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

1. Introduction

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:

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

Definition 2 

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

$$\forall n\in {\mathbb{N}}^{+},G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

where $\mathcal{P}\left({\mathbb{N}}^{+}\right)$ denotes the power set of ${\mathbb{N}}^{+}$.

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 4 

(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:

(1)

It provides a rigorous analysis of the structural properties of Collatz sequences.

(2)

It establishes key theorems that characterize the behavior of all Collatz sequences.

(3)

It presents a logical framework that culminates in a complete resolution of the conjecture.

(4)

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 presents the culminating theorem that resolves the Collatz conjecture.
• Section 9 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. The Inverse Collatz Function: A Key Concept

The fundamental concept that underpins this proof of the Collatz Conjecture is the inverse Collatz function, denoted as G. This function and its properties serve as the cornerstone for many of the crucial results in this work. The significance of G can be summarized as follows:

(1)

Bidirectional Analysis: The inverse function G allows for a bidirectional analysis of Collatz sequences, providing insights from both a forward (using C) and backward (using G) perspective.

(2)

Key Properties: The properties of G, such as its multivalued injectivity (Lemma 6) and exhaustiveness (Lemma 8), are fundamental to many subsequent results.

(3)

Generative Completeness: The Generative Completeness Theorem (Theorem 23), which heavily relies on the properties of G, is crucial for establishing the structure of Collatz sequences.

(4)

Cycle Analysis: Function G enables a deeper analysis of cycles in Collatz sequences, leading to the proof of the uniqueness of the cycle $\{1,4,2\}$ (Theorem 35).

(5)

Bounded Subsequence Property: This key property (Theorem ) is proven using the properties of G and is fundamental to the final argument.

(6)

Equivalence of Properties: Lemma 16 establishes a crucial equivalence between properties of sequences generated by C and those generated by G, allowing for the transfer of results between both perspectives.

(7)

Final Resolution: In the final proof (Theorem 39), the properties derived from G are used to eliminate all possible trajectories that do not converge to 1.

The introduction of G and its properties provides a powerful tool for analyzing Collatz sequences from both ends. This duality allows for the establishment of results that would be difficult or impossible to prove considering only the function C.

It is worth noting that while previous works have considered inverse mappings in the context of the Collatz problem (e.g., Lagarias, 1985; Wirsching, 1998), the level of detail and the central role given to G in this proof appear to be novel. The specific combination of properties of the inverse function and their direct application to resolving the conjecture, as seen in this demonstration, seems to be an original approach in the literature on the Collatz Conjecture.

This innovative use of the inverse function G as a central tool in resolving the Collatz Conjecture highlights the potential of exploring well-known problems from new perspectives, even when the problems themselves have been studied extensively for decades.

4. Preliminaries

4.1. Basic Definitions

Definition 5 

(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 6. 

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 7 

(Pigeonhole Principle). Let A and B be finite sets, and let $f:A\to B$ be a function. Then:

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

where $\left|A\right|$ and $\left|B\right|$ denote the cardinalities of sets A and B respectively.

Proof. 

We proceed by contradiction.

Step 1: 1 Suppose the statement is false. That is, assume:

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

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

Step 3: 3 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: 4 But this contradicts our assumption that $\left|A\right|>\left|B\right|$.

Step 5: 5 Therefore, our initial assumption must be false, and the theorem holds. □

Theorem 8 

(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. 

We proceed by contradiction.

Step 6: 1 Let $S=\{n\in \mathbb{N}:P(n\left)\mathrm{is}\mathrm{false}\right\}$. We will prove that S is empty.

Step 7: 2 Assume, for the sake of contradiction, that S is non-empty. By the Well-Ordering Principle, S has a least element. Let $m=minS$.

Step 8: 3 $m\ne 1$, because $P\left(1\right)$ is true by the base case.

Step 9: 4 Since m is the least element of S, $P(m-1)$ must be true.

Step 10: 5 By the inductive step, if $P(m-1)$ is true, then $P\left(m\right)$ must be true.

