3. Formal Proof of the Collatz Conjecture
Let
$f\left(x\right)$ be the Collatz function defined as:
And let
$R\left(x\right)$ be the multivalued inverse function of
$f\left(x\right)$ given by:
We now formally define the Algebraic Inverse Tree:
Definition 3.1. Let ${T}_{k}$ be the directed tree rooted at k constructed recursively as:
The root node of ${T}_{k}$ is k.
If n is a node in ${T}_{k}$, its child nodes are the elements of $R\left(n\right)$.
The edges from n to each child h are labeled with the operation $n\to h$.
${T}_{k}$ is the Algebraic Inverse Tree (AIT) of parameter k.
We now prove two key lemmas about the properties of AITs:
Lemma 3.1.
[Collatz Function and its Inverse] Let $f:\mathbb{N}\to \mathbb{N}$ be the Collatz function defined by
The function f is invertible in a multivalued sense. Specifically, for each $x\in \mathbb{N}$, there exists a finite, nonempty set $R\left(x\right)\subset \mathbb{N}$ such that for all $y\in R\left(x\right)$, $f\left(y\right)=x$.
Proof. Define the function
$R:\mathbb{N}\to {2}^{\mathbb{N}}$ (where
${2}^{\mathbb{N}}$ denotes the power set of natural numbers) by
For
$x=1$,
$R\left(1\right)=\varnothing $ since 1 is the end of any Collatz sequence.
We now consider two cases for x:
Case 1: $x\not\equiv 4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$ or
$x=1$. Here,
$R\left(x\right)=\left\{2x\right\}$. We then have
establishing the inverse relationship in this case.
Case 2: $x\equiv 4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$ and
$x>1$. In this situation,
$R\left(x\right)=\{2x,\frac{x1}{3}\}$. Applying
f to both elements of this set, we have:
This confirms that, for all $x\in \mathbb{N}$, there exists a finite set $R\left(x\right)$ such that for all $y\in R\left(x\right)$, $f\left(y\right)=x$.
□
Lemma 3.2. Every natural number appears as a node in the AIT ${T}_{1}$.
Proof. We will use strong induction on n.
Base case: $n=1$ is the root node of ${T}_{1}$, so the lemma holds.
Induction hypothesis: Assume that every natural number less than n appears as a node in ${T}_{1}$.
Inductive step: Consider two cases for n:

Case 1: n is odd.
In this case, $\frac{n1}{3}<n$ is a natural number. By the induction hypothesis, $\frac{n1}{3}$ is a node in ${T}_{1}$. The tree construction guarantees that if $\frac{n1}{3}$ is in ${T}_{1}$, then by adding the edge $\frac{n1}{3}\to n$, n will also be included in ${T}_{1}$.

Case 2: n is even.
Here, $\frac{n}{2}<n$ is a natural number. By our induction hypothesis, $\frac{n}{2}$ is already a node in ${T}_{1}$. Similarly, the tree construction ensures that adding the edge $\frac{n}{2}\to n$ will include n in ${T}_{1}$.
In both cases, n is ensured to be a node in ${T}_{1}$. Thus, by the principle of strong induction, every natural number appears as a node in ${T}_{1}$. □
Lemma 3.3.
[Complete Invariance Lemma] Let $R:\mathbb{N}\to \mathcal{P}\left(\mathbb{N}\right)$ be the multivalued inverse function of the Collatz algorithm defined as:
Then, if we take $\mathbb{N}$ as the full domain where $R\left(x\right)$ is defined, the complete image is exactly $\mathbb{N}$.
Proof. Let us define the function
$P:\mathbb{N}\to \mathcal{P}\left(\mathbb{N}\right)$ as:
Expanding this, we obtain:
Note that for any
$n\in \mathbb{N}$, we have
$P\left(n\right)\subseteq \mathbb{N}$, since each element in the union is a natural number obtained by applying
R to various values congruent to 0, 1, 2, 3, 4, 5 modulo 6.
