A reversible Turing machine ($\mathsf{RTM}$) is a Turing machine ($\mathsf{TM}$) whose transition function is a bijection, so that each instantaneous description ($\mathtt{ID}$) has at most one predecessor.
In [Ben73], Bennett showed that "Each conventional multitape Turing machine running in time $T$ and space $S$ can be simulated by a reversible input-saving machine running in time $O(T)$ and space $O(S + T)$."
In [Ben89], he improved his simulation as "For any $\epsilon>0$, any multitape Turing machine running in time $T$ and space $S$ can be simulated by a reversible input-saving machine using time $O(T^{1+\epsilon})$ and space $O(S\cdot\log{T})$."
To the best of my understanding, this is the proof-sketch of [Ben89] simulation.
For the parameters $m\approx S,n\ge 0,k\ge 2$, divide the entire computation $T=m\cdot k^n$ into $k^n$-segments of steps $m$. For $n=0$, simulate each segment of $m$-steps using [Ben73]. Then, for each $n>0$, there are exactly $k$-number of segments each of $m\cdot k^{n-1}$-steps. So, for each $n>0$, do $k$-number of forward simulation in each of which a checkpoint (the final $\mathtt{ID}$ of the simulation) is stored followed by $(k-1)$-number of backward simulation in each of which a checkpoint is deleted.
So, for all $n,k>1$, any $\mathsf{TM}$ (i.e., computation) with $k^n$-segments may be simulated by an $\mathsf{RTM}$ with $(2k-1)^n$-segments. The number of checkpoints is $\le n\cdot(k-1)=O(\log_k{T})$. As the size of each checkpoint is $O(S)$, the $\mathsf{RTM}$ requires $O(S\cdot\log{T})$-space and $O(S\cdot (2k-1)^n)=O(T^{1+(1/\log{k})})$-time.
My question is would it be possible to validate that the reversal has been done correctly? In particular, if the $\mathsf{TM}$ computes a function $f$ on an input $x$, then from these $\log{T}$- number of checkpoints would it be possible to validate that the $\mathsf{RTM}$ has computed $f(x)$ on the input $x$?
