Consider the language $L_1=\{a^pb^qc^r \mid p,q,r>0\}$ and $L_2=\{a^pb^qc^r \mid p,q,r\geq0 \space\text{and}\space p=r\}$, then which of the following statements are true?

  1. $L_1\cup L_2$ is a context free language
  2. $L_1\cap L_2$ is a context free language
  3. $L_1- L_2$ is not regular
  4. $L_1$ and $L_2$ both are regular languages

My attempt:

Since, language $L_1=\{a^pb^qc^r|p,q,r>0\}$ is regular and $L_2=\{a^pb^qc^r|p,q,r\geq0 \space\text{and}\space p=r\}$ is context free but not regular.

So, here

Statement $(1)$ $L_1\cup L_2=\{a^pb^qc^r|p,q,r \geq0\}$ is regular, so also context free language, statement is true.

Statement $(2)$ $L_1\cap L_2=\{a^pb^qc^r|p,q,r>0 \space\text{and}\space p=r\}$ is context free but not regular, statement is true.

Statement $(4)$ is obviously false as $L_2$ is not regular language.

I'm stuck at statement $(3)$ $L_1- L_2=L_1\cap \overline{L_2}$ it should be context free but not regular language?

Can you explain in formal way, please?


Let $L = L_1 - L_2 = \{a^pb^qc^r \mid p,q,r > 0, p \not=r\}$ and suppose that $L$ is regular. Then the language $a^+bc^+ - L = \{a^pbc^p \mid p > 0\}$ would also be regular, which is not the case. Thus $L$ is not regular.


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