# Recent questions tagged regular

can we make DFA for a language where there is comparion between symbols but lanuage is finite a^nb^n;n<=3
Can the following grammar: L = $\{a^mb^n, n>=3, m<=4\}$ be written as the concatenation of 2 languages L = L1 . L2 where, L1 = $\{a^n, n>=3\}$ and L2 = $\{b^m, m<=4\}$
Suppose L1 is a regular language and L2 is a Context-free language. As all regular languages are context-free languages, L1 can be considered as Context-free language. L1 $\cap$ L2 would become <Context-free> $\cap$ <Context-free>. We know that Context free ... lot of places where regular is considered as Context-free and the problems were solved but for this case the result seems to be wrong.