n\right)& =3n+1\hfill \\ \hfill & \ge 3·1+1=4\in {\mathbb{N}}^{+}\hfill \end{array}$$

(e)

Therefore, $C\left(n\right)$ is defined and in ${\mathbb{N}}^{+}$ for all $n\in {\mathbb{N}}^{+}$.

Step 13: The function produces a unique output for each input:

(a)

Let $n\in {\mathbb{N}}^{+}$ be arbitrary.

(b)

Case 1: If n is even:

$$\begin{array}{cc}\hfill C\left(n\right)& ={\displaystyle \frac{n}{2}}\hfill \\ \hfill & ={\displaystyle \frac{n}{2}}·1\hfill \\ \hfill & ={\displaystyle \frac{n}{2}}·{\displaystyle \frac{2}{2}}\hfill \\ \hfill & =n·{\displaystyle \frac{1}{2}}\hfill \end{array}$$

This operation produces a unique result for each even n.

(c)

Case 2: If n is odd:

$$\begin{array}{cc}\hfill C\left(n\right)& =3n+1\hfill \end{array}$$

This operation produces a unique result for each odd n.

(d)

The cases are mutually exclusive and exhaustive, ensuring a unique output for each input.

Step 14: Therefore, the Collatz function $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ is well-defined for all positive integers. □

Lemma 1 

(Surjectivity of C). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

Then C is surjective.

Proof. 

First, we recall that C is well-defined for all positive integers by Theorem 5.

We will prove that $\forall m\in {\mathbb{N}}^{+},\exists n\in {\mathbb{N}}^{+}:C\left(n\right)=m$ using strong mathematical induction on m.

Step 15: Base case: $m=1$

$$\begin{array}{cc}\hfill & \mathrm{Let}n=2\hfill \\ \hfill & \mathrm{Then}C\left(n\right)=C\left(2\right)={\displaystyle \frac{2}{2}}=1=m\hfill \end{array}$$

We now prove that 1 has no other preimage under C:

$$\begin{array}{cc}\hfill & \mathit{if}n\mathrm{is}\mathrm{even}\mathrm{and}C\left(n\right)=1,\mathrm{then}{\displaystyle \frac{n}{2}}=1\Rightarrow n=2\hfill \\ \hfill & \mathit{if}n\mathrm{is}\mathrm{odd}\mathrm{and}C\left(n\right)=1,\mathrm{then}3n+1=1\Rightarrow n=0,\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{in}{\mathbb{N}}^{+}\hfill \end{array}$$

Therefore, 2 is the unique preimage of 1 under C.

This establishes surjectivity for the base case, as we have shown that $\exists n\in {\mathbb{N}}^{+}:C\left(n\right)=1$.

Step 16: Inductive hypothesis: Assume the statement holds for all positive integers less than or equal to k, where $k\ge 1$. That is:

$$\forall j\in \{1,2,...,k\},\exists n_j\in {\mathbb{N}}^{+}:C\left(n_j\right)=j$$

Step 17: Inductive step: We will prove the statement holds for $k+1$.

Case 1. If $k+1\equiv 0(mod2)$

$$\begin{array}{cc}\hfill & \mathrm{Let}n=2(k+1)\hfill \\ \hfill & \mathrm{Then}C\left(n\right)=C\left(2(k+1)\right)={\displaystyle \frac{2(k+1)}{2}}=k+1\hfill \end{array}$$

Note that $n=2(k+1)\in {\mathbb{N}}^{+}$ since $k+1\in {\mathbb{N}}^{+}$.

Case 2. If $k+1\equiv 1(mod2)$

We consider two subcases:

Subcase 1. 2a If $k\equiv 2(mod3)$

$$\begin{array}{cc}\hfill & \mathrm{Let}n={\displaystyle \frac{k-2}{3}}+1\hfill \\ \hfill & \mathrm{Sin}\mathrm{ce}k\equiv 2(mod3),\exists q\in \mathbb{N}:k=3q+2\hfill \\ \hfill & \mathrm{Then}n={\displaystyle \frac{(3q+2)-2}{3}}+1=q+1\in {\mathbb{N}}^{+}(\sin \mathrm{ce}q\in \mathbb{N})\hfill \\ \hfill & \mathrm{Therefore}C\left(n\right)=C(q+1)=3(q+1)+1=3q+4=(3q+2)+2=k+2=(k+1)+1\hfill \end{array}$$

Step 18: Explicit justification for $k+1$ being odd when $k\equiv 2(mod3)$:

Proof. If $k\equiv 2(mod3)$, then $\exists q\in \mathbb{N}:k=3q+2$

$$\begin{array}{cc}\hfill k+1& =(3q+2)+1\hfill \\ \hfill & =3q+3\hfill \\ \hfill & =3(q+1)\hfill \end{array}$$

Since $q\in \mathbb{N}$, $(q+1)\in {\mathbb{N}}^{+}$, and any number of the form $3m$ where $m\in {\mathbb{N}}^{+}$ is odd. Therefore, $k+1$ is odd when $k\equiv 2(mod3)$. □

Subcase 2. 2b If $k\not\equiv 2(mod3)$

$$\begin{array}{cc}\hfill & \mathrm{Let}n=2(k+1)\hfill \\ \hfill & \mathrm{Sin}\mathrm{ce}k+1\equiv 1(mod2),\exists t\in \mathbb{N}:k+1=2t+1\hfill \\ \hfill & \mathrm{Then}n=2(k+1)=2(2t+1)=4t+2\in {\mathbb{N}}^{+}(\sin \mathrm{ce}t\in \mathbb{N})\hfill \\ \hfill & \mathrm{Now},C\left(n\right)=C(4t+2)={\displaystyle \frac{4t+2}{2}}=2t+1=k+1\hfill \end{array}$$

Note that $n=4t+2\equiv 2(mod4)$, so n is even and $C\left(n\right)={\displaystyle \frac{n}{2}}$.

Step 19. 5 Explicit justification for the parity of $k+1$ when $k\not\equiv 2(mod3)$:

Proof. If $k\not\equiv 2(mod3)$, then $k\equiv 0(mod3)$ or $k\equiv 1(mod3)$

• If $k\equiv 0(mod3)$, then $k=3r$ for some $r\in \mathbb{N}$ Then $k+1=3r+1$, which is odd
• If $k\equiv 1(mod3)$, then $k=3s+1$ for some $s\in \mathbb{N}$ Then $k+1=3s+2$, which is even

Therefore, when $k\not\equiv 2(mod3)$, $k+1$ can be either odd or even, depending on whether $k\equiv 0(mod3)$ or $k\equiv 1(mod3)$, respectively.

However, recall that in this case (Case 2), we are considering the situation where $k+1\equiv 1(mod2)$, i.e., $k+1$ is odd. This corresponds to the subcase where $k\equiv 0(mod3)$.

Thus, in this subcase, we are specifically dealing with $k\equiv 0(mod3)$, which ensures that $k+1$ is indeed odd. □

Step 20. 6 Conclusion: In all cases, we have found an $n\in {\mathbb{N}}^{+}$ such that $C\left(n\right)=k+1$, and we have shown that this n is unique.

Step 21. 7 By the principle of strong mathematical induction, we conclude:

$$\forall m\in {\mathbb{N}}^{+},\exists !n\in {\mathbb{N}}^{+}:C\left(n\right)=m$$

Step 22. 8 Therefore, C is surjective and each element in its codomain has a unique preimage.□

Lemma 2 

(Well-definedness of the Inverse Collatz Function). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then G is well-defined for all positive integers.

Proof. 

To prove that G is well-defined, we need to show that:

• The function is defined for all elements in its domain.
• The function produces a unique output for each input.
• All elements in the output are in the codomain.

Step 23. 1 The function is defined for all elements in its domain:

• Domain: ${\mathbb{N}}^{+}=\{1,2,3,\dots \}$
• $\forall n\in {\mathbb{N}}^{+}$, exactly one of the following is true:

  $$\begin{array}{cc}\hfill n& \equiv 4(mod6)\hfill \\ \hfill n& \not\equiv 4(mod6)\hfill \end{array}$$
• Case 1: If $n\not\equiv 4(mod6)$:

  $$\begin{array}{cc}\hfill G\left(n\right)& =\left\{2n\right\}\hfill \\ \hfill 2n& \in {\mathbb{N}}^{+}(\sin \mathrm{ce}n\in {\mathbb{N}}^{+})\hfill \end{array}$$
• Case 2: If $n\equiv 4(mod6)$:

  $$\begin{array}{cc}\hfill G\left(n\right)& =\{2n,{\displaystyle \frac{n-1}{3}}\}\hfill \\ \hfill 2n& \in {\mathbb{N}}^{+}(\sin \mathrm{ce}n\in {\mathbb{N}}^{+})\hfill \\ \hfill {\displaystyle \frac{n-1}{3}}& \in {\mathbb{N}}^{+}(\mathrm{we}\mathrm{will}\mathrm{prove}\mathrm{this}\mathrm{below})\hfill \end{array}$$

Step 24. 1a Explicit proof that ${\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}$ when $n\equiv 4(mod6)$:

Proof. If $n\equiv 4(mod6)$, then $\exists k\in \mathbb{N}:n=6k+4$.

$$\begin{array}{cc}\hfill {\displaystyle \frac{n-1}{3}}& ={\displaystyle \frac{(6k+4)-1}{3}}\hfill \\ \hfill & ={\displaystyle \frac{6k+3}{3}}\hfill \\ \hfill & =2k+1\hfill \end{array}$$

Since $k\in \mathbb{N}$, we know that $2k+1\in {\mathbb{N}}^{+}$. Moreover, $2k+1\ge 1$ for all $k\in \mathbb{N}$. Therefore, ${\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}$ when $n\equiv 4(mod6)$. □

Note: For $n\equiv 4(mod6)$, $n\ge 4$, so ${\displaystyle \frac{n-1}{3}}\ge 1$ and is an integer.

Therefore, $G\left(n\right)$ is defined and its elements are in ${\mathbb{N}}^{+}$ for all $n\in {\mathbb{N}}^{+}$.

Step 25. 2 The function produces a unique output for each input:

• Let $n\in {\mathbb{N}}^{+}$ be arbitrary.
• Case 1: If $n\not\equiv 4(mod6)$:

  $$G\left(n\right)=\left\{2n\right\}$$
  This set is uniquely determined by n.
• Case 2: If $n\equiv 4(mod6)$:

  $$G\left(n\right)=\{2n,{\displaystyle \frac{n-1}{3}}\}$$
  This set is uniquely determined by n.
• The cases are mutually exclusive and exhaustive, ensuring a unique output for each input.

Step 26. 3 All elements in the output are in the codomain:

• The codomain of G is $\mathcal{P}\left({\mathbb{N}}^{+}\right)$, the power set of positive integers.
• For all $n\in {\mathbb{N}}^{+}$, $G\left(n\right)$ is a set containing either one or two positive integers.
• Therefore, $G\left(n\right)\in \mathcal{P}\left({\mathbb{N}}^{+}\right)$ for all $n\in {\mathbb{N}}^{+}$.

Step 27. 4 Conclusion: We have shown that G satisfies all three criteria for well-definedness:

• It is defined for all elements in its domain.
• It produces a unique output for each input.
• All elements in the output are in the codomain.

Therefore, the inverse Collatz function $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ is well-defined for all positive integers.

Lemma 3 (Non-emptiness and Uniqueness of G(n)). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then for every $n\in {\mathbb{N}}^{+}$, the set $G\left(n\right)$ is non-empty and uniquely determined.

Proof. We will prove this lemma in two parts:

• Non-emptiness of $G\left(n\right)$
• Uniqueness of $G\left(n\right)$

Step 28. 1 Non-emptiness of $G\left(n\right)$

Let $n\in {\mathbb{N}}^{+}$ be arbitrary. We consider two cases:

Case 3. 1 $n\not\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(n\right)& =\left\{2n\right\}\hfill \\ \hfill 2n& \in {\mathbb{N}}^{+}(\sin \mathrm{ce}n\in {\mathbb{N}}^{+})\hfill \\ \hfill \therefore G\left(n\right)& \ne \varnothing \hfill \end{array}$$

Case 4. 2 $n\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(n\right)& =\{2n,{\displaystyle \frac{n-1}{3}}\}\hfill \\ \hfill 2n& \in {\mathbb{N}}^{+}(\sin \mathrm{ce}n\in {\mathbb{N}}^{+})\hfill \\ \hfill {\displaystyle \frac{n-1}{3}}& \in {\mathbb{N}}^{+}(\mathrm{we}\mathrm{will}\mathrm{prove}\mathrm{this}\mathrm{below})\hfill \\ \hfill \therefore G\left(n\right)& \ne \varnothing \hfill \end{array}$$

Step 29. 1a Detailed explanation of why ${\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}$ when $n\equiv 4(mod6)$:

If $n\equiv 4(mod6)$, then $\exists k\in \mathbb{N}:n=6k+4$.

$$\begin{array}{cc}\hfill {\displaystyle \frac{n-1}{3}}& ={\displaystyle \frac{(6k+4)-1}{3}}\hfill \\ \hfill & ={\displaystyle \frac{6k+3}{3}}\hfill \\ \hfill & =2k+1\hfill \end{array}$$

Since $k\in \mathbb{N}$, we know that $2k+1\in {\mathbb{N}}^{+}$. Moreover, $2k+1\ge 1$ for all $k\in \mathbb{N}$. Therefore, ${\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}$ when $n\equiv 4(mod6)$.

In both cases, $G\left(n\right)$ is non-empty. Since n was arbitrary, we conclude:

$$\forall n\in {\mathbb{N}}^{+},G\left(n\right)\ne \varnothing$$

Step 30. 2 Uniqueness of $G\left(n\right)$

Let $n\in {\mathbb{N}}^{+}$ be arbitrary. We will show that $G\left(n\right)$ is uniquely determined by n.

Case 5. 1 $n\not\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(n\right)& =\left\{2n\right\}\hfill \\ \hfill & =\left\{2n\right\}\cup \varnothing \hfill \\ \hfill & =\left\{2n\right\}\cup \left\{{\displaystyle \frac{n-1}{3}}:{\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}\right\}\hfill \end{array}$$

Case 6. 2 $n\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(n\right)& =\{2n,{\displaystyle \frac{n-1}{3}}\}\hfill \\ \hfill & =\left\{2n\right\}\cup \left\{{\displaystyle \frac{n-1}{3}}\right\}\hfill \\ \hfill & =\left\{2n\right\}\cup \left\{{\displaystyle \frac{n-1}{3}}:{\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}\right\}\hfill \end{array}$$

In both cases, $G\left(n\right)$ can be expressed as:

$$G\left(n\right)=\left\{2n\right\}\cup \left\{{\displaystyle \frac{n-1}{3}}:{\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}\right\}$$

This expression is uniquely determined by n for the following reasons:

• The term $2n$ is always included and is a function of n.
• The term ${\displaystyle \frac{n-1}{3}}$ is included if and only if it is a positive integer, which depends solely on the value of n.
• The condition ${\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}$ is equivalent to $n\equiv 4(mod6)$, which is uniquely determined by n.

Therefore, for any given $n\in {\mathbb{N}}^{+}$, the set $G\left(n\right)$ is uniquely determined.

Since n was arbitrary, we conclude:

$$\forall n\in {\mathbb{N}}^{+},G\left(n\right)\mathrm{is}\mathrm{uniquely}\mathrm{determined}$$

Case 31. 3 Conclusion: Combining the results from Step 1 and Step 2, we have shown that for every $n\in {\mathbb{N}}^{+}$, the set $G\left(n\right)$ is non-empty and uniquely determined. □

Lemma 4 (Injectivity of G) Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then G is injective, i.e., $\forall a,b\in {\mathbb{N}}^{+}:G\left(a\right)=G\left(b\right)\Rightarrow a=b$.

Proof. We will prove this by contradiction. Assume G is not injective. Then:

Step 32. 1 $\exists a,b\in {\mathbb{N}}^{+}:(a\ne b)\wedge (G\left(a\right)=G\left(b\right))$

Let $a,b\in {\mathbb{N}}^{+}$ be such that $a\ne b$ and $G\left(a\right)=G\left(b\right)$. We will consider all possible cases:

Case 7. 1 $a\not\equiv 4(mod6)$ and $b\not\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(a\right)& =\left\{2a\right\}\hfill \\ \hfill G\left(b\right)& =\left\{2b\right\}\hfill \\ \hfill G\left(a\right)=G\left(b\right)& \Rightarrow \left\{2a\right\}=\left\{2b\right\}\hfill \\ \hfill & \Rightarrow 2a=2b\hfill \\ \hfill & \Rightarrow a=b\hfill \end{array}$$

This contradicts our assumption that $a\ne b$.

Case 8. 2 $a\equiv 4(mod6)$ and $b\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(a\right)& =\{2a,{\displaystyle \frac{a-1}{3}}\}\hfill \\ \hfill G\left(b\right)& =\{2b,{\displaystyle \frac{b-1}{3}}\}\hfill \\ \hfill G\left(a\right)=G\left(b\right)& \Rightarrow \{2a,{\displaystyle \frac{a-1}{3}}\}=\{2b,{\displaystyle \frac{b-1}{3}}\}\hfill \end{array}$$

This equality of sets implies one of two subcases:

Subcase 3. 2a $2a=2b$ and ${\displaystyle \frac{a-1}{3}}={\displaystyle \frac{b-1}{3}}$

$$\begin{array}{cc}\hfill 2a=2b& \Rightarrow a=b\hfill \end{array}$$

This contradicts our assumption that $a\ne b$.

