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Let $L_1,L_2\in$ CFL $-$ REG, with $L_1\subset L_2$. Which of the following always holds?

  1. $L_1-L_2\in$ CFL $-$ REG and $L_1-L_2\in$ REG.
  2. $L_1-L_2\in$ REG and $L_2-L_1\in$ CFL $-$ REG.
  3. $L_1-L_2\in$ REG. $L_2-L_1\in$ REG.
  4. $L_1-L_2\in$ REG. As to $L_2-L_1$, it may be in REG or not.
  5. None of the above.
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  • $\begingroup$ What do you think? $\endgroup$ – Yuval Filmus Feb 11 at 15:24
  • $\begingroup$ First statement states that L1, L2 are context free languages which are not regular and L1 is a subset (a part) of L2. L1 - L2 will then be null. L2 - L1 will be those strings which are in L2 but not in L1. Answer according to me should be (b) $\endgroup$ – kiner_shah Feb 11 at 15:35
  • $\begingroup$ I thought it might be non of the above $\endgroup$ – Yair Ayalon Feb 11 at 17:37
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    $\begingroup$ @kiner_shah I understand what you're saying and it seems reasonable $\endgroup$ – Yair Ayalon Feb 11 at 17:38
  • $\begingroup$ Don't use images as main content of your post. This makes your question impossible to search and inaccessible to the visually impaired; we don't like that. Please transcribe text and mathematics (note that you can use LaTeX). Don't forget to give proper attribution to your sources! $\endgroup$ – D.W. Feb 11 at 20:54
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Since $L_1-L_2=\emptyset\in$ REG, we only need to consider $L_2-L_1$.

Consider $L_1=\{a^nb^n\mid n\text{ is a positive integer}\}, L_2=\{a^nb^n\mid n\text{ is a non-negative integer}\}$, then $L_2-L_1=\{\epsilon\}\in$ REG.

Consider $L_1=\{a^nb^n\mid n\text{ is a positive odd integer}\}, L_2=\{a^nb^n\mid n\text{ is a positive integer}\}$, then $L_2-L_1=\{a^nb^n\mid n\text{ is a positive even integer}\}\in$ CFL $-$ REG.

Hence, the correct answer is 4.

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