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(2) Suppose conversely that we have $\tau \geq \eta_0$. Then that means that there is a specific ITTM program with index $e \in \mathbb{N}$ such that when we define $\beta=\sup\{H_e(r)|r\in \mathbb{R}\}$, we get $\beta \geq \eta_0$ and $\beta<\omega_1$. Now the point is that we want to build an OTM program (or similar) which simulates this ITTM program with index $e$ for arbitrary real numbers.

(3) So let's suppose that $\tau \geq \eta_0$ was false. That means $\tau<\eta_0$. Now take any arbitrary ordinal $x<\eta_0$. We can show that there exists a program (no parameters, input etc.) such that it has a non-decreasing variable $v$ with initially $v:=0$ and the "final" value of $v$ is $<\eta_0$ and $\geq x$ (after reaching this final value the value of variable $v$ never changes). Further this final value of $v$ will be achieved at some time strictly below $\eta_0$.

So again consider two real numbers $r_1$, $r_2$ which encode any linear-order on $\mathbb{N}$ such that the ordinal corresponding to their initial well-founded segment are of length $\alpha_1$, $\alpha_2$ respectively (also we intend $\alpha_1<\alpha_2$ under normal input). Now $e \in \mathbb{N}$ be the index of an OTM program (or similar) such that the final value of the variable $v$ (mentioned in previous paragraph) is both countable and $\geq \tau$. Let's call this final value $\beta$.

(2) Suppose conversely that we have $\tau > \eta_0$. Then that means that there is a specific ITTM program with index $e \in \mathbb{N}$ such that when we define $\beta=\sup\{H_e(r)|r\in \mathbb{R}\}$, we get $\beta \geq \eta_0$ and $\beta<\omega_1$. Now the point is that we want to build an OTM program (or similar) which simulates this ITTM program with index $e$ for arbitrary real numbers.

(3) We want to show that $\tau<\eta_0$ is impossible. Let's assume it was true.

Consider any arbitrary ordinal $x<\eta_0$. We can show that there exists a program (no parameters, input etc.) such that it has a non-decreasing variable $v$ with initially $v:=0$ and the "final" value of $v$ is $<\eta_0$ and $\geq x$ (after reaching this final value the value of variable $v$ never changes). Further this final value of $v$ will be achieved at some time strictly below $\eta_0$.

So again consider two real numbers $r_1$, $r_2$ which encode any linear-order on $\mathbb{N}$ such that the ordinal corresponding to their initial well-founded segment are of length $\alpha_1$, $\alpha_2$ respectively (also we intend $\alpha_1<\alpha_2$ under normal input). Now let $e \in \mathbb{N}$ be the index of an OTM program (or similar) such that the final value of the variable $v$ (mentioned in previous paragraph) is both countable and $\geq \tau$. Let's call this final value $\beta$ (we have $\tau \leq \beta < \eta_0$).

EDIT NOTE: After some thinking (prompted by @DmytroTaranovsky's comment and answer) I thought some more about this. I think I understand why $\tau <\eta_0$ is not possible.

EDIT NOTE: After some thinking (prompted by @DmytroTaranovsky's comment and answer) I thought some more about this. I think I understand why $\tau <\eta_0$ is not possible (note that I am assuming V=L).

So again consider two real numbers $r_1$, $r_2$ which encode any linear-order on $\mathbb{N}$ such that the ordinal corresponding to their initial well-founded segment are of length $\alpha_1$, $\alpha_2$ respectively (also we intend $\alpha_1<\alpha_2$ under normal input). Now $e \in \mathbb{N}$ be an OTM program (or similar) such that the final value of the variable $v$ (mentioned in previous paragraph) is both countable and $\geq \tau$. Let's call this final value $\beta$.

Once again, suppose we are given $r_1$ and $r_2$ encoded as single real number $r$. Now build an ITTM program which, loosely speaking, does the following (the whole simulation the follows can be carried using $r_2$). Find the (ordinal) length of the initial well-founded segment of the linear-order given by $r_1$ (essentially figure out $\alpha_1$). If it turns out the $\alpha_1 \geq \alpha_2$, then run forever. If not, then simulate the OTM program with index $e$ (till time/step $\alpha_2$) and keep a check on the variable $v$. If the value of $v$ does become $\geq \alpha_1$ halt (this is the only halting condition). And if the value of $v$ stays strictly below $\alpha_1$ till the exhaustion of initial well-founded segment of $r_2$, then go into a loop.

So again consider two real numbers $r_1$, $r_2$ which encode any linear-order on $\mathbb{N}$ such that the ordinal corresponding to their initial well-founded segment are of length $\alpha_1$, $\alpha_2$ respectively (also we intend $\alpha_1<\alpha_2$ under normal input). Now $e \in \mathbb{N}$ be the index of an OTM program (or similar) such that the final value of the variable $v$ (mentioned in previous paragraph) is both countable and $\geq \tau$. Let's call this final value $\beta$.

Once again, suppose we are given $r_1$ and $r_2$ encoded as single real number $r$. Now build an ITTM program which, loosely speaking, does the following (the whole simulation the follows can be carried using $r_2$). If it turns out the $\alpha_1 \geq \alpha_2$, then run forever. If $\alpha_1<\alpha_2$, then find the (ordinal) length of the initial well-founded segment of the linear-order given by $r_1$ (essentially figure out $\alpha_1$). Now simulate the OTM program with index $e$ (till time/step $\alpha_2$) and keep a check on the variable $v$. If the value of $v$ does become $\geq \alpha_1$ then halt in that case (this is the only halting condition). And if the value of $v$ stays strictly below $\alpha_1$ till the exhaustion of initial well-founded segment of $r_2$, then go into a loop.