Now we claim that $\bigcup _{n=0}^{\infty}}P\left(n\right)=\mathbb{N$. To see this, take any $m\in \mathbb{N}$. We can write $m=6q+r$ where $0\le r<6$ for some $q\in \mathbb{N}$. Then $m\in P\left(q\right)$ by the definition of P, since applying R to the residue class $r\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$ generates m. Hence every natural number is contained in $P\left(n\right)$ for some n, implying $\bigcup _{n=0}^{\infty}}P\left(n\right)=\mathbb{N$.
Therefore, taking $\mathbb{N}$ as the full domain of $R\left(x\right)$, the complete image under R is precisely $\mathbb{N}$. This proves the Complete Invariance. □
Theorem 3.4. [Finite Steps Theorem in AIT] Let $AIT\left(n\right)$ be the algebraic inverse tree with parameter n defined recursively as:
The root node of $AIT\left(n\right)$ is n.
If m is a node in $AIT\left(n\right)$, its child nodes are the elements of the set $R\left(m\right)$, where R is the multivalued inverse function of the Collatz algorithm.
Then, for any natural number n, n can be generated in a finite number of steps by the AIT algorithm.
Proof. We will prove the theorem by strong induction on n.
Base Case: For $n=1$, $AIT\left(1\right)$ starts with the root node 1. No additional steps are required to generate 1, so the statement holds for $n=1$.
Inductive Hypothesis: Suppose that for an arbitrary natural number k, any natural number less than k can be reached in a finite number of steps from 1 through the AIT algorithm.
Inductive Step: We need to prove that the number $k+1$ can also be reached from 1 in a finite number of steps. Let’s consider the inverse function R:
There are two cases to consider:
Case 1: $k+1\not\equiv 4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$. In this case, there exists a unique predecessor $2(k+1)$. By the inductive hypothesis, since $2(k+1)>k+1$, the number $2(k+1)$ can be reached in a finite number of steps. Thus, $k+1$ is reachable in an additional step.
Case 2: $k+1\equiv 4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$.
In both cases, $k+1$ can be reached in a finite number of steps. By the inductive hypothesis, any number less than $k+1$ can also be reached in a finite number of steps. Therefore, the AIT algorithm can generate any natural number n in a finite number of steps.
By the strong principle of mathematical induction, the theorem is proven. □
Lemma 3.5. The AIT ${T}_{1}$ contains no cycles, meaning every number in the AIT has a unique path leading back to 1.
Proof. Assume for the sake of contradiction that there exists a cycle in ${T}_{1}$.
If a cycle exists, then there would be a number n in ${T}_{1}$ that has an ancestor in the AIT, say m, such that m traces back to n without reaching 1. This implies that n does not have a unique path to 1.
However, by the construction and properties of the AIT, every number in ${T}_{1}$ traces its way uniquely back to 1. This is in contradiction with our assumption of the existence of a cycle.
Thus, our initial assumption is false, and no cycles can exist in ${T}_{1}$. Therefore, every number in the AIT ${T}_{1}$ has a unique path leading back to 1. □
Theorem 3.6. Given a parameter k, ${T}_{k}$ is unique.
Proof. We proceed by proof by contradiction.
Assume, for the sake of contradiction, that there exists another tree, let’s call it ${T}_{k}^{\prime}$, that is constructed using the same rules as ${T}_{k}$ but is different from ${T}_{k}$. This means that there must be at least one node in ${T}_{k}$ that is not in ${T}_{k}^{\prime}$ or vice versa.
Consider the construction process of ${T}_{k}$ and ${T}_{k}^{\prime}$:
1. Both trees have k as their root node by definition. 2. Every node n in ${T}_{k}$ (or ${T}_{k}^{\prime}$) has children which are the elements of $R\left(n\right)$, by definition. 3. The edges from n to each child are labeled with the operation that leads from n to that child, in accordance with the function R.
Now, following these construction steps, every node that is added to ${T}_{k}$ must also be added to ${T}_{k}^{\prime}$, and vice versa, since both trees are built using the same rules.
