I am trying to understand the larger problem of the decidability of the equality of two DFAs. I understand that this problem can be solved using minimizing DFAs, but my textbook states this can be done using symmetric difference.

Theorem: Let $EQ$ be a language, where $EQ$ = $\{<A,B> \mid \text{A and B are both DFAs and L(A) = L(B)}\}$. In my notation, $L(X)$ stands for the language that DFA $X$ recognizes.

Prove that $EQ$ is decidable. This relies on the symmetric difference of 2 sets.

Proof: Let $Q$ be a decider defined as:

On input $<A,B>$ do:

  1. Construct a DFA $C$ such that $L(C) = (L(A) \cap \overline{L(B)}) \cup (\overline{L(A)} \cap L(B)) $
  2. Run $<C>$ on the algorithm used to determine if the language recognized by a DFA is $\emptyset$ (We proved this is decidable before, and I have no issues with that).
  3. If that algorithm accepts, then $Q$ accepts
  4. If that algorithm rejects, then $Q$ rejects

For $Q$ to be a decider, we must prove that $Q$ will halt. Since we have proven that the algorithm used in step 2 is decidable, it will halt. So I have no problems with steps 2-4.

But the first step, how can I prove that constructing $C$ in that way will halt? I understand that the symmetric difference will result in a regular language because of the closure properties, but how can a decider actually carry out this process and be guaranteed to halt?

So this is my primary question, what is the algorithm for determining the language that a DFA accepts? I.e. given DFA $X$, how can I determine $L(X)$ (Again, this must halt, so we can't try every string in existence or something like that).

Some follow up questions are: What if the DFA in question accepts an infinite set of strings, how can we possibly represent that on the decider's tape (or if the complement of $L(A)$ is infinite, etc.)? How can a decider determine the compliment of a set? And if $L(C)$ ends up being an infinite set of strings, how can we process that so that we can construct DFA $C$?

My book simply states

These constructions are algorithms that can be carried out by Turing machines.

which drives me nuts! Prove it.


1 Answer 1


Suppose that $A,B$ are two DFAs on a common alphabet $\Sigma$ with states $Q_A,Q_B$, initial states $q_{0A},q_{0B}$, final states $F_A,F_B$, and transition functions $\delta_A,\delta_B$. We define a product DFA as follows:

  • $Q = Q_A \times Q_B$.
  • $q_0 = \langle q_{0A}, q_{0B} \rangle$.
  • $F$ consists of all pairs $(q_A,q_B)$ such that either $q_A \in F_A$ and $q_B \notin F_B$ or $q_A \notin F_A$ and $q_B \in F_B$.
  • $\delta(\langle q_A,q_B \rangle, \sigma) = \langle \delta_A(q_A, \sigma), \delta_B(q_B, \sigma) \rangle$.

You can prove by induction that $\delta(q_0, w) = \langle \delta_A(q_{0A}, w), \delta_B(q_{0B}, w) \rangle$, and so our definition of $F$ guarantees that the language of this DFA is the symmetric difference of $L(A)$ and $L(B)$.

In order to know whether $L(A) = L(B)$, we have to check whether their symmetric difference is empty. This happens exactly when no accepting state is reachable from the initial state in the product DFA, a condition which we can check using BFS/DFS.

  • $\begingroup$ This is exactly what I was looking for. Thank you! I really wish my text book covered this, everything else has been so detailed. $\endgroup$
    – mover333
    Oct 31, 2018 at 7:07
  • 1
    $\begingroup$ @mover96 Your textbook surely gives this exact construction for union and/or intersection and all you have to do to get this one is to modify the accepting set in the appropriate way. Also, your textbook probably shows how to construct an automaton that decides $\overline{C}$ from one that decides $C$, and then you can use $A\setminus B = A\cap\overline{B}$. One of the key differences between study at university and at school is that you're expected to be able to figure out things that are very similar to things that you have been told, without having them spelled out explicitly. $\endgroup$ Oct 31, 2018 at 10:00
  • $\begingroup$ @David Richerby Yes while I do not believe my text book mentions cross product DFAs, it did show how to prove RLs are closed under union / intersection / complement using DFAs. I think my major confusing came from both the book's (and my professor's) focus on determining $L(A)$ without mentioning that in the end you don't even need to determine $L(C)$ you can simply construct a DFA that recognizes $L(C)$. $\endgroup$
    – mover333
    Oct 31, 2018 at 21:11

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