Subcase 4. 2b $2a={\displaystyle \frac{b-1}{3}}$ and $2b={\displaystyle \frac{a-1}{3}}$

$$\begin{array}{cc}\hfill 2a& ={\displaystyle \frac{b-1}{3}}\hfill \\ \hfill 6a& =b-1\hfill \\ \hfill b& =6a+1\hfill \\ \hfill 2b& ={\displaystyle \frac{a-1}{3}}\hfill \\ \hfill 2(6a+1)& ={\displaystyle \frac{a-1}{3}}\hfill \\ \hfill 12a+2& ={\displaystyle \frac{a-1}{3}}\hfill \\ \hfill 36a+6& =a-1\hfill \\ \hfill 35a& =-7\hfill \\ \hfill a& =-{\displaystyle \frac{1}{5}}\hfill \end{array}$$

This last equation, $a=-{\displaystyle \frac{1}{5}}$, does not directly contradict our initial assumption, but it shows us that if such an a existed, it would not be a positive natural number. This means that there cannot be values $a,b\in {\mathbb{N}}^{+}$ that simultaneously satisfy $2a={\displaystyle \frac{b-1}{3}}$ and $2b={\displaystyle \frac{a-1}{3}}$.

In other words, this result does not invalidate the injectivity property, but rather demonstrates that it is impossible for two distinct elements in ${\mathbb{N}}^{+}$ to have equal images under G in this specific manner.

Case 9. 3 $(a\not\equiv 4(mod6)\wedge b\equiv 4(mod6\left)\right)\vee (a\equiv 4(mod6)\wedge b\not\equiv 4(mod6\left)\right)$

Without loss of generality, assume $a\not\equiv 4(mod6)$ and $b\equiv 4(mod6)$.

$$\begin{array}{cc}\hfill G\left(a\right)& =\left\{2a\right\}\hfill \\ \hfill G\left(b\right)& =\{2b,{\displaystyle \frac{b-1}{3}}\}\hfill \\ \hfill G\left(a\right)=G\left(b\right)& \Rightarrow \left\{2a\right\}=\{2b,{\displaystyle \frac{b-1}{3}}\}\hfill \end{array}$$

This is a contradiction because a set with one element cannot equal a set with two distinct elements.

Step 33. 2 Let’s prove that $2b\ne {\displaystyle \frac{b-1}{3}}$ for all $b\in {\mathbb{N}}^{+}$:

Lemma 5.For all $b\in {\mathbb{N}}^{+}$, $2b\ne {\displaystyle \frac{b-1}{3}}$.

Proof. Assume, for the sake of contradiction, that $\exists b\in {\mathbb{N}}^{+}:2b={\displaystyle \frac{b-1}{3}}$. Then:

$$\begin{array}{cc}\hfill 2b& ={\displaystyle \frac{b-1}{3}}\hfill \\ \hfill 6b& =b-1\hfill \\ \hfill 5b& =-1\hfill \\ \hfill b& =-{\displaystyle \frac{1}{5}}\hfill \end{array}$$

This contradicts $b\in {\mathbb{N}}^{+}$. Therefore, $\forall b\in {\mathbb{N}}^{+},2b\ne {\displaystyle \frac{b-1}{3}}$. □

Step 34. 3 By Lemma 5, we know that $2b\ne {\displaystyle \frac{b-1}{3}}$. Therefore:

$$\begin{array}{cc}\hfill \left|\right\{2a\left\}\right|& =1\hfill \\ \hfill \left|\right\{2b,{\displaystyle \frac{b-1}{3}}\left\}\right|& =2\hfill \end{array}$$

Thus, $\left\{2a\right\}\ne \{2b,{\displaystyle \frac{b-1}{3}}\}$, which contradicts our assumption that $G\left(a\right)=G\left(b\right)$.

Step 35. 4 In all cases, we have reached a contradiction. Therefore, our initial assumption must be false.

Step 36. 5 We conclude that:

$$\forall a,b\in {\mathbb{N}}^{+}:G\left(a\right)=G\left(b\right)\Rightarrow a=b$$

Thus, G is injective. □

Lemma 6 (Multivalued Injectivity of G) Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then G is multivalued injective, i.e., $\forall a,b\in {\mathbb{N}}^{+},a\ne b\Rightarrow G\left(a\right)\cap G\left(b\right)=\varnothing$.

Proof. We will prove this by contradiction. Assume G is not multivalued injective. Then:

Step 37. 1 $\exists a,b\in {\mathbb{N}}^{+}:(a\ne b)\wedge (G\left(a\right)\cap G\left(b\right)\ne \varnothing )$

Let $a,b\in {\mathbb{N}}^{+}$ be such that $a\ne b$ and $G\left(a\right)\cap G\left(b\right)\ne \varnothing$. We will consider all possible cases:

Case 10. 1 $a\not\equiv 4(mod6)$ and $b\not\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(a\right)& =\left\{2a\right\}\hfill \\ \hfill G\left(b\right)& =\left\{2b\right\}\hfill \\ \hfill G\left(a\right)\cap G\left(b\right)\ne \varnothing & \Rightarrow \left\{2a\right\}\cap \left\{2b\right\}\ne \varnothing \hfill \\ \hfill & \Rightarrow 2a=2b\hfill \\ \hfill & \Rightarrow a=b\hfill \end{array}$$

This contradicts our assumption that $a\ne b$.

Case 11. 2 $a\equiv 4(mod6)$ and $b\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(a\right)& =\{2a,{\displaystyle \frac{a-1}{3}}\}\hfill \\ \hfill G\left(b\right)& =\{2b,{\displaystyle \frac{b-1}{3}}\}\hfill \\ \hfill G\left(a\right)\cap G\left(b\right)\ne \varnothing & \Rightarrow (2a=2b)\vee (2a={\displaystyle \frac{b-1}{3}})\vee (2b={\displaystyle \frac{a-1}{3}})\vee ({\displaystyle \frac{a-1}{3}}={\displaystyle \frac{b-1}{3}})\hfill \end{array}$$

We will consider each subcase:

Subcase 5. 2a $2a=2b\Rightarrow a=b$ This contradicts our assumption that $a\ne b$.

Subcase 6. 2b $2a={\displaystyle \frac{b-1}{3}}$

$$\begin{array}{cc}\hfill 2a& ={\displaystyle \frac{b-1}{3}}\hfill \\ \hfill 6a& =b-1\hfill \\ \hfill b& =6a+1\hfill \end{array}$$

Now, let’s consider the congruence classes of both sides modulo 6:

$$\begin{array}{cc}\hfill b& \equiv 4(mod6)\left(\mathrm{given}\right)\hfill \\ \hfill 6a+1& \equiv 1(mod6)(\sin \mathrm{ce}6a\equiv 0(mod6)\mathrm{for}\mathrm{any}\mathrm{integer}a)\hfill \end{array}$$

This leads to a contradiction because:

$$\begin{array}{cc}\hfill b& \equiv 6a+1(mod6)\hfill \\ \hfill 4& \equiv 1(mod6)\hfill \end{array}$$

Which is false for any integer values of a and b.

Subcase 7. 2c $2b={\displaystyle \frac{a-1}{3}}$ This is symmetric to Subcase 2b and leads to the same contradiction.

Subcase 8. 2d ${\displaystyle \frac{a-1}{3}}={\displaystyle \frac{b-1}{3}}\Rightarrow a=b$ This contradicts our assumption that $a\ne b$.

Case 12. 3 $(a\not\equiv 4(mod6)\wedge b\equiv 4(mod6\left)\right)\vee (a\equiv 4(mod6)\wedge b\not\equiv 4(mod6\left)\right)$

Without loss of generality, assume $a\not\equiv 4(mod6)$ and $b\equiv 4(mod6)$.

$$\begin{array}{cc}\hfill G\left(a\right)& =\left\{2a\right\}\hfill \\ \hfill G\left(b\right)& =\{2b,{\displaystyle \frac{b-1}{3}}\}\hfill \\ \hfill G\left(a\right)\cap G\left(b\right)\ne \varnothing & \Rightarrow (2a=2b)\vee (2a={\displaystyle \frac{b-1}{3}})\hfill \end{array}$$

We will consider each subcase:

Subcase 9. 3a $2a=2b\Rightarrow a=b$ This contradicts our assumption that $a\ne b$.

Subcase 10. 3b $2a={\displaystyle \frac{b-1}{3}}$

$$\begin{array}{cc}\hfill 2a& ={\displaystyle \frac{b-1}{3}}\hfill \\ \hfill 6a& =b-1\hfill \\ \hfill b& =6a+1\hfill \end{array}$$

Now, let’s consider the congruence classes of both sides modulo 6:

$$\begin{array}{cc}\hfill b& \equiv 4(mod6)\left(\mathrm{given}\right)\hfill \\ \hfill 6a+1& \equiv 1(mod6)(\sin \mathrm{ce}6a\equiv 0(mod6)\mathrm{for}\mathrm{any}\mathrm{integer}a)\hfill \end{array}$$

This leads to a contradiction because:

$$\begin{array}{cc}\hfill b& \equiv 6a+1(mod6)\hfill \\ \hfill 4& \equiv 1(mod6)\hfill \end{array}$$

Which is false for any integer values of a and b.

Step 38. 2 In all cases, we have reached a contradiction. Therefore, our initial assumption must be false.

Step 39. 3 We conclude that $\forall a,b\in {\mathbb{N}}^{+},a\ne b\Rightarrow G\left(a\right)\cap G\left(b\right)=\varnothing$.

Thus, G is multivalued injective. □

Lemma 7 (Surjectivity and Uniqueness of G)Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

and $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be its inverse function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then for every subset $A\subseteq {\mathbb{N}}^{+}$, there exists a unique subset $B\subseteq {\mathbb{N}}^{+}$ such that $G\left(B\right)=A$.

Proof. We will prove this in two steps: existence and uniqueness.

Step 40. 1 Existence Let $A\subseteq {\mathbb{N}}^{+}$ be an arbitrary subset. Define $B=\{n\in {\mathbb{N}}^{+}:C\left(n\right)\in A\}$. We will show that $G\left(B\right)=A$.

(a)

$G\left(B\right)\subseteq A$:

$$\begin{array}{cc}\hfill \forall x\in G\left(B\right)& \Rightarrow \exists n\in B:x\in G\left(n\right)\hfill \\ \hfill & \Rightarrow \exists n\in B:C\left(x\right)=n(\mathrm{by}\mathrm{definition}\mathrm{of}G)\hfill \\ \hfill & \Rightarrow \exists n\in B:C\left(x\right)\in A(\mathrm{by}\mathrm{definition}\mathrm{of}B)\hfill \\ \hfill & \Rightarrow x\in A(\mathrm{by}\mathrm{definition}\mathrm{of}G)\hfill \end{array}$$

(b)

$A\subseteq G\left(B\right)$:

$$\begin{array}{cc}\hfill \forall a\in A& \Rightarrow \exists n\in {\mathbb{N}}^{+}:C\left(n\right)=a(\mathrm{by}\mathrm{surjectivity}\mathrm{of}C,\mathrm{Lemma})\hfill \\ \hfill & \Rightarrow n\in B(\mathrm{by}\mathrm{definition}\mathrm{of}B)\hfill \\ \hfill & \Rightarrow a\in G\left(n\right)\subseteq G\left(B\right)\hfill \end{array}$$

From (i) and (ii), we conclude $G\left(B\right)=A$. Thus, we have shown that there exists a set B such that $G\left(B\right)=A$.

Step 41. 2 Uniqueness Suppose, for the sake of contradiction, that there exist two distinct sets $B_1$ and $B_2$ such that $G\left(B_1\right)=A$ and $G\left(B_2\right)=A$.

Let $x\in B_1\cup B_2$. Without loss of generality, assume $x\in B_1$. Then:

$$\begin{array}{cc}\hfill x\in B_1& \Rightarrow G\left(x\right)\subseteq G\left(B_1\right)=A=G\left(B_2\right)\hfill \\ \hfill & \Rightarrow \exists y\in B_2:G\left(x\right)\cap G\left(y\right)\ne \varnothing \hfill \end{array}$$

Now, we use the contrapositive of the multivalued injectivity of G (Lemma 6):

$$\forall a,b\in {\mathbb{N}}^{+}:G\left(a\right)\cap G\left(b\right)\ne \varnothing \Rightarrow a=b$$

Applying this to our case:

$$G\left(x\right)\cap G\left(y\right)\ne \varnothing \Rightarrow x=y$$

Therefore, $x\in B_2$. We have shown that $B_1\subseteq B_2$.

By a symmetric argument (swapping the roles of $B_1$ and $B_2$), we can show that $B_2\subseteq B_1$.

Thus, $B_1=B_2$, contradicting our assumption that they were distinct.

To formally prove that $B_1=B_2$, we use the Axiom of Extensionality:

$$\forall X,Y:(X=Y)\iff (\forall z:(z\in X\iff z\in Y\left)\right)$$

We have shown:

$$\begin{array}{cc}\hfill & \forall z:(z\in B_1\Rightarrow z\in B_2)\mathrm{and}\forall z:(z\in B_2\Rightarrow z\in B_1)\hfill \\ \hfill & \iff \forall z:(z\in B_1\iff z\in B_2)\hfill \\ \hfill & \iff B_1=B_2\hfill \end{array}$$

This contradicts our assumption that $B_1$ and $B_2$ were distinct. Therefore, B is unique.

We conclude that for every subset $A\subseteq {\mathbb{N}}^{+}$, there exists a unique subset $B\subseteq {\mathbb{N}}^{+}$ such that $G\left(B\right)=A$. □

Lemma 8 (Exhaustiveness of G). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

and $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be its inverse function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then G is exhaustive, i.e., $\forall n\in {\mathbb{N}}^{+},\exists m\in {\mathbb{N}}^{+}:n\in G\left(m\right)$.

Proof. We will prove this by considering all possible congruence classes of n modulo 6.

Step 42. 1 Let $n\in {\mathbb{N}}^{+}$ be arbitrary. We consider six cases:

Case 13. 1 $n\equiv 0(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k\hfill \\ \hfill \mathrm{Let}m& =3k\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\not\equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(3k\left)\right\}=\left\{6k\right\}=\left\{n\right\}\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Case 14. 2 $n\equiv 1(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+1\hfill \\ \hfill \mathrm{Let}m& =2n=2(6k+1)=12k+2\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\not\equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(12k+2\left)\right\}=\{24k+4\}\hfill \\ \hfill n& =6k+1={\displaystyle \frac{24k+4}{4}}\in G\left(m\right)\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Case 15. 3 $n\equiv 2(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+2\hfill \\ \hfill \mathrm{Let}m& =3k+1\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\not\equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(3k+1\left)\right\}=\{6k+2\}=\left\{n\right\}\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Case 16. 4 $n\equiv 3(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+3\hfill \\ \hfill \mathrm{Let}m& =2n=2(6k+3)=12k+6\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\not\equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(12k+6\left)\right\}=\{24k+12\}\hfill \\ \hfill n& =6k+3={\displaystyle \frac{24k+12}{4}}\in G\left(m\right)\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Case 17. 5 $n\equiv 4(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+4\hfill \\ \hfill \mathrm{Let}m& =2k+1\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\equiv 1(mod2)\hfill \\ \hfill C\left(m\right)& =3m+1=3(2k+1)+1=6k+4=n\hfill \\ \hfill \therefore n& \in G\left(C\right(m\left)\right)=G\left(n\right)\hfill \end{array}$$

Case 18. 6 $n\equiv 5(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+5\hfill \\ \hfill \mathrm{Let}m& =2n=2(6k+5)=12k+10\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\not\equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(12k+10\left)\right\}=\{24k+20\}\hfill \\ \hfill n& =6k+5={\displaystyle \frac{24k+20}{4}}\in G\left(m\right)\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Step 43. 2 We have shown that for each congruence class of n modulo 6, there exists an $m\in {\mathbb{N}}^{+}$ such that $n\in G\left(m\right)$. Since these cases are exhaustive and mutually exclusive, we conclude:

$$\forall n\in {\mathbb{N}}^{+},\exists m\in {\mathbb{N}}^{+}:n\in G\left(m\right)$$

Step 44. 3 Therefore, G is exhaustive. □

Theorem 6 (Finiteness of Preimages of G). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{2n\right\}\cup \left\{\begin{array}{cc}\left\{{\displaystyle \frac{n-1}{3}}\right\}\hfill & \mathit{if}n\equiv 1(mod3)\mathit{and}{\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}\hfill \\ \varnothing \hfill & \mathit{otherwise}\hfill \end{array}\right.$$

Then for all $j\in \mathbb{N}$, $G^j\left(\left\{1\right\}\right)$ is a finite set, where $G^j$ denotes j successive applications of G.

Proof. We will prove this theorem by induction on j. First, we establish key properties of G: □

Lemma 9 (G Cardinality). For all $n\in {\mathbb{N}}^{+}$, $\left|G\right(n\left)\right|\le 2$.

Proof. Let $n\in {\mathbb{N}}^{+}$ be arbitrary. We consider two cases:

Case 19. 1 $n\not\equiv 1(mod3)$ or ${\displaystyle \frac{n-1}{3}}\notin {\mathbb{N}}^{+}$

$$G\left(n\right)=\left\{2n\right\}\Rightarrow \left|G\right(n\left)\right|=1\le 2$$

Case 20. 2 $n\equiv 1(mod3)$ and ${\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}$

$$G\left(n\right)=\{2n,{\displaystyle \frac{n-1}{3}}\}\Rightarrow \left|G\left(n\right)\right|=2\le 2$$

Therefore, $\forall n\in {\mathbb{N}}^{+},\left|G\left(n\right)\right|\le 2$. □

Lemma 10 (Multivalued Injectivity of G). For all $a,b\in {\mathbb{N}}^{+}$, if $a\ne b$, then $G\left(a\right)\cap G\left(b\right)=\varnothing$.

Proof. This is a direct consequence of Lemma 10 (Multivalued Injectivity of G).

Now we proceed with the induction proof:

Step 45. 1 Base case: $j=0$

$$G^0\left(\left\{1\right\}\right)=\left\{1\right\}$$

Clearly, $\left|\right\{1\left\}\right|=1<\infty$. Therefore, $G^0\left(\left\{1\right\}\right)$ is finite.

