Two questions about a parenthesis language Consider the alphabet consisting of the two parenthesis $($ and $)$. Let $L$ be the language that is inductively defined by the rules: 1\. The empty string is in the language. 2\. If $s$ and $t$ are two strings in the language, so is $(st)$. Now, let $L'$ be the language $L$ after the outermost parenthesis have been deleted. Is it true that for every string $s$ in $L'$, there are strings $t$ and $u$ in $L$ such that $s$=$tu$? Also, if the first question has an affirmative answer, is it true that if $s$ is a string that is in $L'$ but not in $L$, there is a unique pair $(t,u)$ of non-empty strings in $L$ such that $s$=$tu$?

The answer to the first question is yes, since $L$ is either the empty string or of the form $(st)$. For all $s \in L'$, there is $t \in L$, such that $t = (s)$. Hence $t$ can not be the empty string, $t = (uw)$ where $u,w \in L$, therefore $s = uw$.

For the second question, we have $(s) = (tu)$ for some $t, u \in L$, given that $s \
otin L$, both $t$ and $u$ are not empty, and both $t$ and $u$ contains outermost parenthesis. Now you just need to argue that for all $x \in L$, the number of left and right bracket must be the same, and show that for non-empty $t,u,t',u' \in L$, such that $tu = t'u'$, then $t=t', u=u'$.