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!