Correct Option: E
For starters, the language $L^{’}$ isn’t really the complement of $L$, it is defined by $G^{’}$ defined by adding a new-rule $S\to SS$.
So, for this problem let’s consider $L = \{ a, b, c \}$ (RG is a CFG) so we’ve a simple derivation $S\to a/b/c$, for $L^{‘}$ use the added rule which generates $L^{‘}=\{a,b,c,aa,ab…,cc,aaa,aab….ccc...\}$.
- Concatenation of every string of $L$ with every string of $L$ itself will produce $\{aa,ab,…,cc\}$ which is only a subset of the language $L^{‘}$.
- B.If this was true, we can’t generate $aa$ for instance, the $L^{‘}$ doesn’t only comprise the original elements but newly-generated too.C.
- The original language needn’t necessarily contain $\epsilon$, which in-turn can’t be used generated a $\epsilon$ int the new-language. The above Language provides the counter-example for this statement. So, not true...D.
- All strings generated by the above grammar do belong in $L^{‘}$ but, there are other elements such as $\{ab,ac,bc\}$ which aren’t generated so it isn’t true.E.
- None of the above options are true so?