[Solved]: Formal construction of PDA intersecting a DFA

Problem Detail: 

Given the PDA $P = (Q_P,\Sigma,\Gamma_P,\delta_P,F_P)$ and the DFA $D = (Q_D, \Sigma, \delta_D,q_D,F_D)$

What is the 6-tuple definition of the PDA such that: $L(P') = L(P) \cap L(D)$

I don't understand what the intersection of these two FAs represent, and my only guess at what the 6-tuple is would be something along the lines of: $P'=(Q,\Sigma,\Gamma,\delta,q_0,F)$ such that:

$P=Q_P$ x $Q_D$
$\Sigma = \Sigma$
$\Gamma =..?$
$\delta =..?$
$F =..?$

The only other resource on this site I found that is remotely similar to this question is:
Cartesian construction of PDA and DFA

However, I didn't find the answer to be sufficient.

Asked By : milkywayz

Answered By : vonbrand

The idea of the intersection is to construct a PDA that keeps track of the original PDA's state in the first component of the state (in the $Q_P$ component) together with the stack, and keep track of the DFA's state in the $Q_D$ component. If both accept by final state, you just need to accept if both automata end up in a final state.

Best Answer from StackOverflow

Question Source : http://cs.stackexchange.com/questions/49091