Step 11: 6 But this contradicts the fact that $m\in S$.

Step 12: 7 Therefore, our assumption must be false, and S must be empty.

Step 13: 8 Thus, $P\left(n\right)$ is true for all $n\in \mathbb{N}$. □

Theorem 9 

(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. 

We proceed by contradiction.

Step 14: 1 Let $S=\{n\in \mathbb{N}:P(n\left)\mathrm{is}\mathrm{false}\right\}$. We will prove that S is empty.

Step 15: 2 Assume, for the sake of contradiction, that S is non-empty.

Step 16: 3 By the Well-Ordering Principle, S has a least element. Let $m=minS$.

Step 17: 4 $m\ne 1$, because $P\left(1\right)$ is true by the base case.

Step 18: 5 Since m is the least element of S, $P\left(j\right)$ is true for all $j<m$.

Step 19: 6 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 20: 7 But this contradicts the fact that $m\in S$.

Step 21: 8 Therefore, our assumption must be false, and S must be empty.

Step 22: 9 Thus, $P\left(n\right)$ is true for all $n\in \mathbb{N}$. □

Definition 10 

(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 11 

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

$$\left\{\begin{array}{c}{a}_0=n\hfill \\ a_{k+1}=C\left(a_k\right)\mathrm{for}k\in \mathbb{N}\hfill \end{array}\right.$$

where C is the Collatz function as defined in Definition 1.

Definition 12 

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

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

where $\mathcal{P}\left({\mathbb{N}}^{+}\right)$ denotes the power set of ${\mathbb{N}}^{+}$.

4.2. Fundamental Properties

Theorem 13 

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

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

is well-defined for all positive integers.

Proof. 

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

(1)

The function is defined for all elements in its domain.

(2)

The function produces a unique output for each input.

Step 23: 1 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 24: 2 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 25: 3 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:

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

Then C is surjective. Formally:

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

Proof. 

We will prove this by strong mathematical induction on m.

Step 26: 1 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:

$$\begin{array}{cc}\hfill & \forall n\in {\mathbb{N}}^{+},(n\mathrm{is}\mathrm{even}\wedge C\left(n\right)=1)\Rightarrow {\displaystyle \frac{n}{2}}=1\Rightarrow n=2\hfill \\ \hfill & \forall n\in {\mathbb{N}}^{+},(n\mathrm{is}\mathrm{odd}\wedge C\left(n\right)=1)\Rightarrow 3n+1=1\Rightarrow n=0\notin {\mathbb{N}}^{+}\hfill \end{array}$$

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

Step 27: 2 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 28: 3 Inductive step: We will prove the statement holds for $k+1$.

Case 1.1 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 2.2 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 29:4 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 30:5 Therefore, C is surjective. □

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:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then G is well-defined for all positive integers.

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

(1)

The function is defined for all elements in its domain.

(2)

The function produces a unique output for each input.

(3)

All elements in the output are in the codomain.

Step 31: 1 The function is defined for all elements in its domain:

(1)

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

(2)

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

$$\begin{array}{cc}\hfill n& \equiv 4(mod6)\hfill \\ \hfill n& \neg \equiv 4(mod6)\hfill \end{array}$$

(3)

Case 1: If $n\neg \equiv 4(mod6)$:

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

(4)

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 32: 2 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 33: 3 The function produces a unique output for each input:

(1)

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

(2)

Case 1: If $n\neg \equiv 4(mod6)$:

$$G\left(n\right)=\left\{2n\right\}$$

This set is uniquely determined by n.

(3)

Case 2: If $n\equiv 4(mod6)$:

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

This set is uniquely determined by n.

(4)

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

Step 34: 4 All elements in the output are in the codomain:

(1)

The codomain of G is $\mathcal{P}\left({\mathbb{N}}^{+}\right)$, the power set of positive integers.

(2)

For all $n\in {\mathbb{N}}^{+}$, $G\left(n\right)$ is a set containing either one or two positive integers.

(3)

Therefore, $G\left(n\right)\in \mathcal{P}\left({\mathbb{N}}^{+}\right)$ for all $n\in {\mathbb{N}}^{+}$.

