# Proving that L is not regular using closure properties

I need to show that the following language is not regular.

$$L = \{\ ab^jc^j\ |\ j \geq 0\ \}\ \cup\ \{\ a^ib^jc^k\ |\ i, j, k \geq 0 \ and\ i \neq 1\ \}$$

There is also a hint that it cannot be proven just with using the Pumping Lemma. Closure properties must also be used.

I'm having a hard time with this one. Any help would be appreciated.

Let $$L_1 = \{\ ab^jc^j\ |\ j \geq 0\ \}$$ and $$L_2 = \{\ a^ib^jc^k\ |\ i, j, k \geq 0 \ and\ i \neq 1\ \}$$. Then

$$L = L_1 \cup L_2.$$

If we take \begin{align} L\ \cap\ ab^*c^* &= (L_1\cup L_2) \cap ab^*c^* \\ &= (L_1\cap ab^*c^*)\ \cup\ (L_2 \cap ab^*c^*)\\ \end{align}

Since $$L_1\cap ab^*c^* = L_1$$ and $$L_2 \cap ab^*c^* = \emptyset$$, we have

$$L\ \cap ab^*c^* = L_1$$

Now, $$ab^*c^*$$ is regular, and $$L_1$$ is not regular (can be shown using the Pumping Lemma). If $$L$$ was regular, by the closure properties, $$L_1$$ would be to, which is a contradiction. Hence, $$L$$ is not regular.

Here is a further hint. $$L$$ consists of two parts. One of them as a language by itself is not regular. Can you see which part?

Can you carve out that part as the intersection of $$L$$ with another regular language?

• I posted a solution. Got it right??? Apr 23, 2019 at 16:51
• Yes. $\quad$. Apr 23, 2019 at 17:07