Decidability of $E_{TM}$ and $A_{TM}$ for "erasing" Turing machines Why is the $A_{ETM}$ for a variant of a Turing machine (an erasing Turing machine), where changing a tape symbol to a nonblank symbol is prohibited, decidable? Why does the following diagonalization argument not work:


  Assume $A_{ETM}$ is decidable, and $R$ decides $A_{ETM}$. Then, $R$
  accepts $\langle M, w \rangle$ when ETM $M$ accepts $w$. Let $D$ be
  an ETM that accepts exactly when $M$ does not accept $w$. Then, $D$
  rejects $\langle D \rangle$ when $D$ accepts $\langle D \rangle$, a
  contradiction.


Why does this not work?

Alternatively, if $A_{ETM}$ is decidable, why is $E_{ETM}$  undecidable? The conventional contradiction argument would not work in that case, and I don't know how to reduce it any other way.

Thanks in advance!
An ETM (Erasing Turing machine) can be safely simulated due to its erasing-only nature.

It can only pass the right-end of the input to a finite limit or else would plunge off to infinity. Adding this to the input itself, we can conclude that the languages accepted by an ETM must be context-sensitive.

So, an ETM can only accept some context-sensitive language. As a result, $A_{ETM}$ is decidable.

Where is the source that says $E_{ETM}$ is undecidable?