Show that this language is decidable? Let A = {  | M is a DFA which doesn't accept any string containing an odd number of 1s}. Show that A is decidable.
The questions seems simple so I designed the following TM D that decides whether there exists a DFA with this property or not.
D = &quot;On input  w:

Construct a DFA B that accept any string containing an odd number of 1s.
Construct a DFA C s. t. $L(C) = L(B) \cap L(M)$
Call $E_{DFA}$ with input C.
If T accepts, then reject. If T rejects, then accept.&quot;

$E_{DFA}$ is defined as following:
$E_{DFA} = \{ &lt;A&gt; |$ A is a DFA and L(A) = $\phi \}$ So assume there is a TM T that decides language $E_{DFA}$. I.e., T accepts if A is empty, otherwise rejects.
Now, my questions is related to the step 4. In Sipser's textbook, he states that if T accepts, then accept and if T rejects, then reject. I don't understand why it says so. For example, If $L(M) \cap L(B) = \phi$, then this means that there are different. So, this is what I said in step 4 above. But in Sipser's answer for this exercise in page 187, it says that if they are empty, then accept. Can you explain to me where is my mistake?
If $E_{DFA}$ with input $C$ accepts, then $C = L(M) \cap B = \emptyset$.
In other words, no word containing an odd number of 1s is also in $L(M)$.
Then, by definition of $A$ you should accept.
Conversely, if  $E_{DFA}$ with input $C$ rejects, then $C = L(M) \cap B  \neq \emptyset$.
That is, there is some word $w \in L(M)$ such that $w$ contains an odd number of 1s. Then, by definition of $A$, you should reject.