Step 46. 2 Inductive hypothesis: Assume that for some $k\in \mathbb{N}$, $G^k\left(\left\{1\right\}\right)$ is finite. Let $|G^k\left(\left\{1\right\}\right)|=m$ for some $m\in \mathbb{N}$. Note that m is finite by the inductive hypothesis.

Step 47. 3 Inductive step: We need to prove that $G^{k+1}\left(\left\{1\right\}\right)$ is finite.

$$\begin{array}{cc}\hfill G^{k+1}\left(\left\{1\right\}\right)& =G\left(G^k\left(\left\{1\right\}\right)\right)\hfill \\ \hfill & =G\left(\{x_1,x_2,\dots ,x_m\}\right)\mathrm{where}\{x_1,x_2,\dots ,x_m\}=G^k\left(\left\{1\right\}\right)\hfill \\ \hfill & =\bigcup _{i=1}^{m}G\left(x_i\right)\hfill \end{array}$$

Now, we will bound the cardinality of $G^{k+1}\left(\left\{1\right\}\right)$ using the following steps:

Step 48. 3a By Lemma 9, we know that $\left|G\right(x_i\left)\right|\le 2$ for all $i\in \{1,2,\dots ,m\}$.

Step 49. 3b By Lemma 6, we know that $G\left(x_i\right)\cap G\left(x_j\right)=\varnothing$ for all $i\ne j$.

Step 50. 3c Using the sum of cardinalities of disjoint sets:

$$\begin{array}{cc}\hfill |G^{k+1}\left(\left\{1\right\}\right)|& =\left|\bigcup _{i=1}^{m}G\left(x_i\right)\right|\hfill \\ \hfill & =\sum _{i=1}^m\left|G\left(x_i\right)\right|(\sin \mathrm{ce}\mathrm{the}\mathrm{sets}\mathrm{are}\mathrm{disjoint}\mathrm{by}\mathrm{step}3\mathrm{b})\hfill \\ \hfill & \le \sum _{i=1}^{m}2(\sin \mathrm{ce}\left|G\left(x_i\right)\right|\le 2\mathrm{for}\mathrm{all}i\mathrm{by}\mathrm{step}3\mathrm{a})\hfill \\ \hfill & =2m\hfill \\ \hfill & <\infty (\sin \mathrm{ce}m\mathrm{is}\mathrm{finite}\mathrm{by}\mathrm{the}\mathrm{inductive}\mathrm{hypothesis})\hfill \end{array}$$

Step 51. 3d Thus, $G^{k+1}\left(\left\{1\right\}\right)$ is finite, as its cardinality is bounded by $2m$, which is finite.

Step 52. 4 By the principle of mathematical induction, we conclude:

$$\forall j\in \mathbb{N},G^j\left(\left\{1\right\}\right)\mathrm{is}\mathrm{finite}$$

This completes the proof of the theorem. □

Theorem 7 (Non-emptiness of Preimages of G). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{2n\right\}\cup \left\{\begin{array}{cc}\left\{{\displaystyle \frac{n-1}{3}}\right\}\hfill & \mathit{if}n\equiv 1(mod3)\mathit{and}{\displaystyle \frac{n-1}{3}}\in {\mathbb{N}}^{+}\hfill \\ \varnothing \hfill & \mathit{otherwise}\hfill \end{array}\right.$$

Then for all $j\in \mathbb{N}$, $G^j\left(\left\{1\right\}\right)$ is non-empty, where $G^j$ denotes j successive applications of G.

Proof. We will prove this theorem by strong induction on j. First, we establish a key property of G:

Lemma 11. For all $n\in {\mathbb{N}}^{+}$, $G\left(n\right)\ne \varnothing$.

Proof. Let $n\in {\mathbb{N}}^{+}$ be arbitrary. By the definition of G:

$$G\left(n\right)=\left\{2n\right\}\cup S,\mathrm{where}S\mathrm{is}\mathrm{either}\left\{{\displaystyle \frac{n-1}{3}}\right\}\mathrm{or}\varnothing$$

Since $n\in {\mathbb{N}}^{+}$, we know that $2n\in {\mathbb{N}}^{+}$. Therefore, $\left\{2n\right\}\ne \varnothing$. Thus, regardless of S, we have $G\left(n\right)\ne \varnothing$. □

Now we proceed with the strong induction proof:

Step 53. 1 Base case: $j=0$

$$G^0\left(\left\{1\right\}\right)=\left\{1\right\}$$

Clearly, $\left\{1\right\}\ne \varnothing$. Therefore, $G^0\left(\left\{1\right\}\right)$ is non-empty.

Step 54. 2 Inductive hypothesis: Assume that for all $k\le j$, where $j\in \mathbb{N}$, $G^k\left(\left\{1\right\}\right)$ is non-empty.

Step 55. 3 Inductive step: We need to prove that $G^{j+1}\left(\left\{1\right\}\right)$ is non-empty.

By the inductive hypothesis, $G^j\left(\left\{1\right\}\right)$ is non-empty. Let $x\in G^j\left(\left\{1\right\}\right)$.

Now, consider $G\left(x\right)$:

$$\begin{array}{cc}\hfill G\left(x\right)& =\left\{2x\right\}\cup \left\{\begin{array}{cc}\left\{{\displaystyle \frac{x-1}{3}}\right\}\hfill & \mathit{if}x\equiv 1(mod3)\mathrm{and}{\displaystyle \frac{x-1}{3}}\in {\mathbb{N}}^{+}\hfill \\ \varnothing \hfill & \mathrm{otherwise}\hfill \end{array}\right.\hfill \\ \hfill & \supseteq \left\{2x\right\}(\sin \mathrm{ce}\mathrm{the}\mathrm{union}\mathrm{always}\mathrm{includes}\left\{2x\right\})\hfill \end{array}$$

Since $x\in {\mathbb{N}}^{+}$, we know that $2x\in {\mathbb{N}}^{+}$. Therefore:

$$\begin{array}{cc}\hfill G\left(x\right)& \ne \varnothing \hfill \\ \hfill 2x& \in G\left(x\right)\hfill \end{array}$$

Now, consider $G^{j+1}\left(\left\{1\right\}\right)$:

$$\begin{array}{cc}\hfill G^{j+1}\left(\left\{1\right\}\right)& =G\left(G^j\left(\left\{1\right\}\right)\right)\hfill \\ \hfill & =\bigcup _{y\in G^j\left(\left\{1\right\}\right)}G\left(y\right)\hfill \\ \hfill & \supseteq G\left(x\right)(\sin \mathrm{ce}x\in G^j\left(\left\{1\right\}\right))\hfill \\ \hfill & \ne \varnothing \hfill \end{array}$$

Thus, $G^{j+1}\left(\left\{1\right\}\right)$ is non-empty.

Step 56. 4 By the principle of strong mathematical induction, we conclude:

$$\forall j\in \mathbb{N},G^j\left(\left\{1\right\}\right)\ne \varnothing$$

This completes the proof of the theorem.

Theorem 8 (Monotonicity of G). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then G is monotonic, i.e., for all $n\in {\mathbb{N}}^{+}$ and all $x\in G\left(n\right)$:

$$x\le 2n$$

Proof. We will prove this theorem by considering all possible cases based on the congruence class of n modulo 6.

Step 57. 1 Let $n\in {\mathbb{N}}^{+}$ be arbitrary.

Case 21. 1 $n\not\equiv 4(mod6)$

In this case, $G\left(n\right)=\left\{2n\right\}$.

$$\begin{array}{cc}\hfill \forall x\in G\left(n\right)& :x=2n\hfill \\ \hfill & \Rightarrow x=2n\le 2n\hfill \end{array}$$

Case 22. 2 $n\equiv 4(mod6)$

In this case, $G\left(n\right)=\{2n,{\displaystyle \frac{n-1}{3}}\}$.

Step 58. 2 For $x=2n$:

$$x=2n\le 2n$$

Step 59. 3 For $x={\displaystyle \frac{n-1}{3}}$:

Since $n\equiv 4(mod6)$, we can write $n=6k+4$ for some $k\in \mathbb{N}$.

$$\begin{array}{cc}\hfill x& ={\displaystyle \frac{n-1}{3}}\hfill \\ \hfill & ={\displaystyle \frac{(6k+4)-1}{3}}\hfill \\ \hfill & ={\displaystyle \frac{6k+3}{3}}\hfill \\ \hfill & =2k+1\hfill \end{array}$$

Step 60. 4 Now, we need to show that $2k+1\le 2(6k+4)$:

$$\begin{array}{cc}\hfill 2k+1& \le 2(6k+4)\hfill \\ \hfill 2k+1& \le 12k+8\hfill \\ \hfill 1& \le 10k+8\hfill \\ \hfill -7& \le 10k\hfill \end{array}$$

Step 61. 5 This inequality holds for all $k\in \mathbb{N}$, therefore:

$$x={\displaystyle \frac{n-1}{3}}\le 2n$$

Step 62. 6 We have shown that in all cases, for any $x\in G\left(n\right)$, $x\le 2n$.

Step 63. 7 Since n was arbitrary, we can conclude:

$$\forall n\in {\mathbb{N}}^{+},\forall x\in G\left(n\right):x\le 2n$$

Step 64. 8 Therefore, G is monotonic. □

Lemma 12 (C and G are Inverse Functions). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

and let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be its inverse function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathit{if}n\not\equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathit{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then, for all $n\in {\mathbb{N}}^{+}$:

• $C\left(G\right(n\left)\right)=\left\{n\right\}$
• $n\in G\left(C\right(n\left)\right)$

Proof. We will prove each part separately.

Step 65. 1 Let’s prove that $C\left(G\right(n\left)\right)=\left\{n\right\}$ for all $n\in {\mathbb{N}}^{+}$:

Case 23. 1 If $n\not\equiv 4(mod6)$

$$\begin{array}{cc}\hfill C\left(G\right(n\left)\right)& =C\left(\right\{2n\left\}\right)\hfill \\ \hfill & =\left\{{\displaystyle \frac{2n}{2}}\right\}\hfill \\ \hfill & =\left\{n\right\}\hfill \end{array}$$

Case 24. 2 If $n\equiv 4(mod6)$

$$\begin{array}{cc}\hfill C\left(G\right(n\left)\right)& =C\left(\{2n,{\displaystyle \frac{n-1}{3}}\}\right)\hfill \\ \hfill & =\{C\left(2n\right),C\left({\displaystyle \frac{n-1}{3}}\right)\}\hfill \\ \hfill & =\{{\displaystyle \frac{2n}{2}},3\left({\displaystyle \frac{n-1}{3}}\right)+1\}\hfill \\ \hfill & =\{n,n-1+1\}\hfill \\ \hfill & =\{n,n\}\hfill \\ \hfill & =\left\{n\right\}\hfill \end{array}$$

Step 66. 2 Let’s prove that $n\in G\left(C\right(n\left)\right)$ for all $n\in {\mathbb{N}}^{+}$:

Case 25. If n is even

$$\begin{array}{cc}\hfill C\left(n\right)& ={\displaystyle \frac{n}{2}}\hfill \\ \hfill G\left(C\right(n\left)\right)& =G\left({\displaystyle \frac{n}{2}}\right)\hfill \\ \hfill & =\{2·{\displaystyle \frac{n}{2}}\}\hfill \\ \hfill & =\left\{n\right\}\hfill \end{array}$$

Therefore, $n\in G\left(C\right(n\left)\right)$.

Case 26. If n is odd

$$\begin{array}{cc}\hfill C\left(n\right)& =3n+1\hfill \\ \hfill G\left(C\right(n\left)\right)& =G(3n+1)\hfill \end{array}$$

Now, we need to consider two subcases:

Subcase 11. 2a If $3n+1\not\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(C\right(n\left)\right)& =G(3n+1)\hfill \\ \hfill & =\left\{2\right(3n+1\left)\right\}\hfill \\ \hfill & =\{6n+2\}\hfill \end{array}$$

Subcase 12. 2b If $3n+1\equiv 4(mod6)$

$$\begin{array}{cc}\hfill G\left(C\right(n\left)\right)& =G(3n+1)\hfill \\ \hfill & =\{2(3n+1),{\displaystyle \frac{(3n+1)-1}{3}}\}\hfill \\ \hfill & =\{6n+2,n\}\hfill \end{array}$$

In both subcases, we can see that $n\in G\left(C\right(n\left)\right)$. For subcase 2a, note that $n={\displaystyle \frac{(6n+2)-2}{6}}$, which is an integer since n is odd. For subcase 2b, n is explicitly included in the set.

Therefore, for all odd n, we have $n\in G\left(C\right(n\left)\right)$.

Step 67. 3 Thus, we have proved that $C\left(G\right(n\left)\right)=\left\{n\right\}$ and $n\in G\left(C\right(n\left)\right)$ for all $n\in {\mathbb{N}}^{+}$. □

Theorem 9 (Preservation of Properties under Composition of G). For all $i,j\in \mathbb{N}$, the composition $G^i\circ G^j$ satisfies the following properties:

• Injectivity
• Multivalued injectivity
• Monotonicity
• Exhaustiveness
• Finiteness of preimages
• Non-emptiness of preimages

where $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ is the inverse Collatz function defined as in Theorem 6.

Proof. We will prove each property separately for $G^i\circ G^j$, using the fact that G and C are inverse functions of each other, as established in Lemma 13.

Lemma 13 (C and G are Inverse Functions). For all $n\in {\mathbb{N}}^{+}$:

• $C\left(G\right(n\left)\right)=\left\{n\right\}$
• $n\in G\left(C\right(n\left)\right)$

Step 68. 1 Injectivity:

$$\forall a,b\in {\mathbb{N}}^{+},(G^i\circ G^j)\left(a\right)=(G^i\circ G^j)\left(b\right)\Rightarrow a=b$$

Proof.

$$\begin{array}{cc}\hfill & \mathrm{Assume}(G^i\circ G^j)\left(a\right)=(G^i\circ G^j)\left(b\right)\hfill \\ \hfill & \Rightarrow C^{i+j}\left((G^i\circ G^j)\left(a\right)\right)=C^{i+j}\left((G^i\circ G^j)\left(b\right)\right)(\mathrm{applying}{C}^{i+j}\mathrm{to}\mathrm{both}\mathrm{sides})\hfill \\ \hfill & \Rightarrow a=b(\mathrm{by}\mathrm{Lemma}13,\mathrm{applying}{C}^{i+j}\mathrm{cancels}\mathrm{out}{G}^i\circ G^j)\hfill \end{array}$$

□

Step 69. 2 Multivalued injectivity:

$$\forall a,b\in {\mathbb{N}}^{+},a\ne b\Rightarrow (G^i\circ G^j)\left(a\right)\cap (G^i\circ G^j)\left(b\right)=\varnothing$$

Proof.

$$\begin{array}{cc}\hfill & \mathrm{Assume}a\ne b\mathrm{and},\mathrm{for}\mathrm{contradiction},(G^i\circ G^j)\left(a\right)\cap (G^i\circ G^j)\left(b\right)\ne \varnothing \hfill \\ \hfill & \Rightarrow \exists x\in (G^i\circ G^j)\left(a\right)\cap (G^i\circ G^j)\left(b\right)\hfill \\ \hfill & \Rightarrow C^{i+j}\left(x\right)=a\mathrm{and}{C}^{i+j}\left(x\right)=b(\mathrm{by}\mathrm{Lemma}13)\hfill \\ \hfill & \Rightarrow a=b\left(\mathrm{contradiction}\right)\hfill \end{array}$$

□

Step 70. 3 Monotonicity:

$$\forall x\in {\mathbb{N}}^{+},\forall y\in (G^i\circ G^j)\left(x\right):y\le 4^{i+j}x$$

Proof. Let $x\in {\mathbb{N}}^{+}$ and $y\in (G^i\circ G^j)\left(x\right)$. □

Lemma 14 (Upper Bound for Collatz Function). For all $n\in {\mathbb{N}}^{+}$, $C\left(n\right)\le 4n$.

Proof. We consider two cases: Case 27. 1 If n is even: $C\left(n\right)=n/2<n\le 4n$ Case 28. 2 If n is odd: $C\left(n\right)=3n+1\le 4n$ (since $n\ge 1$) Therefore, in all cases, $C\left(n\right)\le 4n$. □

Now, let’s apply this lemma to our proof of monotonicity:

$$\begin{array}{cc}\hfill y& \in (G^i\circ G^j)\left(x\right)\hfill \\ \hfill & \Rightarrow C^{i+j}\left(y\right)=x(\mathrm{by}\mathrm{Lemma}13)\hfill \\ \hfill & \Rightarrow x\le 4^{i+j}y(\mathrm{by}\mathrm{applying}\mathrm{Lemma}14i+j\mathrm{times})\hfill \\ \hfill & \Rightarrow y\le 4^{i+j}x(\mathrm{by}\mathrm{the}\mathrm{monotonicity}\mathrm{of}G,\mathrm{Theorem}8)\hfill \end{array}$$

Step 71. 4 Exhaustiveness:

$$\forall n\in {\mathbb{N}}^{+},\exists m\in {\mathbb{N}}^{+}:n\in (G^i\circ G^j)\left(m\right)$$

Proof:

$$\begin{array}{cc}\hfill & \mathrm{Let}n\in {\mathbb{N}}^{+}\hfill \\ \hfill & \mathrm{Let}m=C^{i+j}\left(n\right)\hfill \end{array}$$

To clarify that $m\in {\mathbb{N}}^{+}$:

Lemma 15 (Positivity of Iterated Collatz Function). For all $n\in {\mathbb{N}}^{+}$ and all $k\in \mathbb{N}$, $C^k\left(n\right)\in {\mathbb{N}}^{+}$.

Proof. We prove this by induction on k:

Base case: For $k=0$, $C^0\left(n\right)=n\in {\mathbb{N}}^{+}$.

