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DFArefers to deterministic finite automata. In DFA, there is only one path for a specific input from a current state to the next state.

Consider the language $C=\left\{w鈭�鈭�*|w鈥�鈥�\text{consists}鈥�鈥�111鈥�鈥�\text{as}鈥�鈥�\text{a}鈥�鈥�\text{substring}\right\}$. Consider that C is a regular language with the regular expression $\left(0鈭�1\right)*111\left(0鈭�1\right)*$ and is recognized by the following DFA, D_c.

Now, consider any regular expression with the alphabet $鈭�.$If $C鈭�L\left(R\right)=鈭�$, then R generates a string having 111 as a substring, so $\left(R\right)鈭�A$. Also, if $C鈭�L\left(R\right)=鈭�$, then R does not generate any string having 111 as a substring, thus . By the Kleen鈥檚 Theorem, since L(R) is described by the regular expression R, L(R) must be a regular language. Since C and L(R) are regular languages, $C鈭�L\left(R\right)$is regular since the class of regular languages is closed under intersection. Thus, $C鈭�L\left(R\right)$has some DFA $D_{C鈭�L\left(R\right)}.$The theorem 4.4 shows that is decidable, so there is a Turing machine H that decides the $E_{DFA}$. Apply TM H to $D_{C鈭�L\left(R\right)}$to determine if $C鈭�L\left(R\right)=鈭�$. Putting this all together gives us the following Turing machine T to decide A : T = 鈥淥n input

1. Convert Rinto a DFA using the algorithm in the proof of the Kleene鈥檚 Theorem.

2. Construct a DFA, $D_{C鈭�L\left(R\right)}$ for language $C鈭�L\left(R\right)$ from the DFAs D_c and D_R.

3. Run TM H that decides $E_{DFA}$on input .

4. If Haccepts, reject. If Hrejects, accept.鈥�

Therefore, the Turing machine H decides A . Hence, it has been shown that A is decidable.