So the question:Any regular language is context-free

So here is my proof by construction :

Let M=(Q,Σ,δ,q_0,F) be some DFA that accepts a regular language L. We construct a grammar G=(V,Σ,R,S) such that L(G)= L(M) as follows:

V={ $\displaystyle X_q $ | q ∈ Q }

S=$\displaystyle X_q0 $;where $\displaystyle q_0 $is the start symbol of M

R = $\displaystyle R_1$∪$\displaystyle R_2 $where

$\displaystyle R_1= X_q$→ ϵ iff δ(q,a) = f,where f∈F}

R={ $\displaystyle X_q$→a $\displaystyle X_r $iff δ(q,a)= r }

The above construction works because if the DFA goes from a state $\displaystyle q_i $to a state $\displaystyle q_j $for an input x∈Σ then the CFG does something similar. That is, the CFG generates x and the goes onto the next variable that is, it $\displaystyle X_i $generates $\displaystyle xX_j $thus effectively, going to the next variable. We also add a rule the handle accepting states, that is if $\displaystyle q_k $goes to the state $\displaystyle q_l $where $\displaystyle q_l$∈F, then our CFG just terminates outputting epsilon. So we see that our CFG generates the input read from the DFA M, thusL(G)= L(M).

Is the construction correct? Is the above explanation sufficient to show that the construction is correct? Please check it for technical loop holes. Thanks for helping me again. Seems like your the only one that knows their stuff