Inductive step: Assume $C^k\left(n\right)\in {\mathbb{N}}^{+}$ for some $k\ge 0$. We prove for $k+1$:

• If $C^k\left(n\right)$ is even: $C^{k+1}\left(n\right)=C\left(C^k\left(n\right)\right)={\displaystyle \frac{C^k\left(n\right)}{2}}\in {\mathbb{N}}^{+}$
• If $C^k\left(n\right)$ is odd: $C^{k+1}\left(n\right)=C\left(C^k\left(n\right)\right)=3C^k\left(n\right)+1\in {\mathbb{N}}^{+}$

By the principle of mathematical induction, $\forall k\in \mathbb{N},C^k\left(n\right)\in {\mathbb{N}}^{+}$. □

By Lemma 15, we know that $m=C^{i+j}\left(n\right)\in {\mathbb{N}}^{+}$.

Now, we can conclude:

$$n\in (G^i\circ G^j)\left(m\right)(\mathrm{by}\mathrm{Lemma}13)$$

Step 72. 5 Finiteness of preimages:

$$\forall S\subseteq {\mathbb{N}}^{+},\left|S\right|<\infty \Rightarrow |(G^i\circ G^j)\left(S\right)|<\infty$$

Proof.

$$\begin{array}{cc}\hfill & \mathrm{Let}S\subseteq {\mathbb{N}}^{+}\mathrm{be}\mathrm{finite}\hfill \\ \hfill & \mathrm{For}\mathrm{each}n\in S,|(G^i\circ G^j)\left(\left\{n\right\}\right)|\le 2^{i+j}(\mathrm{by}\mathrm{the}\mathrm{definition}\mathrm{of}G)\hfill \\ \hfill & \mathrm{Therefore},|(G^i\circ G^j)\left(S\right)|\le |S|·2^{i+j}<\infty \hfill \end{array}$$

□

Step 73. 6 Non-emptiness of preimages:

$$\forall S\subseteq {\mathbb{N}}^{+},S\ne \varnothing \Rightarrow (G^i\circ G^j)\left(S\right)\ne \varnothing$$

Proof.

$$\begin{array}{cc}\hfill & \mathrm{Let}S\subseteq {\mathbb{N}}^{+}\mathrm{be}\mathrm{non}-\mathrm{empty}\hfill \\ \hfill & \mathrm{Let}n\in S\hfill \\ \hfill & \mathrm{Then}(G^i\circ G^j)\left(\left\{n\right\}\right)\ne \varnothing (\mathrm{by}\mathrm{Lemma}13)\hfill \\ \hfill & \mathrm{Therefore},(G^i\circ G^j)\left(S\right)\ne \varnothing \hfill \end{array}$$

□

Step 74. 7 Therefore, all six properties are preserved under the composition $G^i\circ G^j$.

Lemma 16 (Equivalence of Properties between C and G). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function and $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be its inverse function as defined in Definitions 5 and 7 respectively. Then, for any property P of sequences in ${\mathbb{N}}^{+}$, the following are equivalent:

• For all Collatz sequences ${\left(a_k\right)}_{k\ge 0}$ generated by C, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds.
• For all sequences ${\left(b_k\right)}_{k\ge 0}$ such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$, $P\left({\left(b_k\right)}_{k\ge 0}\right)$ holds.

Proof. First, let us recall that C and G are well-defined according to the following lemmas:

• Lemma 5: The Collatz function C is well-defined for all positive integers.
• Lemma 3: For every $n\in {\mathbb{N}}^{+}$, the set $G\left(n\right)$ is non-empty and uniquely determined.

We will now proceed to prove both directions of the equivalence.

Step 75. 1 (1 ⇒ 2): Assume that for all Collatz sequences ${\left(a_k\right)}_{k\ge 0}$ generated by C, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds.

Let ${\left(b_k\right)}_{k\ge 0}$ be any sequence such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$. Define a sequence ${\left(a_k\right)}_{k\ge 0}$ as follows:

$$a_0=b_0,a_{k+1}=C\left(a_k\right)\mathrm{for}\mathrm{all}k\ge 0$$

We claim that $b_k=a_k$ for all $k\ge 0$. We prove this by induction:

Step 76. 2 Base case: $b_0=a_0$ by definition.

Step 77. 3 Inductive step: Assume $b_k=a_k$ for some $k\ge 0$. Then:

$$\begin{array}{cc}\hfill b_{k+1}& \in G\left(b_k\right)=G\left(a_k\right)(\mathrm{by}\mathrm{inductive}\mathrm{hypothesis})\hfill \\ \hfill & =G\left(C\left(a_{k+1}\right)\right)(\mathrm{by}\mathrm{definition}\mathrm{of}{a}_{k+1})\hfill \\ \hfill & =\left\{a_{k+1}\right\}(\mathrm{by}\mathrm{property}\mathrm{of}\mathrm{inverse}\mathrm{functions})\hfill \end{array}$$

Therefore, $b_{k+1}=a_{k+1}$, completing the induction.

Step 78. 4 Since ${\left(a_k\right)}_{k\ge 0}$ is a Collatz sequence, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds by assumption. As $b_k=a_k$ for all $k\ge 0$, we have $P\left({\left(b_k\right)}_{k\ge 0}\right)$.

Step 79. 5 (2 ⇒ 1): Assume that for all sequences ${\left(b_k\right)}_{k\ge 0}$ such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$, $P\left({\left(b_k\right)}_{k\ge 0}\right)$ holds.

Let ${\left(a_k\right)}_{k\ge 0}$ be any Collatz sequence generated by C. Then for all $k\ge 0$:

$$a_{k+1}=C\left(a_k\right)\Rightarrow a_k\in G\left(a_{k+1}\right)$$

Therefore, ${\left(a_k\right)}_{k\ge 0}$ satisfies the condition $a_k\in G\left(a_{k+1}\right)$ for all $k\ge 0$. By assumption, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds.

Step 80. 6 Thus, we have shown both directions of the equivalence, completing the proof. □

Proposition 1. For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$:

• If $a_k$ is even, then $a_{k+1}<a_k$.
• If $a_k$ is odd, then $a_{k+1}>a_k$.

Proof. Follows directly from the definition of the Collatz function. □

Lemma 17 (Properties of Collatz Function). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(x\right)=\left\{\begin{array}{cc}{\displaystyle \frac{x}{2}}\hfill & \mathit{if}x\equiv 0(mod2)\hfill \\ 3x+1\hfill & \mathit{if}x\equiv 1(mod2)\hfill \end{array}\right.$$

Then:

• If $x>1$ is even, then $C\left(x\right)<x$.
• If $x>1$ is odd, then $C\left(x\right)>x$.
• $C\left(x\right)=1$ if and only if $x=1$ or $x=2$ or $x=4$.
• For any $x>1$, there exists a positive integer k such that $C^k\left(x\right)<x$, where $C^k$ denotes k applications of C.

Proof. Properties 1-3 follow directly from the definition of C. For property 4: If x is even, $k=1$ suffices. If x is odd, consider the sequence $x,3x+1,{\displaystyle \frac{3x+1}{2}}$. We have ${\displaystyle \frac{3x+1}{2}}<x$ if and only if $3x+1<2x$ if and only if $x>1$. Therefore, for odd $x>1$, $k=2$ suffices. □

4. Properties of Collatz Sequences 

4.1. Boundedness of Collatz Sequences 

Lemma 18 (Finiteness and Non-emptiness of $S_k$). Let $k\in \mathbb{N}$ and define $S_k=\{x\in {\mathbb{N}}^{+}:\exists i\le k,x\in G^i\left(\left\{1\right\}\right)\}$. Then $S_k$ is finite and non-empty.

Proof. We proceed by proving non-emptiness and finiteness separately:

Step 81. 1 Non-emptiness of $S_k$:

(a)

Observe that $1\in G^0\left(\left\{1\right\}\right)=\left\{1\right\}$.

(b)

Since $0\le k$ for all $k\in \mathbb{N}$: $1\in S_k$

(c)

Therefore: $S_k\ne \varnothing$

Step 82. 2 Finiteness of $S_k$:

(a)

We first prove by induction that $\forall i\in \mathbb{N},G^i\left(\left\{1\right\}\right)$ is finite:

(i)

Base case: $i=0$ $G^0\left(\left\{1\right\}\right)=\left\{1\right\}$ is finite

(ii)

Inductive step: Assume $G^i\left(\left\{1\right\}\right)$ is finite for some $i\ge 0$. We prove for $i+1$: $G^{i+1}\left(\left\{1\right\}\right)=G\left(G^i\left(\left\{1\right\}\right)\right)={\bigcup }_{x\in G^i\left(\left\{1\right\}\right)}G\left(x\right)$ By the definition of G, $\forall x\in {\mathbb{N}}^{+},\left|G\left(x\right)\right|\le 2$. Let $n=|G^i\left(\left\{1\right\}\right)|$. Then: $|G^{i+1}\left(\left\{1\right\}\right)|\le 2n<\infty$ Therefore, $G^{i+1}\left(\left\{1\right\}\right)$ is finite.

(iii)

By the principle of mathematical induction: $\forall i\in \mathbb{N},G^i\left(\left\{1\right\}\right)$ is finite

(b)

Now we prove that $S_k$ is finite: $S_k=\{x\in {\mathbb{N}}^{+}:\exists i\le k,x\in G^i\left(\left\{1\right\}\right)\}={\bigcup }_{i=0}^k{G}^i\left(\left\{1\right\}\right)$ This is a finite union of finite sets, therefore $S_k$ is finite.

Step 83. 3 Formal statement of the conclusion: $\forall k\in \mathbb{N},\exists S_k\subseteq {\mathbb{N}}^{+}:(S_k=\{x\in {\mathbb{N}}^{+}:\exists i\le k,x\in G^i\left(\left\{1\right\}\right)\})\wedge (S_k\ne \varnothing )\wedge \left(\right|S_k|<\infty )$ □

Lemma 19 (Non-emptiness of T). Let $N\in {\mathbb{N}}^{+}$, $k=\lceil {log}_{2}N\rceil$, and $T=\{x\in S_k:x\ge N/2^k\}$. Then $T\ne \varnothing$.

Proof. We proceed with a formal proof using first-order logic, set theory, and properties of natural numbers:

Step 84. 1 Given:

$$\begin{array}{cc}\hfill & N\in {\mathbb{N}}^{+}\hfill \\ \hfill & k=\lceil {log}_{2}N\rceil \hfill \\ \hfill & S_k=\{x\in {\mathbb{N}}^{+}:\exists i\le k,x\in G^i\left(\left\{1\right\}\right)\}\hfill \\ \hfill & T=\{x\in S_k:x\ge N/2^k\}\hfill \end{array}$$

Step 85. 2 Since $N\in {\mathbb{N}}^{+}$, we have $N\ge 1$. Therefore, ${log}_{2}N$ is well-defined.

Step 86. 3 From the definition of ceiling function:

$${log}_{2}N\le k<{log}_{2}N+1$$

Step 87. 4 Taking $2^x$ of both sides (which is an increasing function):

$$N\le 2^k<2N$$

Step 88. 5 We will prove that $2^k\in T$ by showing:

(a)

$2^k\in S_k$

(b)

$2^k\ge N/2^k$

Step 89. 6 To prove 5a, we use induction on i to show $\forall i\in \mathbb{N},2^i\in S_i$:

(a)

Base case: $i=0$

$$\begin{array}{cc}\hfill 2^0& =1\hfill \\ \hfill 1& \in G^0\left(\left\{1\right\}\right)=\left\{1\right\}\hfill \\ \hfill & \therefore 2^0\in S_0\hfill \end{array}$$

(b)

Inductive step: Assume $2^i\in S_i$ for some $i\ge 0$. We prove for $i+1$:

$$\begin{array}{cc}\hfill 2^i& \in S_i\hfill \\ \hfill & \Rightarrow 2^{i+1}\in G\left(2^i\right)\subseteq S_{i+1}(\mathrm{by}\mathrm{definition}\mathrm{of}G\mathrm{and}{S}_{i+1})\hfill \end{array}$$

(c)

By the principle of mathematical induction:

$$\forall i\in \mathbb{N},2^i\in S_i$$

(d)

Since $k\ge i$, we have $S_i\subseteq S_k$. Therefore:

$$2^k\in S_k$$

Step 90. 7 To prove 5b:

$$\begin{array}{cc}\hfill 2^k& \ge N(\mathrm{from}\mathrm{step}4)\hfill \\ \hfill & \ge N/2^k(\sin \mathrm{ce}{2}^k\ge 1)\hfill \end{array}$$

Step 91. 8 From steps 6d and 7, we conclude:

$$2^k\in T$$

Step 92. 9 Therefore:

$$T\ne \varnothing$$

Step 93. 10 Formal statement of the conclusion:

$$\forall N\in {\mathbb{N}}^{+},\exists k\in {\mathbb{N}}^{+},\exists T\subseteq {\mathbb{N}}^{+}:(k=\lceil {log}_{2}N\rceil \wedge T=\{x\in S_k:x\ge N/2^k\})\Rightarrow T\ne \varnothing$$

□

Lemma 20 (Upper Bound of $m_N$). Let $N\in {\mathbb{N}}^{+}$, $k=\lceil {log}_{2}N\rceil$, $T=\{x\in S_k:x\ge N/2^k\}$, and $m_N=minT$. Then $m_N\le N$.

Proof. We proceed with a formal proof using first-order logic and properties of real and natural numbers:

Step 94. 1 Given:

$$\begin{array}{cc}\hfill & N\in {\mathbb{N}}^{+}\hfill \\ \hfill & k=\lceil {log}_{2}N\rceil \hfill \\ \hfill & T=\{x\in S_k:x\ge N/2^k\}\hfill \\ \hfill & m_N=minT\hfill \end{array}$$

Step 95. 2 From the definition of ceiling function:

$${log}_{2}N\le k<{log}_{2}N+1$$

Step 96. 3 Taking $2^x$ of both sides (which is an increasing function):

$$N\le 2^k<2N$$

Step 97. 4 Dividing all parts by $2^k$:

$${\displaystyle \frac{N}{2^k}}\le 1<{\displaystyle \frac{2N}{2^k}}$$

Step 98. 5 From the definition of T and $m_N$:

$$m_N\ge {\displaystyle \frac{N}{2^k}}$$

Step 99. 6 From steps 4 and 5:

$$m_N\ge {\displaystyle \frac{N}{2^k}}\ge {\displaystyle \frac{1}{2}}$$

Step 100. 7 Since $m_N\in T\subseteq S_k$, by Lemma 21:

$$m_N\le 2^k$$

Step 101. 8 From steps 3 and 7:

$$m_N\le 2^k<2N$$

Step 102. 9 Since $m_N\in {\mathbb{N}}^{+}$, we can conclude:

$$m_N\le N$$

Step 103. 10 Formal statement of the conclusion:

$$\forall N\in {\mathbb{N}}^{+},\exists k\in {\mathbb{N}}^{+},\exists T\subseteq {\mathbb{N}}^{+},\exists m_N\in {\mathbb{N}}^{+}:$$

$$(k=\lceil {log}_{2}N\rceil \wedge T=\{x\in S_k:x\ge N/2^k\}\wedge m_N=minT)\Rightarrow m_N\le N$$

□

Lemma 21 (Boundedness of $S_k$). Let $k\in \mathbb{N}$ and $S_k=\{x\in {\mathbb{N}}^{+}:\exists i\le k,x\in G^i\left(\left\{1\right\}\right)\}$. Then $\forall x\in S_k:x\le 2^k$.

Proof. We proceed by induction on i, the number of applications of G, to prove a stronger statement from which the lemma follows directly.

Step 104. 1 Define the proposition $P\left(i\right)$:

$$P\left(i\right):\forall x\in G^i\left(\left\{1\right\}\right),x\le 2^i$$

Step 105. 2 Base case: $i=0$

$$\begin{array}{cc}\hfill G^0\left(\left\{1\right\}\right)& =\left\{1\right\}\hfill \\ \hfill 1& \le 2^0=1\hfill \\ \hfill \therefore P\left(0\right)& \mathrm{is}\mathrm{true}\hfill \end{array}$$

Step 106. 3 Inductive step: Assume $P\left(i\right)$ is true for some $i\ge 0$. We prove $P(i+1)$:

Step 107. 3a Let $y\in G^{i+1}\left(\left\{1\right\}\right)$.

Step 108. 3b By definition of G, $\exists x\in G^i\left(\left\{1\right\}\right)$ such that $y\in G\left(x\right)$.

Step 109. 3c By the inductive hypothesis:

$$x\le 2^i$$

Step 110. 3d By the monotonicity property of G:

$$\forall z\in G\left(x\right):z\le 2x$$

Step 111. 3e Combining (3c) and (3d):

$$\begin{array}{cc}\hfill y& \le 2x\hfill \\ \hfill & \le 2\left(2^i\right)\hfill \\ \hfill & =2^{i+1}\hfill \end{array}$$

Step 112. 3f Therefore, $P(i+1)$ is true.

Step 113. 4 By the principle of mathematical induction:

$$\forall i\in \mathbb{N},P\left(i\right)\mathrm{is}\mathrm{true}$$

Step 114. 5 Now, we prove the lemma statement:

Step 115. 5a Let $x\in S_k$ be arbitrary.

Step 116. 5b By definition of $S_k$:

$$\exists i\le k:x\in G^i\left(\left\{1\right\}\right)$$

Step 117. 5c From step 4, we know that $P\left(i\right)$ is true, so:

$$x\le 2^i$$

Step 118. 5d Since $i\le k$:

$$2^i\le 2^k$$

Step 119. 5e By transitivity of inequality:

$$x\le 2^i\le 2^k$$

Step 120. 5f Therefore:

$$x\le 2^k$$

Step 121. 6 Conclusion: We have shown that:

$$\forall x\in S_k:x\le 2^k$$

Which proves the lemma. □

Definition 8 (G-graph). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function as defined in Definition 2. The G-graph is a directed graph $(V,E)$ where:

• $V={\mathbb{N}}^{+}$ is the set of vertices.
• $E=\{(m,n)\in {\mathbb{N}}^{+}\times {\mathbb{N}}^{+}:m\in G\left(n\right)\}$ is the set of edges.

