# Contradiction in regularity of a language

Lets say we have $L_1$ which contains all binary numbers divisivle by 2 but not by 4. I would say this language contains all words with a 10 at the end. Ive found a regular grammar $G$ with $L(G) = L_1$ so that $L_1$ must be regular.

Now I have the problem that my teacher told me that $L_2 = \{ww| w \in \{0,1\}*$ and $w$ ends with 10$\}$ is not context free.

But if $L_1$ is regular or context-free, so should $L_1L_1=L_2$ be also regular and context free.

Where is the mistake?

• Let's say $G$ is a grammar for a regular language $L_1$. Can you use it to make a regular grammar for $L_1.L_1$? If you start with $S_0\rightarrow GG$ you may accept some words which do not belong to $L_1.L_1$. Neither regular Grammars not Context-Free Grammars are not able to determine $ww$ for $w\in\{0,1\}^*$. – Doralisa Aug 4 '18 at 12:37

There is a difference between $L_2 = \{ww\mid w \in L_1\}$ and $L_3 = L_1 L_1$. $L_3$ is $\{wx\mid w, x \in L_1\}$; in, other words all strings consisting of two sentences from $L_1$, which is regular / context-free if $L_1$ is regular / context-free.
$L_2$ is the set of all strings consisting of one repeated string from $L_1$, which is not in general context-free. (It is regular if $L_1$ is finite or is a regular language over a singleton alphabet. But your example does not fall into either of these categories.)