If $L$ is context-free and $R$ is regular, then $L / R$ is context-free?

I'm am stuck solving the next exercise:

Argue that if $L$ is context-free and $R$ is regular, then $L / R = \{ w \mid \exists x \in R \;\text{s.t}\; wx \in L\} $ (i.e. the right quotient) is context-free.

I know that there should exist a PDA that accepts $L$ and a DFA that accepts $R$. I'm now trying to combine these automata to a PDA that accepts the right quotient. If I can build that I proved that $L/R$ is context-free. But I'm stuck building this PDA.

This is how far I've made it:


In the combined PDA the states are a cartesian product of the states of the seperate automata. And the edges are the edges of the DFA but only the ones for which in the future a final state of the original PDA of L can be reached. But don't know how to write it down formally.

Here is a hint.

You need your machine to initially accept part of a word from $L$, consuming the tape as it goes. Then, without consuming anything, you need to find some word from $R$ that will push the machine into a final state. The chosen word from $R$ plays the role of the input word for the second half of the computation.

Clearly, non-determinism will have a role, as will the product between the two machines. The trick in formalising this is adjusting the product to deal with the fact that the input comes from $R$ not from the input.

StackExchange Computer Science Q#1886, answer score: 9