Collatz Conjecture: Am I understanding the Chain Equation from Simons & de Weger (2005) correctly? Is this inequality valid?

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I was thinking through the Chain Equation from this paper by Simons & de Weger on the Collatz Conjecture.

The Chain Equation is an equation which is only true if a nontrivial cycle (a cycle other than $1, 4, 2, 1, \dots, 1, 4, 2, 1$) exists.

I was attempting to restate the equation as a simpler inequality. I reached a conclusion that appears to me to be incorrect. I am unable to find my mistake.

Let:

  • $n$ be a natural number.

  • $T(n) = \begin{cases} \frac{1}{2}(3n + 1), && \text{if }n\text{ is odd}\\ \frac{1}{2}n, && \text{if }n\text{ is even}\\ \end{cases}$

  • sequence be an increasing subsequence of odd integers followed by a decreasing subsequence of even integers

  • a cycle be an $m$-cycle if it consists of $m$ sequences with a total of $K$ odd numbers and a total of $L$ even numbers.

  • a non-trivial cycle be any cycle that contains natural numbers greater than $2$.

  • a sequence is periodic if there exists an integer $p \ge 1$ in the sequence $\{ n, T(n), T^2(n), \dots, T^{p}(n) \}$ where:

    • $T^0(n) = n$
    • $T^{i+1}(n) = T(T^i(n))$
    • $T^p(n) = n$
  • $t_0, t_1, \dots, t_{m-1}$ be the indices of the $m$ local minima in an $m$-cycle such that:

    • $t_0 = 0$
    • $t_0 < t_1 < t_2 < \dots < t_{m-1} < p$
  • $s_0, s_1, \dots, s_{m-1}$ be the indices of the $m$ local maxima in an $m$-cycle such that:

    • $t_0 < s_0 < t_1 < s_1 < \dots < t_{m-1} < s_{m-1} \le p-1$
  • $x_i, y_i$ be the values of the local minima and maxima so that:

    • $x_i = T^{t_i}(n)$
    • $y_i = T^{s_i}(n)$
  • $k_i, l_i$ be defined so that:

    • $k_i = s_i - t_i$ for $i = 0, \dots, m-1$
    • $l_i = t_{i+1} - s_i$ for $i = 0, \dots, m-2$ and $l_{m-1} = p + t_0 - s_{m-1}$
    • $K = \sum\limits_{i=0}^{m-1}k_i$
    • $L = \sum\limits_{i=0}^{m-1}l_i$

Here's the statement of the Chain Equation (2.1) on page 56:

$$3^{k_i}a_i - 1 = 2^{k_{i+1} + l_i}a_{i+1} - 2^{l_i}$$

Here's my restatement of the Chain Equation as an inequality:

(1) $3^{k_i}a_i + 2^{l_i} = 2^{k_{i+1}+l_i}a_{i+1} + 1$

(2) $3^{k_i}a_i + 2^{l_i} > 2^{k_{i+1}+l_i}a_{i+1}$

(3) $\left(\dfrac{3^{k_i}}{2^{l_i}}\right)a_i + 1 > 2^{k_{i+1}}a_{i+1}$

(4) $\left(\dfrac{3^{k_i}}{2^{l_i}}\right)a_i \ge 2^{k_{i+1}}a_{i+1}$

Step (4) appears to me to be unreasonable.

Am I understanding the chain equation correctly? Did I make a mistake in one of my steps?