Question

Show using a cross-product construction that the class of regular languages is closed under set difference. You do not need an inductive proof, but you should convincingly explain why your construction works.

Answer 1

Let be two regular languages with their corresponding DFAs being . Let the description of the DFAs be:

D1 = (Qι, Σ, qo, δι, Fi) D) = (Q2, Σ, 45, 52, F2)
.

To construct DFA for $L_1 \setminus L_2$ , do the following cross product construction. Let the DFA be D, which will have the following description:
$D = (Q, \Sigma, q_0, \delta, F)$ where
$Q = Q_1 \times Q_2$ i.e. the set of states is the set product of the two set of states.
i.e. the starting state is the tuple of starting states of the two DFAs.
$F = F_1 \times (Q_2 \setminus F_2)$ i.e. the set of final states is the tuple of states where the first state is final but the second is not.
$\delta((q_1, q_2), a) = (q_3,q_4) \iff \delta_1(q_1, a) = q_3 \wedge \delta_2(q_2,a) = q_4$ i.e. a transition happens in D if it happens in both the DFAs respectively.

D will accept the language $L_1 \setminus L_2$ . To argue why, note that a DFA takes exactly one path on reading a word and reaches exactly one state at the end. A word is in $L_1 \setminus L_2$ if and only if it is in $L_1$ but not in $L_2$ . Therefore, the first DFA will reach a final state and the second one will not. Hence in D, the word will reach a final state in the first state of the tuple and a non-final state in second state of the tuple. Therefore it will be accepted by D, as this is exactly what final states in D look like.

Similarly, a word is not in $L_1 \setminus L_2$ if it is either in both $L_1$ and $L_2$ , or if it's not in $L_1$ . In this case, the word will either reach a final state in both DFAs or reach a non-final state in $D_1$ . Therefore it will reach a non-final state in D, making it reject the word, as it should.

Therefore D accepts the correct language as desired.

Comment in case of any doubts.