Step 35: 5 Conclusion: We have shown that G satisfies all three criteria for well-definedness:

(1)

It is defined for all elements in its domain.

(2)

It produces a unique output for each input.

(3)

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:

$$\forall n\in {\mathbb{N}}^{+},G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

Then:

$$\forall n\in {\mathbb{N}}^{+},(G\left(n\right)\ne \varnothing )\wedge (\exists !S\subseteq {\mathbb{N}}^{+}:S=G\left(n\right))$$

Proof. We will prove this lemma in two parts:

(1)

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

(2)

Uniqueness of $G\left(n\right)$

Step 36: 1 Non-emptiness of $G\left(n\right)$

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

Case 3.1 $n\neg \equiv 4(mod6)$

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

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 37:1a 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, we have shown $G\left(n\right)\ne \varnothing$. Since n was arbitrary, we conclude:

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

Step 38:2 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 $n\neg \equiv 4(mod6)$

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

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:

(1)

The term $2n$ is always included and is a function of n.

(2)

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.

(3)

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:

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

Step 39:3 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:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

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 40: 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 $a\neg \equiv 4(mod6)$ and $b\neg \equiv 4(mod6)$

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

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 2.2a $2a=2b$ 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 3.2b $2a={\displaystyle \frac{b-1}{3}}$ 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, $a=-{\displaystyle \frac{1}{5}}$, contradicts our initial assumption that $a\in {\mathbb{N}}^{+}$. Let’s explain this contradiction more explicitly:

Explanation 1.The equation $a=-{\displaystyle \frac{1}{5}}$ contradicts $a\in {\mathbb{N}}^{+}$ for two reasons:

(1)

$-{\displaystyle \frac{1}{5}}$ is negative, but all elements in ${\mathbb{N}}^{+}$ are positive.

(2)

$-{\displaystyle \frac{1}{5}}$ is not an integer, but all elements in ${\mathbb{N}}^{+}$ are integers.

Therefore, there cannot be values $a,b\in {\mathbb{N}}^{+}$ that simultaneously satisfy $2a={\displaystyle \frac{b-1}{3}}$ and $2b={\displaystyle \frac{a-1}{3}}$.

Case 9.3 $(a\neg \equiv 4(mod6)\wedge b\equiv 4(mod6\left)\right)\vee (a\equiv 4(mod6)\wedge b\neg \equiv 4(mod6\left)\right)$

Without loss of generality, assume $a\neg \equiv 4(mod6)$ 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 41:2 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 42:3 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 43:4 In all cases, we have reached a contradiction. Therefore, our initial assumption must be false.

Step 44:5 We conclude that:

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

Thus, G is injective. □

Remark 14 (Transition to Multivalued Injectivity)The injectivity of G, as proved in this lemma, lays the foundation for the concept of multivalued injectivity. Here’s how we transition from injectivity to multivalued injectivity:

1. Injectivity (proved here): If $G\left(a\right)=G\left(b\right)$, then $a=b$.

2. Multivalued injectivity: If $a\ne b$, then $G\left(a\right)\cap G\left(b\right)=\varnothing$.

The connection between these concepts is as follows:

- If G is injective, then distinct inputs a and b must produce distinct outputs $G\left(a\right)$ and $G\left(b\right)$. - Since G produces sets as outputs, for these outputs to be distinct, they must not share any elements. - Therefore, if $a\ne b$, the sets $G\left(a\right)$ and $G\left(b\right)$ must be disjoint, i.e., $G\left(a\right)\cap G\left(b\right)=\varnothing$.

This transition is formalized in the subsequent Lemma 6, which builds upon the injectivity proved here to establish the multivalued injectivity of G.

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

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

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 45: 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 $a\neg \equiv 4(mod6)$ and $b\neg \equiv 4(mod6)$

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

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 4.2a $2a=2b\Rightarrow a=b$ This contradicts our assumption that $a\ne b$.

