Basic definitions

Harut-element is a fundamental unit of the model. You may think of it as a simple black dot.

Harut-couple is a pair of two Harut-elements.

Harut-system is a set that may consist of Harut-elements and Harut-couples.

Notation

The Harut-element is denoted by the letter h.

The Harut-couple is denoted by the letter H.

The Harut-system is denoted using curly braces.

Classification of Harut-systems

Harut-systems can be stable or unstable.

Stable Harut-system is a system that consists of only Harut-couples. It is denoted as (H, …).

Unstable Harut-system is a system in which there is one Harut-element. It is denoted as (H, …, h).

Fundamental principles

1. A Harut-element tends to form a pair with another Harut-element.
2. A stable Harut-system tends to split into two equal parts.

Examples:

(h, h) = (H)

(H, H) = (H) (H)

(H, H, H) = (H, H, h, h) = (H, h) (H, h)

Harut-system dimension

It is the number of Harut-couples in the system. It is denoted as (H, ...)_N, where N represents the dimension of the system.

Mutation

A mutation is any change in a Harut-system. By change, we mean that elements may be added to or removed from the system. The system may also be divided into equal parts, or parts may be repeated.

Mutations may have a stabilizing or destabilizing effect on the system.

Examples:

(H, …, h)_N + (h) = (H, ...)_N + (h, h) = (H, ...)_N + (H) = (H, ...)_N+1

3 x (H, …, h)_N = (H, …, h)_N + (H, …, h)_N + (H, …, h)_N = (H, ...)_3N + (h, h) + (h) = (H, ...)_3N + (H) + (h) = (H, …, h)_3N+1

3 x (H, …, h)_N + (h) = (H, …, h)_3N+1 + (h) = (H, ...)_3N+1 + (h, h) = (H, ...)_3N+2

Decomposability

Decomposability is the property of a stable Harut-system to be divided into an equal number of smaller systems without breaking its stability.

This property is denoted as ^m(H, ...)_N, where m indicates the number of equal parts into which the system can be decomposed while preserving stability.

Any decomposable stable Harut-system is denoted by the letter S (e.g. ²S – 2-decomposable stable Harut-system)

Harut-mutation

We will refer to a mutation of the form 3 x (²S, h) + h as a Harut-mutation.

Theorem: The Harut-mutation has a stabilizing effect on the Harut-system; however, if applied recursively to a system that exists according to its own fundamental life cycle rules, the system degrades. This means that, as a result of its division, all stable Harut-couples will ultimately exist outside the framework of the system, becoming self-contained.

Proof: We will describe the chain of system stabilization and division in accordance with the fundamental principles of Harut-system lifecycle.

²(H, …, h)_N

→ (stabilization) 3 x ²(H, …, h)_N + (h)

→ ²(H, …, h)_N + ²(H, …, h)_N + ²(H, …, h)_N + (h)

→ ²(H, …)_N + ²(H, …)_N + ²(H, …)_N + (h, h, h, h)

→ ²(H, …)_N + ²(H, …)_N + ²(H, …)_N + (H, H)

→ (division) ²(H, …)_N + ²(H, …)_N/2 + H

→ (H, …)_N+N/2+1

After stabilization and division, the Harut-system ²(H, …, h)_N mutates into (H, …)_N+N/2+1.

Since the Harut-system initially had the 2-decomposable property, we cannot guarantee that the system still possesses this property.

We are sure, though, that it is still stable.

Therefore, in accordance with the fundamental principles, the Harut-system must divide into two equal parts; however, after division, it may break down into two stable systems or two unstable ones.

(H, …)_N+N/2+1 → (H, …)_(N+N/2+1)/2

Or

(H, …)_N+N/2+1 → (H, …, h)_{⌊(N+N/2+1)/2⌋}

After all mutations, the dimension of the system will stably be ⌊(N+^N/₂+1)/2⌋ and, in the longest mutation scenario, will enter the life cycle of N → ⌊(N+^N/₂+1)/2⌋

We will prove that the recursive function converges.

N = ⌊(N+^N/₂+1)/2⌋

Proof using the deviation from the fixed point method. The stationary point is L = 2.

Let’s consider the difference D_i = N_i – 2.

N_i+1 – 2 = ⌊(3N_i + 2) / 4⌋ – 2 = ⌊(3N_i + 2 – 8) / 4⌋ = ⌊(3N_i – 6) / 4⌋ = ⌊3(N_i – 2) / 4⌋ = ⌊3D_i / 4⌋

D_i+1 = ⌊3D_i / 4⌋, D_i > 0 => D_i+1 < D_i Q.E.D.

Thus, the Harut-mutation leads to the degradation of the system, starting from ²(H, …, h)_N.

Now let us go backward from this system in the life cycle and see what leads to it.

Initially, the system is unstable, which means it was formed as a result of division.

²(H, …, h)_N = (H, …, h)_2n <= (H, …, )_4n (H) = (H, ...)_4n+1

It is impossible to form (H, ...)_4n+1 through stabilization, which means it was also formed as a result of division.

(H, ...)_4n+1 <= (H, ...)_8n+2

Thus, two divisions occurred in a row before the initial system from which the degradation begins.

The Collatz Conjecture

The Collatz Conjecture states that if you take any positive integer n and apply the following rules repeatedly, you will eventually reach the number 1.

The rules are:

If n is even, divide it by 2

If n is odd, multiply it by 3 and add 1

Let us draw an analogy between numbers and Harut-systems.

²(H, …, h)_N = (H, …, h)_2N

The (H, …, h)_2N system represents a number A, such that A mod 4 = 1.

The (H, …)_8n+2 system represents a number B, such that B mod 4 = 2.

The two divisions in a row indicate that the degradation actually begins with a number C, such that C mod 4 = 0.

Harut-mutation represents the 3x+1 function

Thus, for all numbers where the remainder modulo 4 is 0, 1 or 2, the Collatz Conjecture is verified.

Let us prove that for any number D such that D mod 4 = 3, the Collatz Conjecture also holds.

D mod 4 = 3 => D = 4k + 3

4k + 3 is an odd number => C(D) = 3(4k + 3) + 1 = 12k + 10

(12k + 10) mod 4 = (12k mod 4) + (10 mod 4) = 10 mod 4 = (8 + 2) mod 4 = 2 mod 4 = 2

Thus, any number D such that D mod 4 = 3 will lead to a number E such that E mod 4 = 2.

The hypothesis for such numbers E has already been proven.

Q.E.D.