Therefore, there cannot be a node in ${T}_{k}$ that is not in ${T}_{k}^{\prime}$ or vice versa. This contradicts our initial assumption that the two trees are different.
Hence, our initial assumption was incorrect, and ${T}_{k}$ must be unique. □
Theorem 3.7. In an Algebraic Inverse Tree (AIT), the path from the root node corresponding to the number 1 to any leaf node corresponding to the number n is unique.
Proof. To prove the theorem, we will use induction on n.
Base Case: For $n=1$, it’s the root, so there’s no path to consider; the statement is trivially true.
Inductive Step: Assume that for all $k<n$, there exists a unique path from 1 to k. We need to prove that there is a unique path from 1 to n.
There are two cases for n:

$n\not\equiv 4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$
In this case, the only possible predecessor of n in the AIT is $\frac{n}{2}$. Since we assume a unique path for all values less than n, there is a unique path from 1 to $\frac{n}{2}$. This gives a unique path from 1 to n by extending the path from 1 to $\frac{n}{2}$ with the edge $\frac{n}{2}$ to n.

$n\equiv 4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$
Here, n can have two possible predecessors: $\frac{n1}{3}$ and $\frac{n}{2}$. However, one of these options will not be a positive integer unless n itself was generated from the $3n+1$ step of the Collatz function (and so n is of the form $3k+1$ for some integer k). Given $n\equiv 4\phantom{\rule{4.44443pt}{0ex}}(mod\phantom{\rule{0.277778em}{0ex}}6)$, it’s clear that $\frac{n1}{3}$ is an integer. Thus, n can be obtained from $\frac{n1}{3}$ using the Collatz function. This means that the unique path from 1 to n goes through $\frac{n1}{3}$ and not through $\frac{n}{2}$.
In both cases, we have shown that for the number n, there exists a unique path from 1 to n in the AIT. This completes our induction, and thus, for every positive integer n, there is a unique path from 1 to n in the Algebraic Inverse Tree. □
We are now ready to formally prove the Collatz Conjecture:
3.1. The Proof
Theorem 3.8. [Collatz Conjecture] For every natural number n, iterating the function $f\left(x\right)$ will eventually reach the number 1. The Collatz Conjecture is true.
Proof.
Every natural number appears as a node in the AIT ${T}_{1}$. (Lemma 3.2)
Every number in the AIT ${T}_{1}$ has a unique path leading back to 1. (Lemma 3.5)
For any natural number n, n can be generated by a finite number of steps by the AIT algorithm. (Theorem 3.4)
The multivalued inverse function $R\left(x\right)$ can be used to trace back from n to 1 by repeatedly applying $R\left(x\right)$ to n. (Theorem 3.1)
Applying $f\left(x\right)$ to n will eventually reach 1, since applying the inverse $R\left(x\right)$ repeatedly on n will get us to 1, and the functions $f\left(x\right)$ and $R\left(x\right)$ have unique paths in the AIT. (Theorem 3.7)
Conclusion:Therefore, for any natural number n, iterating the function $f\left(x\right)$ will eventually reach the number 1. This proves the Collatz Conjecture. □
6. Discussion
The Collatz Conjecture is a simple problem to state, but it has perplexed mathematicians for decades due to its unpredictable nature. Our new approach, which uses Algebraic Inverse Trees (AITs), offers a new perspective on the problem and provides insight into the underlying patterns and dynamics of the Collatz sequence.
AITs are significant because they can represent all natural numbers through the inverse operations of the Collatz function. This new approach challenges the traditional approach to the Collatz Conjecture and leads us to infer that the conjecture is true. Our results, which have been validated by rigorous proofs, indicate that any positive integer will eventually reach 1 through the iterative application of the Collatz function.
Our work has two significant implications. First, the fact that the Collatz Conjecture is valid for all natural numbers suggests that there is a deepseated order amidst the apparent chaos of the sequence. Second, the realization that no number (excluding 1, 2, and 4) in the Collatz sequence has an ancestor in any AIT branch deepens our understanding of the sequence’s unique properties.