Subcase 5.2b $2a={\displaystyle \frac{b-1}{3}}$

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

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 6.2c $2b={\displaystyle \frac{a-1}{3}}$ This is symmetric to Subcase 2b and leads to the same contradiction.

Subcase 7.2d ${\displaystyle \frac{a-1}{3}}={\displaystyle \frac{b-1}{3}}\Rightarrow a=b$ This contradicts our assumption that $a\ne b$.

Case 12.3 $(a\neg \equiv 4(mod6)\wedge b\equiv 4(mod6\left)\right)\vee (a\equiv 4(mod6)\wedge b\neg \equiv 4(mod6\left)\right)$

Without loss of generality, assume $a\neg \equiv 4(mod6)$ 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 8.3a $2a=2b\Rightarrow a=b$ This contradicts our assumption that $a\ne b$.

Subcase 9.3b $2a={\displaystyle \frac{b-1}{3}}$

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

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 46:2 In all cases, we have reached a contradiction. Therefore, our initial assumption must be false.

Step 47:3 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:

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

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

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

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 48: 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 $G\left(B\right)=A$.

(i)

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

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

(ii)

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

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

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 49: 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:

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

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

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

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 50: 1 Let $n\in {\mathbb{N}}^{+}$ be arbitrary. We consider six cases:

Case 13.1 $n\equiv 0(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k\hfill \\ \hfill \mathrm{Let}m& =3k\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\neg \equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(3k\left)\right\}=\left\{6k\right\}=\left\{n\right\}\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Case 14.2 $n\equiv 1(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+1\hfill \\ \hfill \mathrm{Let}m& =2n=2(6k+1)=12k+2\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\neg \equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(12k+2\left)\right\}=\{24k+4\}\hfill \\ \hfill n& =6k+1={\displaystyle \frac{24k+4}{4}}\in G\left(m\right)\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Case 15.3 $n\equiv 2(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+2\hfill \\ \hfill \mathrm{Let}m& =3k+1\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\neg \equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(3k+1\left)\right\}=\{6k+2\}=\left\{n\right\}\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Case 16.4 $n\equiv 3(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+3\hfill \\ \hfill \mathrm{Let}m& =2n=2(6k+3)=12k+6\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\neg \equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(12k+6\left)\right\}=\{24k+12\}\hfill \\ \hfill n& =6k+3={\displaystyle \frac{24k+12}{4}}\in G\left(m\right)\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

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 $n\equiv 5(mod6)$

$$\begin{array}{cc}\hfill \exists k\in {\mathbb{N}}^{+}:n& =6k+5\hfill \\ \hfill \mathrm{Let}m& =2n=2(6k+5)=12k+10\hfill \\ \hfill \mathrm{Then}m& \in {\mathbb{N}}^{+}\mathrm{and}m\neg \equiv 4(mod6)\hfill \\ \hfill G\left(m\right)& =\left\{2m\right\}=\left\{2\right(12k+10\left)\right\}=\{24k+20\}\hfill \\ \hfill n& =6k+5={\displaystyle \frac{24k+20}{4}}\in G\left(m\right)\hfill \\ \hfill \therefore n& \in G\left(m\right)\hfill \end{array}$$

Step 51:2 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 52:3 Therefore, G is exhaustive. □

Theorem 15 (Finiteness of Preimages of G)Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

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

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 $n\neg \equiv 1(mod3)$ 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 6 (Multivalued Injectivity of G). □

Now we proceed with the induction proof:

Step 53: 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 54: 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 55: 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 56: 3a By Lemma 9, we know that $\left|G\right(x_i\left)\right|\le 2$ for all $i\in \{1,2,\dots ,m\}$.

Explanation 2.This follows directly from Lemma 9, which states that for any $n\in {\mathbb{N}}^{+}$, $\left|G\right(n\left)\right|\le 2$. Since each $x_i\in {\mathbb{N}}^{+}$, we can apply this lemma to each $G\left(x_i\right)$.

Step 57: 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 58: 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 59: 3d Thus, $G^{k+1}\left(\left\{1\right\}\right)$ is finite, as its cardinality is bounded by $2m$, which is finite.

