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Consider the following language $$EQ_{DFA} = \{ \langle A, B\rangle: A \ and \ B \ are \ DFAs \ and \ L(A) = L(B)\}$$

Given the fact that $EQ_{DFA}$ is decidable, how can I prove that the language $$SEQ_{DFA} = \{ \langle A, B\rangle: A \ and \ B \ are \ DFAs \ and \ L(A)\subseteq L(B)\}$$ is also decidable?

Thank you in advance

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  • $\begingroup$ Hint: $L(\mathcal{A}) \subseteq L(\mathcal{B})$ iff $L(\mathcal{A}) \cap L(\mathcal{B})^C = \varnothing$. $\endgroup$ – ttnick Oct 7 '18 at 10:48
  • $\begingroup$ What does Theorem 4.5 for EQDFA say? Can you write that in the question? $\endgroup$ – Bader Abu Radi Oct 7 '18 at 11:54
  • $\begingroup$ The Theorem 4.5 proves that EQDFA is decidable using the symmetric difference $\endgroup$ – ElDon90 Oct 7 '18 at 13:00
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There are several ways to prove that $SEQ_{DFA}$ is decidable. One of them is by using the fact that $EQ_{DFA}$ is decidable.

Assume that $L_1$ and $L_2$ are languages such that $L_1\leq_m L_2$ (that is, there is a mapping reduction from $L_1$ to $L_2$). It holds that if $L_2$ is decidable, then so is $L_1$. Hence, in order to prove that $SEQ_{DFA}$ is deciadable, we reduce it to $EQ_{DFA}$. As hinted in the comment, we propose the following reduction.

  • The Reduction: Let $C$ be a fixed DFA for the empty language, that is, $L(C) = \emptyset$. The reduction operates as follows. Given instance $\langle A= \langle Q_{A}, \Sigma, q^{A}_0, \delta_{A}, F_{A}\rangle, B= \langle Q_{B}, \Sigma, q^{B}_0, \delta_{B}, F_{B}\rangle\rangle$ of $SEQ_{DFA}$, the reduction computes a product automaton, $D$, for $L(A)\cap L(B)^C$ and outputs $\langle D, C\rangle$.

  • Correctness: $L(A)\subseteq L(B)$ iff $L(A)\cap L(B)^C = \emptyset$ iff $L(D) = \emptyset$ iff $L(D) = L(C)$. (the first equivalence mentioned in the comment is simple enough and thus is left for the reader).

  • Computability: the only non-trivial part is to compute $D$ given $A$ and $B$. We claim that $D$ is defined by $D = \langle Q_{A}\times Q_{B}, \Sigma, (q^{A}_0, q^{B}_0), \delta_{D}, F_{D}\rangle$. Where:

1) For every $(q, s)\in Q_{A}\times Q_{B}$ and $\sigma \in \Sigma$, $\delta_{D}$ is defined by: $$\delta_{D}((q, s),\sigma) = (\delta_{A}(q, \sigma), \delta_{B}(s, \sigma))$$

2) $F_{D} = F_{A}\times (Q_{B}\setminus F_{B})$.

Note that $D$ is a standard product automaton and thus for every finite word $w\in \Sigma^*$, we have that $\delta_{D}((q, s), w) = (\delta_{A}(q, w), \delta_{B}(s, w))$ (this can be proven by induction on $|w|$). Therefore, $w\in L(D)$ iff $\delta_{D}((q^{A}_0, q^{B}_0), w) \in F_{D}$ iff $(\delta_{A}(q^{A}_0, w),\delta_{B}(q^{B}_0, w) ) \in F_{A}\times (Q_{B}\setminus F_{B})$ iff $w\in L(A)$ and $w\in L(B)^C$ iff $w\in L(A)\cap L(B)^C$. Therefore $L(D) = L(A)\cap L(B)^C$ and thus we're done.

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    $\begingroup$ Alternatively, note that $A\subseteq B$ iff $A\cup B =B$. $\endgroup$ – Hendrik Jan Oct 7 '18 at 23:02
  • $\begingroup$ @HendrikJan Indeed, that would have been a cleaner solution :) $\endgroup$ – Bader Abu Radi Oct 8 '18 at 8:53
  • $\begingroup$ We construct a new DFA C from A and B, where C accepts only those strings that are accepted only by A or B but not only by B or both. Thus, if A is a subset of B, C will accept nothing. (using the emptiness test). $\endgroup$ – ElDon90 Jan 16 at 12:54

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