I'm trying to find where does the language $\{\langle M,q,x\rangle$| $M$ is a Turingmachine and $q$ is a state of M and running of $M$ on $w$ passes on $q\}$ belong? whether it's $R,RE$ or none of the above.
At first, I thought it is decidable, since if it halts I can go through the states that it visited, and if it is not stopping I can tell from detecting a specific configuration twice that it's in loop, but it is correct for Linear bounded automate and not for a infinite stripe machine (Am I correct?).
Then I wanted to prove that it's not in $R$.I want to make a reduction from the acceptance problem: Can I do the following? given $(\langle M\rangle,x)$ I give the same $(\langle M\rangle,x)$ and a $q$ that would be the accepting state of $M$. if it doesn't have any, it would be a new isolated state, so I get that if M accepts x it visits $q$, otherwise it's not.
The recursion: $f(\langle M \rangle, x)= (\langle M' \rangle, x',q')$
given $(\langle M\rangle,x)$ I construct the new $M'$ as the following:
If $M$ accepts $x$, $M'$ would be a copy of $M$, $x'=x$ and $q'=q_a$ where $q_a$ is the accepting state from original $M$, if it's not I copy $M=M'$ , $x=x'$ and $q'$ would be a new isolated state where no running reaches. basically I ignore the input of the new $M'$ and use it only with given $x$.
Am I correct?