My goal is to show, that a given language is not a regular one by using the Properties of Regular Languages.

The language is $ A \triangleq\left\{w \in \Sigma^{*} \mid |\left.w\right|_{b} \neq|w|_{c} \wedge|w|>0\right\} $ with $ \Sigma \triangleq\{a, b, c\} $

My idea is to use this clearly non regular language (which is given in the task, so I can/should use it) $ \left\{b^{n} c^{n} | n \in \mathbb{N}\right\} $ to prove $A$ is also no regular language.

Could the following work?

$(A\cap L(b^*c^*)) \cdot \mathrm{L}(\boldsymbol{\epsilon}) = \left\{b^{n} c^{n} | n \in \mathbb{N}\right\}$


1 Answer 1


It looks like you are onto the right track. However, your notation is confusing me.

I would believe the following is what you are coming at.

$$\overline{(A\cap L(b^*c^*))} \cup \{\epsilon\} = \left\{b^{n} c^{n} \mid n \in \mathbb{N}\right\}$$ where $\overline S$ is the language of words not in $S$, a.k.a. the complement of $S$. Here are the facts that you will use.

  • $b^*c^*$ is a regular expression.
  • the intersection of two regular languages is regular.
  • the complement of a regular language is regular.
  • $\{\epsilon\}$ is a regular language.
  • the union of two regular languages is regular.
  • $\{b^{n} c^{n} \mid n \in \mathbb{N}\}$ is not regular.

I will leave the rest of reasoning to you.

  • $\begingroup$ Hey, thanks for your answer. But there are two problems: First, I should not use the complement, because our proof did not mentioned it, so there is no proof for that. Second, this here $(A\cap L(b^*c^*)) $ should look like $L(b^*c^*) \ \{\epsilon\}$ right? So we just need to add the empty word, right? So is there a use for the complement in this case? I believe withe the complement the language is not right. But I'am just i my first year on university, so I'm maybe wrong. $\endgroup$
    – Marie.L
    Feb 20, 2019 at 11:37
  • $\begingroup$ I mean $L(b^*c^*)$ \ $\{\epsilon\}$ $\endgroup$
    – Marie.L
    Feb 20, 2019 at 11:50
  • $\begingroup$ $A$ is the language of words with unequal number of $b$'s and $c$'s. $L(b^*c^*)$ is the language of words that are zero or more $b$'s followed by zero or more $c$'s. The intersection of them is the words that are some number of $b$'s followed by a different number of $c$'s. $\endgroup$
    – John L.
    Feb 20, 2019 at 14:24
  • $\begingroup$ "I should not use the complement". If so, I cannot see the fact "$\{b^{n} c^{n} \mid n \in \mathbb{N}\}$ is not regular." can be useful. There are, of course, many other ways to prove $A$ is non-regular. $\endgroup$
    – John L.
    Feb 20, 2019 at 14:29
  • $\begingroup$ I'm sorry I can use the complement, I saw it 5 minutes ago. But I have a question, can I just use the complement of $A$ and to this : $\overline{(A))} \cap L(b^*c^*) = \{b^{n} c^{n} \mid n \in \mathbb{N}\}$ $\endgroup$
    – Marie.L
    Feb 20, 2019 at 14:41

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