# Is $L / R$ context free?

I was reviewing the post If $L$ is context-free and $R$ is regular, then $L / R$ is context-free?

I completely understand why $L/R$ is context free.

I just tried a different approach, which is not that rigorous according to the answers given. It is just based on the basic understanding.

How I tried :

Let $L = a^N b^N a^M b^M$ and $R = \left (a + b \right )^*$.

Then $wx$ belongs to $L$. Suppose I assume $wx = aaabbbab$.

So, now $x = ab$ which belongs to $R$.

So $L/R$ contains $aaabbb$.

Like this, I will get all strings in $L/R = \left \{ a^N b^N | N \geq 1 \right \}$.

This language is nothing but context free.

This approach is not as rigorous as what I saw in the answers.

I just want to confirm: is this approach right?

Your proof is erroneous since $L/R$ is larger than what you indicate. In fact, whenever $\epsilon \in R$, we always have $L/R \supseteq L$. In your case $$L/R = \{ a^{n_1} b^{n_2} : n_2 \leq n_1 \} \cup \{ a^n b^n a^{m_1} b^{m_2} : m_2 \leq m_1 \}.$$
• Sir, why did you use $<=$ ? Commented Nov 2, 2016 at 17:40
• Since otherwise what I wrote would be incorrect. For example, $ab \in L/R$. Commented Nov 2, 2016 at 17:42