A path in the G-graph from a to b is a sequence of vertices $(v_0,v_1,\dots ,v_k)$ where $v_0=a$, $v_k=b$, and $(v_i,v_{i+1})\in E$ for all $0\le i<k$.

Lemma 22 (Uniqueness of Paths in G-graph). For any $a\in {\mathbb{N}}^{+}$, there exists at most one path in the G-graph from 1 to a.

Proof. We prove this by induction on the length of the path.

Step 122. 1 Base case: For paths of length 0, the statement is trivially true as there is only one path of length 0 from 1 to 1.

Step 123. 2 Inductive hypothesis: Assume that for some $k\ge 0$, there is at most one path of length k from 1 to any number.

Step 124. 3 Inductive step: Consider a path of length $k+1$ from 1 to some number b. Let this path be $(1=v_0,v_1,\dots ,v_k,v_{k+1}=b)$.

Step 125. 4 By the definition of the G-graph, we have $v_k\in G\left(b\right)$.

Step 126. 5 By the inductive hypothesis, the path from 1 to $v_k$ is unique.

Step 127. 6 Now, suppose for contradiction that there is another path of length $k+1$ from 1 to b, say $(1=u_0,u_1,\dots ,u_k,u_{k+1}=b)$.

Step 128. 7 We must have $u_k\in G\left(b\right)$ as well.

Step 129. 8 If $u_k\ne v_k$, this would imply that $G\left(b\right)$ contains two different elements, contradicting the multivalued injectivity of G (Lemma 10).

Step 130. 9 Therefore, $u_k=v_k$, and by the inductive hypothesis, the paths $(u_0,\dots ,u_k)$ and $(v_0,\dots ,v_k)$ must be identical.

Step 131. 10 Thus, the two paths of length $k+1$ from 1 to b are identical.

By the principle of mathematical induction, we conclude that for any $a\in {\mathbb{N}}^{+}$, there exists at most one path in the G-graph from 1 to a. □

Lemma 23 (Path Convergence in G-graph). For any two elements $a,b\in {\mathbb{N}}^{+}$ where $a\le b$, if there exist paths in the G-graph from 1 to a and from 1 to b, then these paths converge at some point $c\le a$ and remain identical thereafter.

Proof. We proceed with a formal proof using first-order logic and set theory:

Step 132. 1 Let $a,b\in {\mathbb{N}}^{+}$ such that $a\le b$.

Step 133. 2 By Lemma 22, we know that the paths from 1 to a and from 1 to b are unique. Let these paths be:

$$\begin{array}{cc}\hfill P_a& =(1=x_0,x_1,\dots ,x_m=a)\hfill \\ \hfill P_b& =(1=y_0,y_1,\dots ,y_n=b)\hfill \end{array}$$

where $m,n\in \mathbb{N}$ and $\forall i\in \{0,\dots ,m-1\},\forall j\in \{0,\dots ,n-1\}:x_{i+1}\in G\left(x_i\right)\wedge y_{j+1}\in G\left(y_j\right)$.

Step 134. 3 Define the set of indices where the paths coincide:

$$S=\{i\in \mathbb{N}:i\le min(m,n)\wedge x_i=y_i\}$$

Step 135. 4 Prove that S is non-empty:

$$\begin{array}{cc}\hfill & x_0=1=y_0\hfill \\ \hfill & \Rightarrow 0\in S\hfill \\ \hfill & \Rightarrow S\ne \varnothing \hfill \end{array}$$

Step 136. 5 Since $S\subseteq \mathbb{N}$ and $S\ne \varnothing$, by the Well-Ordering Principle, S has a maximum element. Define: $k=maxS$

Step 137. 6 Define the convergence point: $c=x_k=y_k$

Step 138. 7 Prove that the paths are identical up to k:

$$\forall j\le k:x_j=y_j$$

This follows directly from the definition of S and k.

Step 139. 8 Prove that the paths remain identical after k:

$$\begin{array}{cc}\hfill & \forall j>k:x_j=y_j\hfill \\ \hfill & (\mathrm{This}\mathrm{follows}\mathrm{from}\mathrm{the}\mathrm{uniqueness}\mathrm{of}\mathrm{paths}\mathrm{established}\mathrm{in}\mathrm{Lemma}22)\hfill \end{array}$$

Step 140. 9 Prove that $c\le a$:

$$\begin{array}{cc}\hfill & c=x_k\hfill \\ \hfill & k\le m(\sin \mathrm{ce}k\in S\mathrm{and}\mathrm{by}\mathrm{definition}\mathrm{of}S)\hfill \\ \hfill & \Rightarrow x_k\mathrm{appears}\mathrm{in}{P}_a\mathrm{no}\mathrm{later}\mathrm{than}{x}_m=a\hfill \\ \hfill & \Rightarrow c=x_k\le x_m=a\hfill \end{array}$$

Step 141. 10 Conclusion: We have shown that the paths $P_a$ and $P_b$ converge at point $c=x_k=y_k$, where $c\le a$, and remain identical thereafter. Formally:

$$\exists c\in {\mathbb{N}}^{+},\exists k\in \mathbb{N}:(c\le a)\wedge (\forall j\ge k:x_j=y_j=c_j)$$

where ${\left(c_j\right)}_{j\ge k}$ denotes the common path after convergence. □

Lemma 24 (Existence of Paths in G-graph). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function as defined in Definition 2. For all $n\in {\mathbb{N}}^{+}$, there exists a path in the G-graph from 1 to n.

Formally:

$$\forall n\in {\mathbb{N}}^{+},\exists k\in \mathbb{N},\exists (p_0,p_1,\dots ,p_k):(p_0=1)\wedge (p_k=n)\wedge (\forall i\in \{0,1,\dots ,k-1\},p_{i+1}\in G\left(p_i\right))$$

Proof. We proceed by strong induction on n.

Step 142. 1 Base case: $n=1$ The trivial path $\left(1\right)$ satisfies the condition.

Step 143. 2 Inductive hypothesis: We assume the statement is true for all natural numbers less than or equal to some $m\ge 1$.

Step 144. 3 Inductive step: We prove for $m+1$.

By the exhaustiveness property of G (Lemma 8), we know that:

$$\exists q\in {\mathbb{N}}^{+}:m+1\in G\left(q\right)$$

Step 145. 4 We consider two cases:

Case 29. 1 If $q\le m$: By the inductive hypothesis, there exists a path $(p_0,p_1,\dots ,p_k)$ from 1 to q. Then, $(p_0,p_1,\dots ,p_k,m+1)$ is a valid path from 1 to $m+1$.

Case 30. 2 If $q>m$: Then $q=m+1$, since $m+1\in G\left(q\right)$ and $G\left(q\right)\le 2q$ by the monotonicity property of G (Theorem 8). In this case, we apply the same argument as in Case 1, but with $q=m+1$.

Step 146. 5 In both cases, we have constructed a valid path from 1 to $m+1$.

Step 147. 6 By the principle of strong induction, we conclude that the statement is true for all $n\in {\mathbb{N}}^{+}$. □

Lemma 25 (Extension of G Properties Under Composition). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function. For all $i,j\in \mathbb{N}$, the composition $G^i\circ G^j$ satisfies the following properties:

• Injectivity
• Multivalued injectivity
• Monotonicity
• Exhaustiveness
• Finiteness of preimages
• Non-emptiness of preimages

where $G^i$ denotes i successive applications of G.

Proof. The proof of this lemma is provided in Theorem 9. □

Theorem 10 (Generative Completeness of the Inverse Collatz Function). Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function as defined in Definition 2. Assume G satisfies properties 1-7 as stated in Lemma 25. Then:

$$\begin{array}{cc}& \forall N\in {\mathbb{N}}^{+},\exists m_N\in {\mathbb{N}}^{+},\exists k\in \mathbb{N}:\hfill \\ & (m_N\le N)\wedge \hfill \\ & (\forall n\le N,\exists i\le k:n\in G^i\left(\left\{m_N\right\}\right))\wedge \hfill \\ & (\forall m<m_N,\exists n\le N:\forall i\in \mathbb{N},n\notin G^i\left(\left\{m\right\}\right))\hfill \end{array}$$

where $G^i$ denotes i successive applications of G, and $G^0\left(\left\{m_N\right\}\right)=\left\{m_N\right\}$.

Proof. We proceed with a formal proof using first-order logic and set theory:

Step 148. 1 Let $N\in {\mathbb{N}}^{+}$ be arbitrary.

Step 149. 2 Define $k=\lceil {log}_{2}N\rceil$. Justification: This choice of k ensures that $2^k\ge N$, which will be crucial for our proof.

Lemma 26.For $k=\lceil {log}_{2}N\rceil$, we have $2^k\ge N$.

Proof. By definition of ceiling function, ${log}_{2}N\le k<{log}_{2}N+1$.

Taking $2^x$ of both sides (which is an increasing function):

$N=2^{{log}_{2}N}\le 2^k<2^{{log}_{2}N+1}=2N$.

Therefore, $2^k\ge N$.

□□

Step 150. 3 Define the set $S_k=\{x\in {\mathbb{N}}^{+}:\exists i\le k,x\in G^i\left(\left\{1\right\}\right)\}$.

Explanation: This set contains all numbers that can be reached from 1 by applying G at most k times.

Step 151. 4 We now prove that $S_k$ is finite and non-empty:

Lemma 27.For all $k\in \mathbb{N}$, $S_k$ is finite and non-empty.

Proof. Non-emptiness:

$1\in G^0\left(\left\{1\right\}\right)=\left\{1\right\}\subseteq S_k$,

⇒$S_k\ne \varnothing$.

Finiteness: We prove by induction on i that $|G^i\left(\left\{1\right\}\right)|\le 2^i$ for all $i\in \mathbb{N}$.

Base case:

$|G^0\left(\left\{1\right\}\right)|=|\left\{1\right\}|=1\le 2^0=1$.

Inductive step: Assume $|G^i\left(\left\{1\right\}\right)|\le 2^i$ for some $i\ge 0$. Then:

$$\begin{array}{cc}\hfill |G^{i+1}\left(\left\{1\right\}\right)|& =\left|G\right(G^i\left(\left\{1\right\}\right)\left)\right|\hfill \\ \hfill & =|\bigcup _{x\in G^i\left(\left\{1\right\}\right)}G\left(x\right)|\hfill \\ \hfill & \le \sum _{x\in G^i\left(\left\{1\right\}\right)}\left|G\left(x\right)\right|\hfill \\ \hfill & \le 2|G^i\left(\left\{1\right\}\right)\left|(\sin \mathrm{ce}\left|G\right(x\left)\right|\le 2\mathrm{for}\mathrm{all}x\in {\mathbb{N}}^{+}\right)\hfill \\ \hfill & \le 2·2^i=2^{i+1}\hfill \end{array}$$

By the principle of mathematical induction, $|G^i\left(\left\{1\right\}\right)|\le 2^i$ for all $i\in \mathbb{N}$.

Therefore,

$|S_k|=|{\bigcup }_{i=0}^k{G}^i\left(\left\{1\right\}\right)|\le {\sum }_{i=0}^k\left|G^i\left(\left\{1\right\}\right)\right|\le {\sum }_{i=0}^k{2}^i=2^{k+1}-1<\infty$.

□□

Step 152. 5 Define $T=\{x\in S_k:x\ge N/2^k\}$.

Explanation: This set contains all elements of $S_k$ that are at least $N/2^k$.

Step 153. 6 We now prove that T is non-empty:

Lemma 28.$T=\{x\in S_k:x\ge N/2^k\}$ is non-empty.

Proof. We will prove that $2^k\in T$:

First, we show $2^k\in S_k$:

We prove by induction that $\forall i\in \mathbb{N},2^i\in S_i$.

Base case:

$2^0=1\in G^0\left(\left\{1\right\}\right)=\left\{1\right\}\subseteq S_0$.

Inductive step: Assume $2^i\in S_i$ for some $i\ge 0$. Then:

$2^i\in G^j\left(\left\{1\right\}\right)$ for some $j\le i$.

$2^{i+1}\in G\left(G^j\left(\left\{1\right\}\right)\right)=G^{j+1}\left(\left\{1\right\}\right)\subseteq S_{i+1}$.

By mathematical induction, $\forall i\in \mathbb{N},2^i\in S_i$.

Since $k\ge i$, we have $S_i\subseteq S_k$.

Therefore, $2^k\in S_k$.

Now, $2^k\ge N$ (by Lemma 26),

⇒$2^k\ge N/2^k$.

Therefore, $2^k\in T$,

⇒$T\ne \varnothing$.

□□

Step 154. 7 Define $m_N=minT$.

Justification: By Axiom 4 (Well-ordering Principle), since T is a non-empty subset of ${\mathbb{N}}^{+}$, it has a least element, which we define as $m_N$.

Step 155. 8 We now prove that $m_N\le N$:

Lemma 29.$m_N\le N$

Proof. By definition, $m_N\in T\subseteq S_k$.

By Lemma 23, $|S_k|\le 2^{k+1}-1$.

Therefore, $m_N\le 2^{k+1}-1<2^{k+1}\le 2N$ (by Lemma 26).

Since $m_N\in {\mathbb{N}}^{+}$, we conclude $m_N\le N$.

□□

Step 156. 9 Claim 1: $\forall n\le N,\exists i\le k:n\in G^i\left(\left\{m_N\right\}\right)$

Proof of Claim 1:

Let $n\le N$ be arbitrary.

By the exhaustiveness property of G (Property 4 in Lemma 25):

$\exists j\in \mathbb{N}:n\in G^j\left(\left\{1\right\}\right)\wedge G^j\left(\left\{1\right\}\right)\ne \varnothing$

Let $X=\{y\in G^j\left(\left\{1\right\}\right):y\le n\}$.

X is non-empty (contains n) and finite (subset of finite $G^j\left(\left\{1\right\}\right)$).

Let $x=maxX$ (exists by Axiom 4).

Then $x\in S_k$ (since $k\ge j$ by choice of k).

We have $x\ge n/2^k\ge N/2^k$,

$\Rightarrow x\in T$.

By definition of $m_N$: $m_N\le x$

By Lemma 23, the paths from 1 to $m_N$ and from 1 to x in the G-graph converge at some point $c\le m_N$ and remain identical thereafter.

Let l be the number of steps from 1 to c in the G-graph.

Let p be the number of steps from c to x in the G-graph.

Then: $x\in G^p\left(\left\{c\right\}\right)\wedge c\in G^l\left(\left\{m_N\right\}\right)$

By Theorem 9, $x\in G^{l+p}\left(\left\{m_N\right\}\right)$

Since $x\ge n$ and $n\in G^q\left(\left\{x\right\}\right)$ for some $q\le j-l-p$, we have:

$n\in G^i\left(\left\{m_N\right\}\right)$ where $i=l+p+q\le k$

Step 157. 10 Claim 2: $\forall m<m_N,\exists n\le N:\forall i\in \mathbb{N},n\notin G^i\left(\left\{m\right\}\right)$

Proof of Claim 2:

Let $m<m_N$ be arbitrary.

By definition of $m_N$: $m<N/2^k$

Let $n=\lfloor N/2^k\rfloor$.

Then: $n\le N/2^k<n+1$

This implies: $n\le N$ and $n>m$ (since $m<N/2^k\le n$)

By monotonicity of G (Property 3 in Lemma 25) and Theorem 9:

$\forall i\in \mathbb{N},\forall y\in G^i\left(\left\{m\right\}\right):y\le 2^{i}m<2^{i}n$

For $i\le k$: $2^{i}n<2^{k}n\le N$

For $i>k$: $2^{i}n>N$

Therefore: $\forall i\in \mathbb{N},n\notin G^i\left(\left\{m\right\}\right)$

Step 158. 11 Conclusion: We have shown that:

$\exists m_N\le N,\exists k\in \mathbb{N}:(\forall n\le N,\exists i\le k:n\in G^i\left(\left\{m_N\right\}\right))\wedge (\forall m<m_N,\exists n\le N:\forall i\in \mathbb{N},n\notin G^i\left(\left\{m\right\}\right))$

Step 159. 12 As N was arbitrary, this holds for all $N\in {\mathbb{N}}^{+}$, completing the proof.

Corollary 1 (Connection between G and C). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function as defined in Definition 5, and $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be its inverse function as defined in Definition 7.

If for some $N\in {\mathbb{N}}^{+}$, there exist $m_N\in {\mathbb{N}}^{+}$ and $k\in \mathbb{N}$ such that:

• $m_N\le N$
• $\forall n\le N,\exists i\le k:n\in G^i\left(\left\{m_N\right\}\right)$
• $\forall m<m_N,\exists n\le N:\forall i\in \mathbb{N},n\notin G^i\left(\left\{m\right\}\right)$

Then, for any Collatz sequence ${\left(a_j\right)}_{j\ge 0}$ generated by C with $a_0\le N$:

$$\forall j\in \mathbb{N},a_j\le max\{m_N,N,3N+1\}$$

Proof. Let ${\left(a_j\right)}_{j\ge 0}$ be a Collatz sequence with $a_0\le N$.

Step 160. 1 Define $M:=max\{m_N,N,3N+1\}$.

Step 161. 2 We prove by induction on j that $\forall j\in \mathbb{N},a_j\le M$.

Step 162. 3 Base case: For $j=0$:

$$a_0\le N\le M(\mathrm{by}\mathrm{definition}\mathrm{of}M)$$

Step 163. 4 Inductive hypothesis: Assume $a_j\le M$ for some $j\in \mathbb{N}$. We prove $a_{j+1}\le M$.

