Here is an incorrect proof that E DFA$=\{$D:D is a DFA, and L$($D$)=\}$E DFA $=\{$D:D is a DFA, and L$($D$)=\}$ is decidable.

proof: Define a TM M$1$ as follows:On input DD:For all w$\{0,1\}$w$\{0,1\}$ : simulate DD on ww$.$ If DD accepts ww$,$ that means L$($D$)$L$($D$)=,$ so reject.If we find no strings that DD accepts, then L$($D$)=$L$($D$)=,$ so accept.

This Turin Machine M$1$ accepts inputs DD with L$($D$)=$L$($D$)=$ and rejects inputs DD with L$($D$)$L$($D$)=,$ so this machine decides E DFA E DFA

Question $1$

The Turing Machine M $1$ M $1$ does prove something about E DFA E DFA $.$ What does it prove?

Choice $1$ of $3$: E DFA E DFA $$ is recognizable

Choice $2$ of $3$: E DFA E DFA $$ is co$-$recognizable

Choice $3$ of $3$: E DFA E DFA $$ is not decidable

Question $2$

Is E DFA E DFA $$ decidable?

Choice $1$ of $2$: Yes

Choice $2$ of $2$: No