Step 60: 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 16 (Non-emptiness of Preimages of G)Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

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

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 61: 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 62: 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 63: 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 $G\left(x\right)$:

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

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 64: 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 17 (Monotonicity of G)Let $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ be the inverse Collatz function defined as:

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

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 65: 1 Let $n\in {\mathbb{N}}^{+}$ be arbitrary.

Case 21.1 $n\neg \equiv 4(mod6)$

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 66:2 For $x=2n$:

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

Step 67:3 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 68:4 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 69:5 This inequality holds for all $k\in \mathbb{N}$, therefore:

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

Step 70:6 We have shown that in all cases, for any $x\in G\left(n\right)$, $x\le 2n$.

Step 71:7 Since n was arbitrary, we can conclude:

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

Step 72:8 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:

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

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

$$G\left(n\right)=\left\{\begin{array}{cc}\left\{2n\right\}\hfill & \mathrm{if}n\neg \equiv 4(mod6)\hfill \\ \{2n,{\displaystyle \frac{n-1}{3}}\}\hfill & \mathrm{if}n\equiv 4(mod6)\hfill \end{array}\right.$$

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

(1)

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

(2)

$n\in G\left(C\right(n\left)\right)$

Proof. We will prove each part separately.

Step 73: 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 $n\neg \equiv 4(mod6)$

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

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 74:2 Let’s prove that $n\in G\left(C\right(n\left)\right)$ for all $n\in {\mathbb{N}}^{+}$:

Case 25.1 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 26.2 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 10.2a If $3n+1\neg \equiv 4(mod6)$

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

Subcase 11.2b 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 75:3 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 18 (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:

(1)

Injectivity

(2)

Multivalued injectivity

(3)

Monotonicity

(4)

Exhaustiveness

(5)

Finiteness of preimages

(6)

Non-emptiness of preimages

where $G:{\mathbb{N}}^{+}\to \mathcal{P}\left({\mathbb{N}}^{+}\right)$ is the inverse Collatz function defined as in Theorem 15.

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 12.

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

(1)

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

(2)

$n\in G\left(C\right(n\left)\right)$

Step 76: 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:

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

Step 77: 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:

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

Step 78: 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:

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

Explanation 3 (Monotonicity Implication)The inequality $y\le 4^{i+j}x$ implies monotonicity for $G^i\circ G^j$ because:

(1)

It provides an upper bound for all elements y in $(G^i\circ G^j)\left(x\right)$ in terms of x.

(2)

This upper bound, $4^{i+j}x$, is a strictly increasing function of x (since $4^{i+j}>0$).

(3)

Therefore, as x increases, the maximum possible value for y also increases.

(4)

This ensures that for any $x_1<x_2$, all elements in $(G^i\circ G^j)\left(x_1\right)$ are less than or equal to all elements in $(G^i\circ G^j)\left(x_2\right)$, which is the definition of monotonicity for set-valued functions.

Thus, $y\le 4^{i+j}x$ guarantees that $G^i\circ G^j$ is monotonic.

Step 79: 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:

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

Step 80: 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 81: 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:

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

Step 82: 7 Therefore, all six properties are preserved under the composition $G^i\circ G^j$.

Remark 19 (Key Properties of G and Their Preservation)This theorem establishes that the crucial properties of G are preserved under composition. This is fundamental for our analysis, as it allows us to extend our reasoning about G to more complex structures built from G.

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 1 and 2 respectively. Then, for any property P of sequences in ${\mathbb{N}}^{+}$, the following are equivalent:

(1)

For all Collatz sequences ${\left(a_k\right)}_{k\in \mathbb{N}}$ generated by C, $P\left({\left(a_k\right)}_{k\in \mathbb{N}}\right)$ holds.

(2)

For all sequences ${\left(b_k\right)}_{k\in \mathbb{N}}$ such that $\forall k\in \mathbb{N},b_{k+1}\in G\left(b_k\right)$, $P\left({\left(b_k\right)}_{k\in \mathbb{N}}\right)$ holds.

