Let then

Let Using Theorem 1, we create the sequence Assuming that the process of binary decomposition according to formula (1) stops at the j-th step, it follows that the remaining terms of the decomposition are zeros, and we immediately reach the truth of the theorem statement. Therefore, we will consider the case when the generation of the decomposition according to formula (1) does not stop, and j reaches n. This means that all ${\sigma }_j>0,j<n$ A more detailed analysis of the number of zeros and ones in our binary representation is conducted. The following notations are introduced:

To solve the following equations we introduce the notations ${\lambda }_k$- the number of ones after the appearance of ${\alpha }_k>0$ and before the next appearance of a zero in the binary decomposition and Consider the set by definition ${\lambda }_1\ge 1$

Define: if the set of k satisfying the condition is not empty. A series of transformations are conducted for understanding the next steps. continuing the transformations we get Considering the definition of $m({\lambda }_{*},i)$, in the case of ${\lambda }_i>1$ Introduce the notations Note that ${\delta }_k,{\sigma }_k$ appear at the points with coordinates $x\left({\delta }_k\right),x\left({\sigma }_k\right),x\left({\delta }_k\right)=x\left({\sigma }_k\right)$ and by definition ${\alpha }_i,{\gamma }_i$ Thus all possible variants with L-zeros will be defined by all possible sets of With corresponding coordinates Let now , , , , , , considering all mentioned above, we get the following estimate From the last estimate, it follows that after a zero, no more than two ones can follow, after which at least one zero will appear, as these arguments can be sequentially applied to ${\sigma }_{i}1,{\sigma }_{i}2....$ where ${\alpha }_{i}1>0,{\alpha }_{i}2>0....$. Note that in our reasoning we used that the first one does not introduce any changes in the process of estimating ${\sigma }_j$ as ${\beta }_j=2^{{\delta }_j}$.

Let’s now move to more precise estimates Consider the following equalities Consider the variant $k=0$ then from which it follows Here we have ${\sigma }_{j+1}$ can approach zero only if ${\sigma }_j$ approaches 1. That is, we will get the number of ones balanced by a larger number of zeros !

Consider the variant $k=1$ then from which it follows that ${\delta }_j=2$ as for ${\delta }_j>2$ we get a contradiction.

Consider the variant $k=2$ then from which it follows that this is not possible. The statement of the theorem is true.  □

Let

then

Let Using Theorem 1, we create the sequence Suppose the binary decomposition process, according to formula (1), stops at the j-th step. Then, it immediately follows that the remaining terms of the decomposition are zeros, and we immediately achieve the truth of the Theorem. Therefore, we will consider the case when the generation of the decomposition according to formula (1) does not stop, and j reaches n. This means that all ${\sigma }_j>0,j<n$. Let’s conduct a more detailed analysis of the number of zeros and ones in our binary representation. Introduce the following designations: To solve the following equations let’s introduce designations ${\lambda }_k$ - the number of ones after the appearance of ${\alpha }_k>0$ and before the next appearance of zero in the binary decomposition, and Consider the set ${\lambda }_1,{\lambda }_2,...,{\lambda }_n$. By definition, ${\lambda }_1\ge 1$. Let’s define: if the set satisfying the condition is not empty. Let’s perform a series of transformations to understand the next steps. Continuing transformations, we get Taking into account the definition of $m({\lambda }_{*},i)$, in case of existence ${\lambda }_i>1$ Let’s introduce designations ${\alpha }_{*}={\inf }_{0\le i\le n}\frac{|{\alpha }_i|}{{\beta }_i},{\alpha }^{*}={sup}_{0\le i\le n}\frac{|{\alpha }_i|}{{\beta }_i}$, ${\beta }_{*}={\inf }_{0\le i\le n,{\delta }_i=1}{\beta }_i$, ${\beta }^{*}={sup}_{0\le i\le n,{\delta }_i=1}{\beta }_i$ Note that ${\delta }_k,{\sigma }_k$ occur at points with coordinates $x\left({\delta }_k\right),x\left({\sigma }_k\right),x\left({\delta }_k\right)=x\left({\sigma }_k\right)$ and by definition ${\alpha }_i,{\gamma }_i$ So all possible options with L-zeros will be determined by all possible options of sets With corresponding coordinates Now suppose that , Taking into account everything said above, we obtain the following estimate From the last estimate, it follows that after zero, only three ones can follow, after which there will be at least one zero, since these arguments can be sequentially applied to ${\sigma }_2,{\sigma }_3,...$ Let’s move on to more accurate estimates. Consider the following equalities Consider the case $k=0$ then from which it follows ${\sigma }_j\ge 1-\frac{2^{-{\delta }_j}-2^{-2{\delta }_j+1}}{ln2}$ Here we have ${\sigma }_{j+1}$ can approach zero only if ${\sigma }_j$ approaches 1. That is, we get the number of ones is balanced by a large number of zeros in! Consider the case $k=1$ then from which it follows that ${\delta }_j=2$ thus for ${\delta }_j>2$ we get a contradiction. Consider the case $k=2$ then from which it follows that this is impossible.

