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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 = "On input w:

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

$E_{DFA}$ is defined as following: $E_{DFA} = \{ <A> |$ 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?

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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.

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  • $\begingroup$ Thank you Steven! I think my mistake was not taken carful look at definition of A. I thought it accept any string with odd number of 1s. $\endgroup$ – user777 Mar 24 at 14:58

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