Formally:

$$\forall P:\mathcal{S}\left({\mathbb{N}}^{+}\right)\to \{\mathrm{true},\mathrm{false}\},$$

$$\left(\forall {\left(a_k\right)}_{k\in \mathbb{N}}\in \mathcal{C},P\left({\left(a_k\right)}_{k\in \mathbb{N}}\right)\right)\iff \left(\forall {\left(b_k\right)}_{k\in \mathbb{N}}\in \mathcal{G},P\left({\left(b_k\right)}_{k\in \mathbb{N}}\right)\right)$$

where $\mathcal{S}\left({\mathbb{N}}^{+}\right)$ is the set of all sequences in ${\mathbb{N}}^{+}$, $\mathcal{C}$ is the set of all Collatz sequences, and $\mathcal{G}$ is the set of all sequences generated by G.

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

• Lemma 13: 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 83: 1 (⇒): Assume that for all Collatz sequences ${\left(a_k\right)}_{k\in \mathbb{N}}$ generated by C, $P\left({\left(a_k\right)}_{k\in \mathbb{N}}\right)$ holds.

Let ${\left(b_k\right)}_{k\in \mathbb{N}}$ be any sequence such that $\forall k\in \mathbb{N},b_{k+1}\in G\left(b_k\right)$. Define a sequence ${\left(a_k\right)}_{k\in \mathbb{N}}$ as follows:

$$a_0=b_0,\forall k\in \mathbb{N},a_{k+1}=C\left(a_k\right)$$

We claim that $\forall k\in \mathbb{N},b_k=a_k$. We prove this by induction:

Step 84: 2 Base case: $b_0=a_0$ by definition.

Step 85: 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 86: 4 Since ${\left(a_k\right)}_{k\in \mathbb{N}}$ is a Collatz sequence, $P\left({\left(a_k\right)}_{k\in \mathbb{N}}\right)$ holds by assumption. As $\forall k\in \mathbb{N},b_k=a_k$, we have $P\left({\left(b_k\right)}_{k\in \mathbb{N}}\right)$.

Step 87: 5 (⇐): Assume that for all sequences ${\left(b_k\right)}_{k\in \mathbb{N}}$ such that $\forall k\in \mathbb{N},b_{k+1}\in G\left(b_k\right)$, $P\left({\left(b_k\right)}_{k\in \mathbb{N}}\right)$ holds.

Let ${\left(a_k\right)}_{k\in \mathbb{N}}$ be any Collatz sequence generated by C. Then $\forall k\in \mathbb{N}$:

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

Therefore, ${\left(a_k\right)}_{k\in \mathbb{N}}$ satisfies the condition $\forall k\in \mathbb{N},a_k\in G\left(a_{k+1}\right)$. By assumption, $P\left({\left(a_k\right)}_{k\in \mathbb{N}}\right)$ holds.

Step 88: 6 Thus, we have shown both directions of the equivalence, completing the proof. □

Remark 20 (Bridging C and G)This lemma provides a critical link between sequences generated by C and those generated by G. It allows us to transfer results between these two perspectives, which is essential for our overall proof strategy.

Proposition 21.For any Collatz sequence ${\left(a_k\right)}_{k\ge 0}$:

(1)

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

(2)

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:

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

Then:

(1)

If $x>1$ is even, then $C\left(x\right)<x$.

(2)

If $x>1$ is odd, then $C\left(x\right)>x$.

(3)

$C\left(x\right)=1$ if and only if $x=1$ or $x=2$ or $x=4$.

(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. □

Lemma 18 (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 1 and 2 respectively. Then, for any property P of sequences in ${\mathbb{N}}^{+}$, the following are equivalent:

(1)

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.

(2)

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 13: 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 89: 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 90: 2 Base case: $b_0=a_0$ by definition.

Step 91: 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 92: 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 93: 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.

Figure 1. Boundnedness of Collatz Sequence

Figure 1. Boundnedness of Collatz Sequence

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 94: 6 Thus, we have shown both directions of the equivalence, completing the proof. □