There is an equivalent definition for the class $NL$ with verifier. Those verifies are deterministic Turing machines that can read the witness tape only once in one way from left to right.

Given a function $f:\mathbb{N}\to\mathbb{N}$ we say that $NL[f(n)]$ is the class obtained by the above definition but the verifier can read the witness $f(n)$ times for an input of size $n$ (i.e. when the verifier finished reading the witness the goes straight to the beginning of it).

We can see of course that $NL=NL[1]$.

The question is whether $NL=NL[2]$.

Clarification: Prove or Disprove that $NL=NL[2]$.

It is clear that $NL\subseteq NL[2]$. For the second part I tried to construct a verifier that can read the witness only once for $L\in NL[2]$. I said that the verifier expects a witness of the form $w\sharp w$ and runs the $NL[2]$ verifier for $L$ with $w$ and then when it finishes and wants to read it again with second copy of $w$. But the major problem with my approach is that maybe someone tricked me and put a non equal sub-witnesses and I won't be able to find out about this with $\log(n)$ space thus it does not work.

There is an equivalent definition for the class $\mathsf{NL}$ with verifier. Those verifiers are deterministic Turing machines that can read the witness tape only once in one way from left to right.

Given a function $f:\mathbb{N}\to\mathbb{N}$ we say that $\mathsf{NL}[f(n)]$ is the class obtained by the above definition but the verifier can read the witness $f(n)$ times for an input of size $n$ (i.e. when the verifier finished reading the witness the goes straight to the beginning of it).

We can see of course that $\mathsf{NL}=\mathsf{NL}[1]$.

The question is whether $\mathsf{NL}=\mathsf{NL}[2]$.

Clarification: Prove or Disprove that $\mathsf{NL}=\mathsf{NL}[2]$.

It is clear that $\mathsf{NL}\subseteq \mathsf{NL}[2]$. For the second part I tried to construct a verifier that can read the witness only once for $L\in \mathsf{NL}[2]$. I said that the verifier expects a witness of the form $w\sharp w$ and runs the $\mathsf{NL}[2]$ verifier for $L$ with $w$ and then when it finishes and wants to read it again with second copy of $w$. But the major problem with my approach is that maybe someone tricked me and put a non equal sub-witnesses and I won't be able to find out about this with $\log(n)$ space thus it does not work.

There is an equivalent definition for the class $NL$ with verifier. Those verifies are deterministic Turing machines that can read the witness tape only once in one way from left to right.

Given a function $f:\mathbb{N}\to\mathbb{N}$ we say that $NL[f(n)]$ is the class obtained by the above definition but the verifier can read the witness $f(n)$ times for an input of size $n$ (i.e. when the verifier finished reading the witness the goes straight to the beginning of it).

We can see of course that $NL=NL[1]$.

The question is whether $NL=NL[2]$.

It is clear that $NL\subseteq NL[2]$. For the second part I tried to construct a verifier that can read the witness only once for $L\in NL[2]$. I said that the verifier expects a witness of the form $w\sharp w$ and runs the $NL[2]$ verifier for $L$ with $w$ and then when it finishes and wants to read it again with second copy of $w$. But the major problem with my approach is that maybe someone tricked me and put a non equal sub-witnesses and I won't be able to find out about this with $\log(n)$ space thus it does not work.

There is an equivalent definition for the class $NL$ with verifier. Those verifies are deterministic Turing machines that can read the witness tape only once in one way from left to right.

Given a function $f:\mathbb{N}\to\mathbb{N}$ we say that $NL[f(n)]$ is the class obtained by the above definition but the verifier can read the witness $f(n)$ times for an input of size $n$ (i.e. when the verifier finished reading the witness the goes straight to the beginning of it).

We can see of course that $NL=NL[1]$.

The question is whether $NL=NL[2]$.

Clarification: Prove or Disprove that $NL=NL[2]$.

It is clear that $NL\subseteq NL[2]$. For the second part I tried to construct a verifier that can read the witness only once for $L\in NL[2]$. I said that the verifier expects a witness of the form $w\sharp w$ and runs the $NL[2]$ verifier for $L$ with $w$ and then when it finishes and wants to read it again with second copy of $w$. But the major problem with my approach is that maybe someone tricked me and put a non equal sub-witnesses and I won't be able to find out about this with $\log(n)$ space thus it does not work.