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Let then

Introduce operators defined as follows:

Consider all possible scenarios of Collatz sequence behavior, which can be written in the following form: We need to estimate each $2n$-th term of the Collatz sequence based on the number of applied operators $P,T,Z$ during n steps. Let $a_n$ have m ones in its binary representation, then we count the number of applications of operator Z using the following formula: and the number of applications of operator P using the following formula: Since each application of Z is accompanied by operator P, and the number of applications of operator P corresponds to the number of zeros in $a_n$, which equals $n-m$. According to the rules of Collatz, after n steps we have: According to the last formula, we see that the growth of each term of the sequence depends on the number of ones in the binary representation. Next, we will show that a large number of ones at the $2n$-th step leads to an increase in the number of zeros at the $3n$-th step for binary representation according to the previous theorems, from which it follows that subsequent terms of the sequence decrease: Repeating the reasoning of Theorem 2, consider the equation From the last equation, to apply the results of theorem 2, we need ${\sigma }_1>\frac{1}{2ln2}$. To satisfy the last inequality, consider $m_j=m-j,$$\theta =(a_n-2^n)·2^{-n}$, Consider $p=(m-j)\frac{ln3}{ln2}=(2k+l)1.5849625007\dots ,$$ϵ=1.5849625007\dots -1.5,$ we get Choosing l from even numbers less than 10, if inequalities $0\le \{\left(2k\right)·ϵ+\frac{ln(1+\theta )}{ln2}\}\le 0.5,$ are true Choosing l from odd numbers less than 10, if inequalities $0.5<\{2k·ϵ+\frac{ln(1+\theta )}{ln2}\}<1,$ are true Using $ϵ<0.1,$ also satisfy the condition ${\sigma }_1=1-\left\{x\right\}>\frac{1}{2ln2}.$ According to theorem 2 we get According to our application of Collatz rules, we have an element $a_{4n-j^{*}}$, and the order of its binary representation is After $3n-j^{*}$ steps of applying Collatz rules we have By definition of $m^{*},l^{*},B_n$ we get Using $n>1000$, it follows that $q_1<1\Rightarrow a_{4n-j^{*}}<a_n$.  □

Let then

Let’s introduce operators defined by the formulas Consider all possible scenarios of the behavior of the Collatz sequence, which can be written in the following form: It is necessary to calculate an estimate for each $2n$-th member of the Collatz sequence based on the number of $P,T,Z$ operators applied during n steps. Let $a_n$ have m units in its binary representation, then calculate the number of applications of the Z operator by the following formula: and calculate the number of applications of the P operator by the following formula: Since each application of Z is accompanied by the P operator, and the number of applications of the P operator corresponds to the number of zeros in $a_n$, which is equal to $n-m$. According to the rules of Collatz after n steps, we have: According to the last formula, we see that the growth of each member of the sequence depends on the number of units in the binary representation. Next, we will show that a large number of units on the $2n$-th step leads to an increase in the number of zeros in the $3n$-th step for the binary representation according to previous theorems, hence the reduction of subsequent members of the sequence: Repeating the reasoning of Theorem 2, consider the equation From the last equation, in order to apply the results of theorem 2, we need ${\sigma }_1=1-\left\{x\right\}>0.5$. To fulfill the last inequality, consider $m_j=m-j,$$\theta =(a_n-2^n)·2^{-n}$, Consider $p=(m-j)\frac{ln3}{ln2}=(2k+l)1.5849625007\dots ,$$ϵ=1.5849625007\dots -1.5,$ we get Choosing $l=0$, if the inequalities $0\le \{\left(2k\right)·ϵ+\frac{ln(1+\theta )}{ln2}\}\le 0.5$ are true, Choosing $l=1$, if the inequalities $0.5<\{2k·ϵ+\frac{ln(1+\theta )}{ln2}\}<1$ are true, Using $ϵ<0.1,$ we also satisfy the condition ${\sigma }_1=1-\left\{x\right\}>0.51.$ According to theorem 2 we get According to our application of the Collatz rules, we have the element $a_{4n-j^{*}}$, and the order of its binary representation is After $3n-j^{*}$ steps of applying the Collatz rules, we have By definition of $m^{*},l^{*},B_n$ we get Using $n>1000$, implies $q_1<1\Rightarrow a_{4n-j^{*}}<a_n$.  □

Let then for $a_n$ the Collatz conjecture is true.

The proof follows from Theorems 1-3.  □