[Solved]: Is an irregular language concatenated with a language with which it has no common symbols irregular?

Problem Detail: 

Here's an example of what I'm talking about. Suppose I have a languages
$$ L_{1} = \{a^ib^i \mid i>0\},\\ L_{2} = \{c^i \mid i>0\} $$

and $$ L_{1}L_{2} = \{a^ib^ic^i \mid i>0\} $$

Is it true that if $L_{1}$ is irregular and $L_{1}$ has no common symbols with $L_{2}$, then $L_{1}L_{2}$ is irregular? How would you prove this?

Asked By : ebracho

Answered By : Terence Hang

First, your $L1L2$ is wrong. $$L1L2 = \{a^ib^ic^j\ | \ i>0, j>0\}$$ Your conclusion is right, $L1L2$ is irregular(as long as $L2\neq\phi$, otherwise $L1L2=\phi$ is clearly regular). This can be proved using Myhill–Nerode theorem.

Suppose $X$,$Y$ are any 2 different equivalence classes of $R_{L1}$, $x\in X, y\in Y$, then there exists string $z$ such that $xz \in L1$ while $yz \notin L1$.

If $L2=\{\epsilon\}$, then $L1L2$=$L1$ is irregular. Otherwise, take any string $w\in L2, w \neq \epsilon$, then $xzw \in L1L2$, but $yzw\notin L1L2$ (consider that, due to no common symbol, no suffix of $yzw$ longer than $|w|$ in L2, and no prefix of $yzw$ shorter than $|yz|$ in $L1$).

This shows that $x$,$y$ belongs to different equivalence classes of $R_{L1L2}$, too. ie. there must be at least the same number of equivalence classes in $R_{L1L2}$ as in $R_{L1}$

From Myhill–Nerode theorem, $L1$ has infinite number of equivalence classes. Thus $L1L2$ also has infinite number of equivalence classes, which means $L1L2$ is irregular also.

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Question Source : http://cs.stackexchange.com/questions/47188