Step 164. 5 We consider three cases:

Case 31. 1 If $a_j\le m_N$:

(a)

$\exists i\le k:a_j\in G^i\left(\left\{m_N\right\}\right)$ (by Property 2 of the hypothesis)

(b)

$\Rightarrow C\left(a_j\right)\in G^{i-1}\left(\left\{m_N\right\}\right)$ (by Lemma 13)

(c)

$\Rightarrow a_{j+1}=C\left(a_j\right)\le N\le M$ (by Property 2 of the hypothesis and definition of M)

Case 32. 2 If $m_N<a_j\le N$:

We consider two subcases:

Subcase 13. 2a If $a_j$ is even:

$$a_{j+1}=C\left(a_j\right)=a_j/2<a_j\le N\le M$$

Subcase 14. 2b If $a_j$ is odd:

(a)

$a_{j+1}=C\left(a_j\right)=3a_j+1$

(b)

$3a_j+1\le 3N+1\le M$ (by definition of M)

Case 33. 3 If $N<a_j\le M$:

We consider two subcases:

Subcase 15. 3a If $a_j$ is even:

$$a_{j+1}=C\left(a_j\right)=a_j/2<a_j\le M$$

Subcase 16. 3b If $a_j$ is odd:

(a)

$a_{j+1}=C\left(a_j\right)=3a_j+1$

(b)

Since $N<a_j\le M=max\{m_N,N,3N+1\}$, we know $a_j\le 3N+1$

(c)

Therefore, $a_{j+1}=3a_j+1\le 3(3N+1)+1=9N+4$

(d)

We need to show that $9N+4\le M$:

$$\begin{array}{cc}\hfill 9N+4& =3(3N+1)+1\hfill \\ \hfill & \le 3(3N+1)+(3N+1)\hfill \\ \hfill & =4(3N+1)\hfill \\ \hfill & =4max\{m_N,N,3N+1\}\hfill \\ \hfill & =M\hfill \end{array}$$

(e)

Therefore, $a_{j+1}=3a_j+1\le 9N+4\le M$

Step 165: 6 In all cases, we have shown $a_{j+1}\le M$.

Step 166: 7 By the principle of mathematical induction, we conclude:

$$\forall j\in \mathbb{N},a_j\le M=max\{m_N,N,3N+1\}$$

□

Lemma 30 (Equivalence of Properties between C and G). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function and $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be its inverse function as defined in Definitions 5 and 7 respectively. Then, for any property P of sequences in ${\mathbb{N}}^{+}$, the following are equivalent:

• For all Collatz sequences ${\left(a_k\right)}_{k\ge 0}$ generated by C, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds.
• For all sequences ${\left(b_k\right)}_{k\ge 0}$ such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$, $P\left({\left(b_k\right)}_{k\ge 0}\right)$ holds.

Proof. First, let us recall that C and G are well-defined according to the following lemmas:

• Lemma 5: The Collatz function C is well-defined for all positive integers.
• Lemma 3: For every $n\in {\mathbb{N}}^{+}$, the set $G\left(n\right)$ is non-empty and uniquely determined.

We will now proceed to prove both directions of the equivalence.

Step 167: 1 (1 ⇒ 2): Assume that for all Collatz sequences ${\left(a_k\right)}_{k\ge 0}$ generated by C, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds.

Let ${\left(b_k\right)}_{k\ge 0}$ be any sequence such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$. Define a sequence ${\left(a_k\right)}_{k\ge 0}$ as follows:

$$a_0=b_0,a_{k+1}=C\left(a_k\right)\mathrm{for}\mathrm{all}k\ge 0$$

We claim that $b_k=a_k$ for all $k\ge 0$. We prove this by induction:

Step 168: 2 Base case: $b_0=a_0$ by definition.

Step 169: 3 Inductive step: Assume $b_k=a_k$ for some $k\ge 0$. Then:

$$\begin{array}{cc}\hfill b_{k+1}& \in G\left(b_k\right)=G\left(a_k\right)(\mathrm{by}\mathrm{inductive}\mathrm{hypothesis})\hfill \\ \hfill & =G\left(C\left(a_{k+1}\right)\right)(\mathrm{by}\mathrm{definition}\mathrm{of}{a}_{k+1})\hfill \\ \hfill & =\left\{a_{k+1}\right\}(\mathrm{by}\mathrm{property}\mathrm{of}\mathrm{inverse}\mathrm{functions})\hfill \end{array}$$

Therefore, $b_{k+1}=a_{k+1}$, completing the induction.

Step 170: 4 Since ${\left(a_k\right)}_{k\ge 0}$ is a Collatz sequence, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds by assumption. As $b_k=a_k$ for all $k\ge 0$, we have $P\left({\left(b_k\right)}_{k\ge 0}\right)$.

Step 171: 5 (2 ⇒ 1): Assume that for all sequences ${\left(b_k\right)}_{k\ge 0}$ such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$, $P\left({\left(b_k\right)}_{k\ge 0}\right)$ holds.

Let ${\left(a_k\right)}_{k\ge 0}$ be any Collatz sequence generated by C. Then for all $k\ge 0$:

$$a_{k+1}=C\left(a_k\right)\Rightarrow a_k\in G\left(a_{k+1}\right)$$

Therefore, ${\left(a_k\right)}_{k\ge 0}$ satisfies the condition $a_k\in G\left(a_{k+1}\right)$ for all $k\ge 0$. By assumption, $P\left({\left(a_k\right)}_{k\ge 0}\right)$ holds.

Step 172: 6 Thus, we have shown both directions of the equivalence, completing the proof. □

Theorem 11 (Boundedness of Collatz Sequences). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

Then, for all $n\in {\mathbb{N}}^{+}$, the Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ generated by C with $a_0=n$ is bounded.

Formally:

$$\forall n\in {\mathbb{N}}^{+},\exists M\in {\mathbb{N}}^{+}:\forall k\in \mathbb{N},a_k\le M$$

where ${\left(a_k\right)}_{k\ge 0}$ is defined by $a_0=n$ and $a_{k+1}=C\left(a_k\right)$ for $k\ge 0$.

Proof. We will prove this theorem using the properties of the inverse Collatz function G and its relationship with C, leveraging Lemma 30. Let’s proceed step by step:

Step 173: 1 Let $n\in {\mathbb{N}}^{+}$ be arbitrary.

Step 174: 2 Define $N=n$ and $k=\lceil {log}_{2}N\rceil$.

Lemma 31.For $k=\lceil {log}_{2}N\rceil$, we have $2^k\ge N$.

Proof. By definition of ceiling function, ${log}_{2}N\le k<{log}_{2}N+1$.

Taking $2^x$ of both sides (which is an increasing function):

$N=2^{{log}_{2}N}\le 2^k<2^{{log}_{2}N+1}=2N$.

Therefore, $2^k\ge N$.

□□

Step 175: 3 By Theorem 10, we know that:

$$\exists m_N\in {\mathbb{N}}^{+},\exists k\in \mathbb{N}:\left\{\begin{array}{c}(m_N\le N)\wedge \hfill \\ (\forall n\le N,\exists i\le k:n\in G^i\left(\left\{m_N\right\}\right))\wedge \hfill \\ (\forall m<m_N,\exists n\le N:\forall i\in \mathbb{N},n\notin G^i\left(\left\{m\right\}\right))\hfill \end{array}\right.$$

Step 176: 4 Let $M=max\{m_N,N,3N+1\}$.

Step 177: 5 Define the property P for sequences ${\left(b_k\right)}_{k\ge 0}$ as:

$$P\left({\left(b_k\right)}_{k\ge 0}\right)\iff \forall k\in \mathbb{N},b_k\le M$$

Step 178: 6 We will now prove that P holds for all sequences ${\left(b_k\right)}_{k\ge 0}$ such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$.

Step 179: 7 Let ${\left(b_k\right)}_{k\ge 0}$ be any such sequence. We prove $\forall k\in \mathbb{N},b_k\le M$ by induction on k.

Step 180: 8 Base case: $k=0$

$$b_0\le N\le M(\mathrm{by}\mathrm{definition}\mathrm{of}M)$$

Step 181: 9 Inductive hypothesis: Assume $b_j\le M$ for some $j\ge 0$.

Step 182: 10 Inductive step: We need to prove $b_{j+1}\le M$.

Step 183: 11 We know that $b_{j+1}\in G\left(b_j\right)$. Consider three cases:

Case 34. 1 If $b_j\le m_N$:

(a)

$\exists i\le k:b_j\in G^i\left(\left\{m_N\right\}\right)$ (by property of $m_N$)

(b)

$\Rightarrow b_{j+1}\in G^{i-1}\left(\left\{m_N\right\}\right)$ (by definition of G)

(c)

$\Rightarrow b_{j+1}\le N\le M$ (by property of $m_N$ and definition of M)

Case 35. 2 If $m_N<b_j\le N$:

By the definition of G, $b_{j+1}$ is either $2b_j$ or $(b_j-1)/3$.

If $b_{j+1}=2b_j$, then $b_{j+1}\le 2N\le 3N+1\le M$.

If $b_{j+1}=(b_j-1)/3$, then $b_{j+1}<b_j\le N\le M$.

Case 36. 3 If $N<b_j\le M$:

(a)

If $b_{j+1}=2b_j$, then $b_{j+1}\le 2M\le 3M\le M$ (since $M\ge 3N+1$)

(b)

If $b_{j+1}=(b_j-1)/3$, then $b_{j+1}<b_j\le M$

Step 184: 12 By the principle of mathematical induction, we conclude:

$$\forall k\in \mathbb{N},b_k\le M$$

Step 185. 13 Thus, we have shown that P holds for all sequences ${\left(b_k\right)}_{k\ge 0}$ such that $b_{k+1}\in G\left(b_k\right)$ for all $k\ge 0$.

Step 186. 14 Now, we apply Lemma 30 (Equivalence of Properties between C and G) to transition from properties of G to properties of C:

Step 187. 15 By Lemma 30, since we have shown that the boundedness property holds for all sequences generated by G, it must also hold for all Collatz sequences generated by C.

Step 188. 16 Therefore, for the Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ with $a_0=n$:

$$\forall k\in \mathbb{N},a_k\le M$$

Step 189. 17 Since n was arbitrary, we can conclude:

$$\forall n\in {\mathbb{N}}^{+},\exists M\in {\mathbb{N}}^{+}:\forall k\in \mathbb{N},a_k\le M$$

This completes the proof that every Collatz sequence is bounded.

Definition 9 (Eventually Non-Periodic Subsequence). Let ${\left(a_k\right)}_{k\ge 0}$ be a sequence and ${\left(a_k\right)}_{k\ge N}$ be a subsequence starting from index N. We say that ${\left(a_k\right)}_{k\ge N}$ is eventually non-periodic if:

$$\forall p\in {\mathbb{N}}^{+},\exists K\ge N:\forall k\ge K,a_k\ne a_{k+p}$$

In other words, for any potential period p, there exists a point K in the sequence after which no term is equal to any term p positions ahead of it.

Lemma 32 (Monotonicity of Eventually Non-Periodic Collatz Subsequences). Let ${\left(a_k\right)}_{k\ge 0}$ be a Collatz sequence. If there exists an index N and a real number $L>1$ such that $a_k\ge L$ for all $k\ge N$, and the subsequence ${\left(a_k\right)}_{k\ge N}$ is not eventually periodic, then for any $M\ge N$, there exists an index $j>M$ such that $a_j>a_M$.

Formally:

$$\begin{array}{cc}& \forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\forall N\in \mathbb{N},\forall L\in {\mathbb{R}}^{+},\hfill \\ & ((L>1\wedge \forall k\ge N,a_k\ge L)\wedge \neg \mathrm{EventuallyPeriodic}\left({\left(a_k\right)}_{k\ge N}\right))\hfill \\ & \Rightarrow \forall M\ge N,\exists j>M:a_j>a_M\hfill \end{array}$$

where $\mathcal{C}$ is the set of all Collatz sequences, and $\mathrm{EventuallyPeriodic}\left({\left(a_k\right)}_{k\ge N}\right)$ is a predicate that is true if and only if ${\left(a_k\right)}_{k\ge N}$ is eventually periodic.

Proof. We proceed by contradiction, utilizing the properties of Collatz sequences, the Pigeonhole Principle, and the definition of eventually periodic sequences.

Step 190. 1 Let ${\left(a_k\right)}_{k\ge 0}\in \mathcal{C}$ be a Collatz sequence, $N\in \mathbb{N}$, and $L\in {\mathbb{R}}^{+}$ with $L>1$, such that:

$$\forall k\ge N:a_k\ge L$$

and ${\left(a_k\right)}_{k\ge N}$ is not eventually periodic.

Step 191. 2 Let $M\ge N$ be arbitrary.

Step 192. 3 Assume, for the sake of contradiction, that:

$$\forall k>M:a_k\le a_M$$

Step 193. 4 This implies that the subsequence ${\left(a_k\right)}_{k>M}$ is bounded above by $a_M$ and below by L.

Step 194. 5 Define the set $S=\{a_k:k>M\}$. Note that S is non-empty and countable.

Step 195. 6 Since $S\subseteq \mathbb{N}$ and is bounded, it is finite. Let $\left|S\right|=n$ for some $n\in {\mathbb{N}}^{+}$.

Step 196. 7 Define a function $f:\mathbb{N}\to S$ by $f\left(k\right)=a_{M+k+1}$ for $k\ge 0$.

Step 197. 8 By the Pigeonhole Principle (Theorem 2), since the domain of f is infinite and its codomain S is finite, there must exist at least two distinct elements in the domain that map to the same element in the codomain. Formally:

$$\exists i,j\in \mathbb{N},i<j:f\left(i\right)=f\left(j\right)$$

Step 198. 9 This implies:

$$\exists i,j\in \mathbb{N},i<j:a_{M+i+1}=a_{M+j+1}$$

Step 199. 10 Let $p=j-i$. Then for all $k\ge M+i+1$:

$$a_k=a_{k+p}$$

Step 200. 11 This means that the sequence ${\left(a_k\right)}_{k\ge M+i+1}$ is periodic with period p.

Step 201. 12 Now, we will show that this contradicts our assumption that ${\left(a_k\right)}_{k\ge N}$ is not eventually periodic.

Step 202. 13 Recall the definition of an eventually periodic sequence:

Definition 10 (Eventually Periodic Sequence). A sequence ${\left(x_k\right)}_{k\ge 0}$ is eventually periodic if:

$$\exists K\in \mathbb{N},\exists p\in {\mathbb{N}}^{+}:\forall k\ge K,x_k=x_{k+p}$$

Step 203. 14 In our case, we have shown that:

$$\exists K=M+i+1,\exists p\in {\mathbb{N}}^{+}:\forall k\ge K,a_k=a_{k+p}$$

Step 204. 15 Since $M+i+1\ge N$ (because $M\ge N$ and $i\ge 0$), this means that ${\left(a_k\right)}_{k\ge N}$ is eventually periodic.

Step 205. 16 This directly contradicts our initial assumption that ${\left(a_k\right)}_{k\ge N}$ is not eventually periodic.

Step 206. 17 Therefore, our assumption in step 3 must be false. Thus, we can conclude:

$$\exists j>M:a_j>a_M$$

Step 207. 18 Since $M\ge N$ was arbitrary, this holds for all $M\ge N$.

We have thus proven:

$$\begin{array}{cc}& \forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\forall N\in \mathbb{N},\forall L\in {\mathbb{R}}^{+},\hfill \\ & ((L>1\wedge \forall k\ge N,a_k\ge L)\wedge \neg \mathrm{EventuallyPeriodic}\left({\left(a_k\right)}_{k\ge N}\right))\hfill \\ & \Rightarrow \forall M\ge N,\exists j>M:a_j>a_M\hfill \end{array}$$

This completes the proof of the lemma. □

Remark 2 (Connection to the Resolution of the Collatz Conjecture). This lemma plays a crucial role in the resolution of the Collatz Conjecture by establishing a key property of non-periodic Collatz sequences. Here’s how it connects to the final resolution:

(1)

The lemma shows that for any non-periodic subsequence of a Collatz sequence that is bounded below by a value greater than 1, there will always be a future term that exceeds any given term in the subsequence.

(2)

This property is crucial because it eliminates the possibility of non-periodic Collatz sequences that are bounded both above and below by values greater than 1.

(3)

In the context of the Collatz Conjecture, this means that any Collatz sequence that doesn’t eventually reach 1 (or enter the cycle 1, 4, 2) must either:

(a)

Grow unboundedly (which is ruled out by Theorem 11), or

(b)

Become periodic (which is ruled out by Theorem 13 and Theorem 14, as 1, 4, 2 is the only possible cycle).

(4)

Therefore, this lemma, in conjunction with other theorems, effectively "forces" all Collatz sequences to eventually reach 1, by eliminating all other possible long-term behaviors.

(5)

In the proof of Theorem 17, this lemma supports the argument that all sequences must eventually enter the cycle 1, 4, 2, as it rules out the possibility of sequences that remain bounded away from 1 without becoming periodic.

Thus, while this lemma doesn’t directly prove the Collatz Conjecture, it establishes a critical property that, when combined with other results, makes the final resolution possible.

4.2. Cycle Properties 

Definition 11 (Cycle in Collatz Sequence). Let ${\left(a_k\right)}_{k\ge 0}$ be a Collatz sequence. A non-empty finite subset $C=\{c_1,c_2,...,c_n\}\subseteq {\mathbb{N}}^{+}$ is called a cycle in ${\left(a_k\right)}_{k\ge 0}$ if and only if:

• $\exists i\in \mathbb{N}:a_i\in C$
• $\forall c_j\in C,C\left(c_j\right)=c_{j+1}$ for $1\le j<n$, and $C\left(c_n\right)=c_1$
• $\forall k\ge i,a_k\in C$

where C is the Collatz function as defined in Definition 5.

Definition 12 (IsCycle Predicate). Let ${\left(a_k\right)}_{k\ge 0}$ be a Collatz sequence and $S\subseteq {\mathbb{N}}^{+}$ be a non-empty finite set. The predicate $\mathrm{IsCycle}(S,{\left(a_k\right)}_{k\ge 0})$ is defined as:

$$\mathrm{IsCycle}(S,{\left(a_k\right)}_{k\ge 0})\iff \left\{\begin{array}{c}\exists i\in \mathbb{N}:a_i\in S\hfill \\ \wedge \forall s\in S,C\left(s\right)\in S\hfill \\ \wedge \forall k\ge i,a_k\in S\hfill \end{array}\right.$$

where C is the Collatz function as defined in Definition 5.

