how can i prove the following statement :1- $L_1\subseteq\Sigma^*$ is CFL and $L_2\subseteq\Sigma^*$ is regular Language, then $L_1$\ $L_2$ is CFL .so i want to know what is the method to prove it ?I'm not sure, but can i use a PDA and DFA or something else ?so i tried the following :$L_2$ is regular and thus is $\overline{\rm  L_2}$ also regular.$L_1 \cap \overline{\rm  L_2}$ is CFL $\iff$ $L_1 \cap \overline{\rm  L_2}$ is recognized by a PDA.$M1=(Z_1,\Sigma,\Gamma,\delta_1,z_{01},$#$,E_1)$ is a PDA for $L_1$$M_2=(Z_1,\Sigma,\delta_2,z_{02},E_2)$ is a DFA for $\overline{\rm  L_2}$$M^{'}$=($Z_1$x$Z_2$,$\Sigma,\Gamma,\delta^{''},(z_{01},z_{02}),$#$,E_1$x$E_2)$, so that $M^{'}$=$M_1\cap M_2$but how should i define $\delta{'' ?}$Many Thanks
CFL Intersection with Regular Language prove
formal languages;context free;closure properties