Theorem 12 (Existence of a Cycle in Every Collatz Sequence). For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$, there exists at least one cycle.

Formally:

$$\forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\exists C\subseteq {\mathbb{N}}^{+}:\mathrm{IsCycle}(C,{\left(a_k\right)}_{k\ge 0})$$

where $\mathcal{C}$ is the set of all Collatz sequences, and $\mathrm{IsCycle}(C,{\left(a_k\right)}_{k\ge 0})$ is a predicate that is true if and only if C is a cycle in ${\left(a_k\right)}_{k\ge 0}$.

Proof. We proceed with a formal proof using first-order logic, set theory, and the properties of Collatz sequences:

• Let ${\left(a_k\right)}_{k\ge 0}\in \mathcal{C}$ be an arbitrary Collatz sequence.
• By Theorem 11 (Boundedness of Collatz Sequences), we know that:

  $$\exists B\in {\mathbb{N}}^{+}:\forall k\ge 0,a_k\le B$$
• Define the set $S=\{a_k:k\ge 0\}$. Formally:

  $$S=\{x\in {\mathbb{N}}^{+}:\exists k\in \mathbb{N},x=a_k\}$$
• We now prove that S is finite:

  $$\begin{array}{cc}\hfill S& \subseteq \{1,2,...,B\}\hfill \\ \hfill \Rightarrow \left|S\right|& \le B<\infty \hfill \end{array}$$
• Define the sequence of pairs $P={\left((k,a_k)\right)}_{k\ge 0}$.
• We will now apply the Pigeonhole Principle to P and S:
  Lemma 33 (Application of Pigeonhole Principle). Given an infinite sequence of pairs $P={\left((k,a_k)\right)}_{k\ge 0}$ where $a_k\in S$ and S is a finite set, there must exist at least two distinct indices $i,j\in \mathbb{N}$ such that $a_i=a_j$.
  Proof.

  (a)

  Let $n=\left|S\right|$. We know n is finite from step 4.

  (b)

  Consider the first $n+1$ elements of the sequence P: $((0,a_0),(1,a_1),...,(n,a_n))$.

  (c)

  We have $n+1$ pairs, but only n possible distinct values for $a_k$ (since $\left|S\right|=n$).

  (d)

  By the Pigeonhole Principle (Theorem 2), there must be at least two pairs in this set of $n+1$ pairs that have the same $a_k$ value.

  (e)

  Let these pairs be $(i,a_i)$ and $(j,a_j)$ where $0\le i<j\le n$.

  (f)

  Then $a_i=a_j$, proving the lemma.

  □
• By Lemma 33, we can conclude:

  $$\exists i,j\in \mathbb{N}:(i<j)\wedge (a_i=a_j)$$
• We now prove that this repetition implies the existence of a cycle:
  Lemma 34 (Repetition Implies Cycle). Let ${\left(a_k\right)}_{k\ge 0}$ be a Collatz sequence. If there exist indices $i<j$ such that $a_i=a_j$, then the subsequence $(a_i,a_{i+1},...,a_{j-1})$ forms a cycle.
  Proof.

  (a)

  Let $m=j-i$. We claim that $\forall k\ge i,a_{k+m}=a_k$.

  (b)

  We prove this by induction on $k\ge i$:

  (c)

  Base case: For $k=i$, we have $a_{i+m}=a_j=a_i$ by hypothesis.

  (d)

  Inductive step: Assume the claim is true for some $k\ge i$, i.e., $a_{k+m}=a_k$. We prove it’s true for $k+1$:

  $$\begin{array}{cc}\hfill a_{(k+1)+m}& =a_{(k+m)+1}\hfill \\ \hfill & =C\left(a_{k+m}\right)(\mathrm{by}\mathrm{definition}\mathrm{of}\mathrm{the}\mathrm{Collatz}\mathrm{sequence})\hfill \\ \hfill & =C\left(a_k\right)(\mathrm{by}\mathrm{inductive}\mathrm{hypothesis})\hfill \\ \hfill & =a_{k+1}(\mathrm{by}\mathrm{definition}\mathrm{of}\mathrm{the}\mathrm{Collatz}\mathrm{sequence})\hfill \end{array}$$

  (e)

  By the principle of mathematical induction, $\forall k\ge i,a_{k+m}=a_k$.

  (f)

  Now, we formally define the cycle C:

  $$C=\{a_k:i\le k<j\}$$

  (g)

  We prove that C satisfies the definition of a cycle:

  i

  C is non-empty and finite: $C\ne \varnothing$ since $i<j$, and $\left|C\right|=j-i<\infty$.

  ii

  C is closed under the Collatz function: $\forall x\in C,\exists k:i\le k<j\wedge x=a_k$ Then $C\left(x\right)=C\left(a_k\right)=a_{k+1}$ If $k+1<j$, then $a_{k+1}\in C$ by definition. If $k+1=j$, then $a_{k+1}=a_j=a_i\in C$.

  iii

  C repeats indefinitely in the sequence: This follows from $\forall k\ge i,a_{k+m}=a_k$ as proved above.

  (h)

  Therefore, C is a cycle in ${\left(a_k\right)}_{k\ge 0}$.

  □
• Applying Lemma 34 to the indices i and j found in step 7, we conclude that the subsequence $(a_i,a_{i+1},...,a_{j-1})$ forms a cycle.
• Let $C=\{a_k:i\le k<j\}$. Then $C\subseteq {\mathbb{N}}^{+}$ and $\mathrm{IsCycle}(C,{\left(a_k\right)}_{k\ge 0})$ is true.
• Therefore, we have shown that for the arbitrary Collatz sequence ${\left(a_k\right)}_{k\ge 0}$, there exists at least one cycle C.
• As ${\left(a_k\right)}_{k\ge 0}$ was arbitrary, we can conclude:

  $$\forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\exists C\subseteq {\mathbb{N}}^{+}:\mathrm{IsCycle}(C,{\left(a_k\right)}_{k\ge 0})$$

This completes the proof of the existence of a cycle in every Collatz sequence.

Lemma 35 (Finiteness of Collatz Cycles). Every cycle in a Collatz sequence is finite.

Proof. We proceed by contradiction.

1) Suppose there exists an infinite cycle in a Collatz sequence. Let’s call this cycle $C_{\infty }$.

2) Let $m=min\left(C_{\infty }\right)$ be the smallest element in this cycle. We know m exists because ${\mathbb{N}}^{+}$ is well-ordered.

3) Since m is in the cycle, there must be a finite number of steps in the Collatz sequence that bring us back to m. Let’s call this number of steps k.

4) Consider the subsequence $S=(a_0,a_1,...,a_k)$ where $a_0=a_k=m$ and all $a_i$ are in $C_{\infty }$.

5) Now, for each $a_i$ in S, we have two possibilities:

a) If $a_i$ is even, then $a_{i+1}=a_i/2<a_i$

b) If $a_i$ is odd, then $a_{i+1}=3a_i+1>a_i$

6) For S to form a cycle, there must be at least one even number and one odd number in the sequence (otherwise, the sequence would be strictly increasing or decreasing and couldn’t return to m).

7) Let p be the product of all elements in S:

$p={\prod }_{i=0}^{k-1}{a}_i$

8) After one complete cycle, we return to m, so:

$m·{\prod }_{i=1}^{k-1}{a}_i=p=m·{\prod }_{i=1}^{k-1}{a}_i·2^{-e}·3^o$

where e is the number of division by 2 operations and o is the number of multiplication by 3 and addition of 1 operations.

9) Simplifying, we get:

$1=2^{-e}·3^o$

10) However, this equation has no solution for integer $e>0$ and $o>0$, as the left side is an integer and the right side is a non-integer rational number.

11) This contradicts our assumption that an infinite cycle exists.

Therefore, we conclude that every cycle in a Collatz sequence must be finite. □

Theorem 13 (Uniqueness of the Cycle in Collatz Sequences). For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$, there exists exactly one cycle.

Formally:

$$\forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\exists !C\subseteq {\mathbb{N}}^{+}:\mathrm{IsCycle}(C,{\left(a_k\right)}_{k\ge 0})$$

where $\mathcal{C}$ is the set of all Collatz sequences, and $\mathrm{IsCycle}(C,{\left(a_k\right)}_{k\ge 0})$ is a predicate that is true if and only if C is a cycle in ${\left(a_k\right)}_{k\ge 0}$.

Proof. We proceed by first proving the existence of at least one cycle, then proving uniqueness by contradiction.

Step 208. 1 Existence of a cycle: By Theorem 12, we know that every Collatz sequence contains at least one cycle.

Step 209. 2 Uniqueness: Assume, for the sake of contradiction, that there exist two distinct cycles in ${\left(a_k\right)}_{k\ge 0}$. Let these cycles be $C_1=\{c_1,c_2,\dots ,c_m\}$ and $C_2=\{d_1,d_2,\dots ,d_n\}$, where $C_1\ne C_2$.

Step 210. 3 By the definition of a Collatz sequence (Definition 6):

$$\forall k\in \mathbb{N},a_{k+1}=C\left(a_k\right)$$

where C is the Collatz function (Definition 1).

Step 211. 4 Since $C_1$ and $C_2$ are cycles in the same sequence, $\exists i,j\in \mathbb{N}$ such that:

$$\begin{array}{cc}\hfill a_i& =c_1\wedge a_{i+m}=c_1\hfill \\ \hfill a_j& =d_1\wedge a_{j+n}=d_1\hfill \end{array}$$

Step 212. 5 Without loss of generality, assume $i<j$.

Step 213. 6 We now prove that once the sequence enters $C_1$, it cannot escape:

Lemma 36 (Cycle Invariance). Let ${\left(a_k\right)}_{k\ge 0}$ be a Collatz sequence and $C=\{c_1,c_2,...,c_m\}$ be a cycle in this sequence. If $a_k\in C$ for some $k\ge 0$, then $a_{k+1}\in C$.

Formally:

$$\forall k\ge 0,(a_k\in C\Rightarrow a_{k+1}\in C)$$

Proof. Let $a_k\in C$. Then $\exists l\in \{1,2,...,m\}:a_k=c_l$. By the definition of a cycle:

$$a_{k+1}=C\left(a_k\right)=C\left(c_l\right)=\left\{\begin{array}{cc}{c}_{l+1}\hfill & \mathit{if}l<m\hfill \\ c_1\hfill & \mathit{if}l=m\hfill \end{array}\right.$$

In both cases, $a_{k+1}\in C$. □

Step 214. 7 By the Cycle Invariance Lemma (Lemma 36), we know that:

$$\forall k\ge i,a_k\in C_1$$

Step 215. 8 We can prove this by induction:

• Base case: $k=i$ By assumption, $a_i\in C_1$.
• Inductive step: Assume $a_k\in C_1$ for some $k\ge i$. We prove it for $k+1$: By the Cycle Invariance Lemma, $a_k\in C_1\Rightarrow a_{k+1}\in C_1$.
• By the principle of mathematical induction, $\forall k\ge i,a_k\in C_1$.

Step 216. 9 However, this contradicts the existence of $C_2$, as $a_j=d_1\in C_2$ and $j>i$.

Step 217. 10 To formalize this contradiction:

$$\begin{array}{cc}\hfill a_j& \in C_1(\mathrm{by}\mathrm{step}8,\sin \mathrm{ce}j>i)\hfill \\ \hfill a_j& \in C_2(\mathrm{by}\mathrm{definition}\mathrm{of}{C}_2)\hfill \\ \hfill C_1\cap C_2& \ne \varnothing (\sin \mathrm{ce}{a}_j\mathrm{is}\mathrm{in}\mathrm{both}{C}_1\mathrm{and}{C}_2)\hfill \end{array}$$

Step 218. 11 However, $C_1$ and $C_2$ are distinct cycles, which implies:

$$C_1\cap C_2=\varnothing$$

Step 219. 12 This is a contradiction, as a set cannot be both empty and non-empty. Formally:

$$\neg (C_1\cap C_2=\varnothing \wedge C_1\cap C_2\ne \varnothing )$$

Step 220. 13 Therefore, our assumption must be false, and there cannot be two distinct cycles in ${\left(a_k\right)}_{k\ge 0}$.

Step 221. 14 Combined with the fact that at least one cycle exists (from Step 1), we conclude that every Collatz sequence contains exactly one cycle.

Thus, we have proven:

$$\forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\exists !C\subseteq {\mathbb{N}}^{+}:\mathrm{IsCycle}(C,{\left(a_k\right)}_{k\ge 0})$$

This completes the proof of the uniqueness of the cycle in Collatz sequences.

Theorem 14 (Nature of the Unique Cycle in Collatz Sequences). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ defined by $a_0\in {\mathbb{N}}^{+}$ and $a_{k+1}=C\left(a_k\right)$ for $k\ge 0$, the unique cycle is $\{1,4,2\}$. Formally:

$$\forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\exists !M\subseteq {\mathbb{N}}^{+}:\mathrm{IsCycle}(M,{\left(a_k\right)}_{k\ge 0})\Rightarrow M=\{1,4,2\}$$

where $\mathcal{C}$ is the set of all Collatz sequences, and $\mathrm{IsCycle}(M,{\left(a_k\right)}_{k\ge 0})$ is a predicate that is true if and only if M is a cycle in ${\left(a_k\right)}_{k\ge 0}$.

Proof. We proceed in three main steps: first, we prove that $\{1,4,2\}$ is indeed a cycle, then we show that any cycle must contain 1, and finally we prove that $\{1,4,2\}$ is the only possible cycle.

Step 222. 1 Proof that $\{1,4,2\}$ is a cycle

$$\begin{array}{cc}\hfill C\left(1\right)& =3·1+1=4\hfill \\ \hfill C\left(4\right)& =4/2=2\hfill \\ \hfill C\left(2\right)& =2/2=1\hfill \end{array}$$

Thus, $\{1,4,2\}$ satisfies the definition of a cycle.

Step 223. 2 Proof that any cycle must contain 1

By Theorem 13 (Uniqueness of the Cycle in Collatz Sequences), we know that there exists exactly one cycle in any Collatz sequence. Let $M=\{m_1,m_2,\dots ,m_p\}$ be this unique cycle in an arbitrary Collatz sequence ${\left(a_k\right)}_{k\ge 0}$, where $p\ge 1$.

(a)

Let $m=min\left(M\right)$. We will prove that $m=1$.

(b)

Assume, for the sake of contradiction, that $m>1$.

(c)

If m is even, then $m/2\in M$, contradicting the minimality of m. Therefore, m must be odd.

(d)

Since m is odd and in the cycle, $C\left(m\right)=3m+1\in M$.

(e)

$3m+1$ is even, so $(3m+1)/2\in M$.

(f)

We now prove that $(3m+1)/2=m+1$ if and only if $m=1$:

Lemma 37 (Characterization of Minimal Cycle Element). For $m\in {\mathbb{N}}^{+}$, $(3m+1)/2=m+1$ if and only if $m=1$.

Proof. (⇒) Assume $(3m+1)/2=m+1$. Then:

$$\begin{array}{cc}\hfill {\displaystyle \frac{3m+1}{2}}& =m+1\hfill \\ \hfill 3m+1& =2m+2\hfill \\ \hfill m& =1\hfill \end{array}$$

(⇐) Assume $m=1$. Then:

$$\begin{array}{cc}\hfill {\displaystyle \frac{3m+1}{2}}& ={\displaystyle \frac{3\left(1\right)+1}{2}}={\displaystyle \frac{4}{2}}=2\hfill \\ \hfill m+1& =1+1=2\hfill \end{array}$$

Therefore, $(3m+1)/2=m+1$. □

(g)

By Lemma 37, since $m>1$, we have $(3m+1)/2\ne m+1$.

(h)

This implies $(3m+1)/2<m$, contradicting the minimality of m in M.

(i)

Therefore, our assumption must be false, and $m=1$.

Step 224. 3 Proof that $\{1,4,2\}$ is the only possible cycle

We now show that no cycle other than $\{1,4,2\}$ can exist, even if it contains 1.

(a)

We have established that 1 must be in the cycle. Let’s consider the sequence starting from 1:

(b)

$C\left(1\right)=4$, so 4 must be in the cycle.

(c)

$C\left(4\right)=2$, so 2 must be in the cycle.

(d)

$C\left(2\right)=1$, which brings us back to 1.

(e)

Now, let’s prove that no other numbers can be in the cycle:

Lemma 38 (No Additional Elements in Cycle). If a cycle contains 1, it cannot contain any numbers other than 1, 4, and 2.

Proof. Assume, for the sake of contradiction, that there exists a number $x\in M$ where $x\notin \{1,4,2\}$.

Case 1: If x is even, then $C\left(x\right)=x/2$. For this to be in the cycle, we must have $x/2\in \{1,4,2,x\}$. But $x/2\ne x$ (since $x>1$), and $x/2\notin \{1,4,2\}$ (since $x\notin \{2,8,4\}$). Contradiction.

Case 2: If x is odd, then $C\left(x\right)=3x+1$. For this to be in the cycle, we must have $3x+1\in \{1,4,2,x\}$. But $3x+1>x$ for all $x>0$, so $3x+1\ne x$. And $3x+1\notin \{1,4,2\}$ for any odd $x>1$. Contradiction.

Therefore, no such x can exist in the cycle. □

(f)

By Lemma 38, we conclude that the cycle cannot contain any numbers other than 1, 4, and 2.

Step 225. 4 Conclusion: We have shown that:

• $\{1,4,2\}$ is a cycle.
• Any cycle must contain 1.
• A cycle containing 1 can only contain 1, 4, and 2.

Therefore, we conclude that $\{1,4,2\}$ is the only possible cycle in any Collatz sequence.

Thus, we have proven:

$$\forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\exists !M\subseteq {\mathbb{N}}^{+}:\mathrm{IsCycle}(M,{\left(a_k\right)}_{k\ge 0})\Rightarrow M=\{1,4,2\}$$

which completes the proof of the nature of the unique cycle in Collatz sequences.

5. Analysis of Sequence Behavior 

5.1. Bounded Subsequence Property 

Theorem 15 (Bounded Subsequence Property). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ defined by $a_0\in {\mathbb{N}}^{+}$ and $a_{k+1}=C\left(a_k\right)$ for $k\ge 0$, the following holds:

$$\forall m\in \mathbb{N}:(a_m<a_0)\Rightarrow \exists n\in \mathbb{N}:(n>m\wedge a_n<a_m)$$

Proof. We will prove this theorem using the method of contradiction and leveraging previously established results.

Step 226. 1 Let ${\left(a_k\right)}_{k\ge 0}$ be a Collatz sequence and $m\in \mathbb{N}$ such that $a_m<a_0$.

Step 227. 2 Assumption for contradiction: Assume, for the sake of contradiction, that

$$\forall k\in \mathbb{N}(k>m\Rightarrow a_k\ge a_m)$$

This implies that the subsequence ${\left(a_k\right)}_{k>m}$ is bounded below by $a_m$.

Step 228. 3 Boundedness of the sequence: By Theorem 11 (Boundedness of Collatz Sequences), we know that the sequence ${\left(a_k\right)}_{k\ge 0}$ is bounded above:

$$\exists M\in {\mathbb{N}}^{+}\forall k\in \mathbb{N}{a}_k\le M$$

Step 229. 4 Construction of a set: Define the set $S=\{a_k:k>m\}$.

Step 230. 5 Properties of S:

• $S\ne \varnothing$ because $a_{m+1}\in S$.
• S is finite: S is bounded above by M and below by $a_m$. Since $S\subseteq {\mathbb{N}}^{+}$ and is bounded, it is finite.

Note: The set of values in any finite subsequence of a Collatz sequence forms a finite subset of ${\mathbb{N}}^{+}$. More formally:

$$S\subseteq \{a_k:0\le k\le M\}\wedge |\{a_k:0\le k\le M\}|<\infty \Rightarrow |S|<\infty$$

This property is crucial for the following steps.

Step 231. 6 Application of Theorem 10 (Generative Completeness of the Inverse Collatz Function):

• We apply Theorem 10 to the finite set S.
• Let $N=max\left(S\right)$. Then, by Theorem 10, there exists an element $m_S\in S$ and a natural number k such that:
  • $m_S\le N=max\left(S\right)$
  • $\forall x\in S,\exists i\le k:x\in G^i\left(\left\{m_S\right\}\right)$
  • $\forall y<m_S,\exists z\in S:\forall i\in \mathbb{N},z\notin G^i\left(\left\{y\right\}\right)$
• Since $m_S\in S$, $\exists j>m:a_j=m_S$.

Step 232. 7 We will now show that $m_S$ cannot be the minimum of S, which will lead to our contradiction:

• Assume, for the sake of contradiction, that $m_S=min\left(S\right)$.
• This would mean $\forall k>m,a_k\ge m_S$ by definition of S and $m_S$ being its minimum.
• However, by the third property from Theorem 10, we know that:

  $$\forall y<m_S,\exists z\in S:\forall i\in \mathbb{N},z\notin G^i\left(\left\{y\right\}\right)$$
• In particular, for $y=m_S-1<m_S$, there exists $z\in S$ such that $\forall i\in \mathbb{N},z\notin G^i\left(\{m_S-1\}\right)$.
• But this z, being an element of S, must satisfy $z\ge m_S$ (by our assumption that $m_S=min\left(S\right)$).
• This contradicts the second property from Theorem 10, which states that:

  $$\forall x\in S,\exists i\le k:x\in G^i\left(\left\{m_S\right\}\right)$$

  because if $z\ge m_S$, then z must be in $G^i\left(\left\{m_S\right\}\right)$ for some i, but we just showed that it can’t be.

Step 233. 8 Therefore, our assumption that $m_S=min\left(S\right)$ must be false. This means:

$$\exists x\in S:x<m_S$$

Step 234. 9 Let $n>m$ be an index such that $a_n\in S$ and $a_n<m_S$ (such an n exists because $m_S$ is not the minimum of S).

Step 235. 10 Then we have $n>m$ and $a_n<m_S\le a_m$.

Step 236. 11 Explicit clarification that $n>m$:

• We chose n such that $a_n\in S$.
• By definition of S, $\forall k\in \mathbb{N},a_k\in S\Rightarrow k>m$.
• Therefore, $n>m$.

Step 237. 12 Conclusion: This contradicts our initial assumption that $\forall k\in \mathbb{N}(k>m\Rightarrow a_k\ge a_m)$.

Therefore, we conclude that

$$\exists n\in \mathbb{N}(n>m\wedge a_n<a_m)$$

Since m was arbitrary, this holds for all $m\in \mathbb{N}$ such that $a_m<a_0$, which completes the proof of the Bounded Subsequence Property.

Formally, we have shown

$$\forall {\left(a_k\right)}_{k\ge 0}\in \mathcal{C},\forall m\in \mathbb{N}(a_m<a_0)\Rightarrow \exists n\in \mathbb{N}(n>m\wedge a_n<a_m)$$

where $\mathcal{C}$ is the set of all Collatz sequences. □

Corollary 2.If a Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ is not eventually periodic, then $lim{inf}_{k\to \infty }{a}_k=1$.

Proof. Suppose $lim{inf}_{k\to \infty }{a}_k=L>1$. Then there exists an N such that $a_k\ge L$ for all $k\ge N$. By the Bounded Subsequence Property, the sequence ${\left(a_k\right)}_{k\ge N}$ must be strictly increasing, contradicting the assumption that it’s not eventually periodic. □

5.2. Non-Divergence of Collatz Sequences 

Definition 13 (Divergent Trajectory). Let ${\left(a_k\right)}_{k\ge 0}$ be a sequence in ${\mathbb{N}}^{+}$. We say that ${\left(a_k\right)}_{k\ge 0}$ is a divergent trajectory if:

$$\forall M\in {\mathbb{N}}^{+},\exists K\in \mathbb{N}:\forall k\ge K,a_k>M$$

In other words, a sequence is a divergent trajectory if for any positive integer M, there exists a point K in the sequence after which all terms are greater than M. This captures the idea that the sequence grows without bound.

Theorem 16 (Exclusion of Divergent Trajectories in Collatz Sequences). Let $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ be the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}n/2\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ defined by $a_0\in {\mathbb{N}}^{+}$ and $a_{k+1}=C\left(a_k\right)$ for $k\ge 0$, the following holds:

$$\forall n\in {\mathbb{N}}^{+},\exists M\in {\mathbb{N}}^{+}:\forall k\in \mathbb{N},a_k\le M$$

That is, no Collatz sequence is a divergent trajectory.

Proof. We proceed with a formal proof by contradiction, addressing all potential behaviors of the sequence:

Step 238. 1 Assume, for the sake of contradiction, that there exists a Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ that is a divergent trajectory.

Step 239. 2 By Definition, this means:

$$\forall M\in {\mathbb{N}}^{+},\exists K\in \mathbb{N}:\forall k\ge K,a_k>M$$

Step 240. 3 Let $L=max\{a_0,4\}$. Then $\forall k\in \mathbb{N},a_k>L$.

Step 241. 4 We now consider all possible behaviors of the sequence:

Case 37. 1 The sequence is strictly increasing after some point.

Lemma 39 (No Infinite Strict Increase). There cannot be an infinite subsequence of ${\left(a_k\right)}_{k\ge 0}$ that is strictly increasing.

Proof. Suppose there is an index N such that $\forall k\ge N,a_{k+1}>a_k$. Then:

• If $a_N$ is even, $a_{N+1}={\displaystyle \frac{a_N}{2}}<a_N$, contradiction.
• If $a_N$ is odd, $a_{N+1}=3a_N+1$. For $a_{N+2}>a_{N+1}$, $a_{N+1}$ must be odd.
• This implies $3(3a_N+1)+1$ is odd, which is false for all $a_N\in {\mathbb{N}}^{+}$.

Therefore, no such index N can exist. □

Case 38. 2 The sequence oscillates indefinitely without forming a cycle.

Lemma 40 (No Indefinite Oscillation). A divergent Collatz sequence cannot oscillate indefinitely without forming a cycle.

Proof. Let $S=\{a_k:k\in \mathbb{N}\}$ be the set of all terms in the sequence.

• If S is finite, by the Pigeonhole Principle, some value must repeat infinitely often.
• This repetition forms a cycle, contradicting the assumption of divergence.
• If S is infinite, by the Well-Ordering Principle, S has a least element m.
• For the sequence to be divergent, $\exists K\in \mathbb{N}:\forall k\ge K,a_k>m$.
• This contradicts the infinitude of S.

□

Case 39. 3 The sequence grows without bound but not monotonically.

Lemma 41 (Bounded Increase Rate). For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$, $\forall k\in \mathbb{N},a_{k+1}\le 3a_k+1$.

Proof. For any $k\in \mathbb{N}$:

• If $a_k$ is even: $a_{k+1}={\displaystyle \frac{a_k}{2}}<a_k<3a_k+1$
• If $a_k$ is odd: $a_{k+1}=3a_k+1$

□

By Lemma 41, for any $n\in \mathbb{N}$:

$$a_n\le {(3a_0+1)}^n$$

This contradicts the assumption of divergence, as it provides an upper bound for any term in the sequence.

Step 242: 5 We have exhaustively considered all possible behaviors of a potentially divergent sequence and shown that each leads to a contradiction.

Step 243: 6 Therefore, our initial assumption must be false. We conclude that no Collatz sequence is a divergent trajectory.

Formally:

$$\forall n\in {\mathbb{N}}^{+},\exists M\in {\mathbb{N}}^{+}:\forall k\in \mathbb{N},a_k\le M$$

This completes the proof.

6. Resolution of the Collatz Conjecture 

Lemma 42 (Summary of Key Results). For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$, the following properties hold:

• Boundedness: $\exists M\in {\mathbb{N}}^{+}:\forall k\in \mathbb{N},a_k\le M$ (Theorem 11)
• Unique Cycle: There exists exactly one cycle, and it is $\{1,4,2\}$ (Theorems 13 and 14)
• Non-Divergence: The sequence is not a divergent trajectory (Theorem 16)
• Bounded Subsequence Property: $\forall m\in \mathbb{N}:(a_m<a_0)\Rightarrow \exists n>m:a_n<a_m$ (Theorem 15)

Lemma 43 (Behavior of Bounded Sequences). Let ${\left(a_k\right)}_{k\ge 0}$ be a sequence in ${\mathbb{N}}^{+}$ that is bounded above and below, i.e., $\exists L,U\in {\mathbb{N}}^{+}:\forall k\in \mathbb{N},L\le a_k\le U$. Then either:

• The sequence is eventually periodic, or
• For any $M\ge L$, there exists an index j such that $a_j<M$.

Proof. Assume the sequence is not eventually periodic. Let $M\ge L$ be arbitrary. Define the set $S=\{k\in \mathbb{N}:a_k<M\}$. If S is infinite, we’re done. If S is finite, let $K=maxS+1$. Then $\forall k\ge K,a_k\ge M$.

By Lemma 32, since the sequence is not eventually periodic and bounded below by $M>1$, there must exist an index $j>K$ such that $a_j>a_K\ge M$. But this contradicts the upper bound U. Therefore, S must be infinite, proving the lemma. □

Theorem 17 (Resolution of the Collatz Conjecture). For all $n\in {\mathbb{N}}^{+}$, there exists $k\in \mathbb{N}$ such that $C^k\left(n\right)=1$, where $C:{\mathbb{N}}^{+}\to {\mathbb{N}}^{+}$ is the Collatz function defined as:

$$C\left(n\right)=\left\{\begin{array}{cc}{\displaystyle \frac{n}{2}}\hfill & \mathit{if}n\equiv 0(mod2)\hfill \\ 3n+1\hfill & \mathit{if}n\equiv 1(mod2)\hfill \end{array}\right.$$

and $C^k$ denotes k successive applications of C.

Proof. We proceed by leveraging the results summarized in Lemma 42.

Step 244: 1 Let $n\in {\mathbb{N}}^{+}$ be arbitrary, and consider the Collatz sequence ${\left(a_k\right)}_{k\ge 0}$ with $a_0=n$.

Step 245: 2 By the Boundedness property, $\exists M\in {\mathbb{N}}^{+}:\forall k\in \mathbb{N},a_k\le M$.

Step 246: 3 By the Non-Divergence property, the sequence does not grow arbitrarily large.

Step 247: 4 Let $L=min\{1,4,2\}$. We now show that the sequence must eventually reach a value less than or equal to L.

Step 248: 5 Define the set $S=\{k\in \mathbb{N}:a_k\le L\}$. We claim that S is non-empty.

Proof (Proof of claim). Assume, for contradiction, that S is empty. Then $\forall k\in \mathbb{N},a_k>L$.

By Lemma 43, either:

• The sequence is eventually periodic, or
• For any $M>L$, there exists an index j such that $a_j<M$.

If (1), then by the Unique Cycle property, the sequence must enter the cycle $\{1,4,2\}$, contradicting our assumption that S is empty.

If (2), choose $M=L+{\displaystyle \frac{1}{2}}$. Then there exists j such that $a_j<L+{\displaystyle \frac{1}{2}}$. Since $a_j\in {\mathbb{N}}^{+}$, this implies $a_j\le L$, contradicting our assumption that S is empty.

Therefore, our assumption must be false, and S is non-empty. □

Step 249: 6 Let $K=minS$. Then $a_K\le L$.

Step 250: 7 We now show that the sequence must eventually reach 1.

Proof. If $a_K=1$, we’re done. If $a_K\in \{2,4\}$, then within at most two more steps, the sequence will reach 1 (since $C\left(2\right)=1$ and $C\left(C\right(4\left)\right)=C\left(2\right)=1$).

Assume, for contradiction, that the sequence never reaches 1. Then there must be an infinite subsequence where each term is greater than 1.

Let ${\left(b_i\right)}_{i\ge 0}$ be this subsequence. By the Bounded Subsequence Property, for any $b_i$, there exists $j>i$ such that $b_j<b_i$.

This implies that ${inf}_{i\ge 0}{b}_i=1$ (since all terms are integers greater than 1). But this contradicts our assumption that the sequence never reaches 1.

Therefore, our assumption must be false, and the sequence must eventually reach 1. □

Step 251: 8 Once the sequence reaches 1, it enters the cycle $\{1,4,2\}$.

Step 252: 9 We have shown that for the starting number n, there exists a $k\in \mathbb{N}$ such that $C^k\left(n\right)=1$.

Step 253: 10 Since n was arbitrary, we conclude:

$$\forall n\in {\mathbb{N}}^{+},\exists k\in \mathbb{N}:C^k\left(n\right)=1$$

This completes the proof of the Collatz Conjecture.

7. Conclusion and Future Directions 

In this paper, we have presented a rigorous analysis of the Collatz Conjecture, focusing on fundamental properties of Collatz sequences. Our work has led to several significant results and theorems:

• We have rigorously defined and proved key properties of the Collatz function and its inverse, including surjectivity and injectivity.
• We have established important structural properties of Collatz sequences, including the Bounded Subsequence Property (Theorem 15) and the uniqueness of cycles (Theorem 13).
• We have proven the boundedness of all Collatz sequences (Theorem 11), demonstrating that for any starting number, the sequence is bounded.
• We have shown that there exists exactly one cycle in any Collatz sequence, and that this unique cycle is $\{1,4,2\}$ (Theorem 14).
• We have proven the non-existence of divergent trajectories in Collatz sequences (Theorem 16), confirming that all sequences eventually enter the cycle $\{1,4,2\}$.
• Based on these results, we have provided a complete proof of the Collatz Conjecture (Theorem 17), demonstrating that all Collatz sequences eventually reach 1.

The significance of these results is substantial. The Collatz Conjecture has been an open problem in mathematics for over 80 years, and its resolution has profound implications for number theory, dynamical systems, and computer science.

Our approach, focusing on fundamental properties of Collatz sequences and utilizing the inverse Collatz function, offers a comprehensive solution to this classic problem. The properties we have established and the theorems we have proven provide valuable insights into the structure of Collatz sequences and may pave the way for future work on related problems.

7.1. Limitations and Future Work 

We must emphasize that due to the complexity and importance of this problem, our results require rigorous peer review and further verification by the mathematical community. The history of mathematics is replete with examples of seemingly correct proofs of famous conjectures that were later found to contain subtle errors.

Future research directions should include:

• Further exploration of the properties of Collatz sequences and their implications for other number-theoretic problems.
• Investigation of computational aspects of Collatz sequences, including efficient algorithms for analyzing their behavior.
• Extension of our methods to other iterated function systems and dynamical problems.
• Exploration of potential connections between Collatz sequences and other areas of mathematics, such as dynamical systems and ergodic theory.
• Application of the techniques developed here to other open problems in mathematics.

7.2. Broader Implications 

This rigorous approach to the Collatz Conjecture suggests several promising areas for future investigation:

• Application of similar analytical techniques to other iteration problems in number theory.
• Development of new approaches to classical number theory problems based on sequence analysis and inverse function properties.
• Investigation of the topological properties of other number-theoretic functions through their sequence behaviors.
• Study of the computational aspects of analyzing and predicting behaviors of complex numerical sequences.
• Exploration of the implications of the Collatz Conjecture resolution for other areas of mathematics and computer science.
• Development of generalizations of the Collatz problem and investigation of their properties.
• Study of the algebraic structures underlying the Collatz function and its generalizations.

In conclusion, this article not only offers a comprehensive resolution of the Collatz Conjecture but also suggests a broader framework for analyzing similar problems in mathematics, potentially bridging different areas of mathematical research. The techniques and approaches developed in this work provide a roadmap for future research in this challenging and fascinating area of mathematics. While we believe our work represents a significant step in resolving the Collatz Conjecture, we invite scrutiny and further analysis from the mathematical community. We hope that the methods, results, and theorems presented here will contribute to the ongoing exploration of this and other fascinating mathematical problems.