I've been asked to solve this problem, but I'm completely stuck now.
Is the set $\{G \in\text{CFG} \mid L(G)\supseteq L(A) \}$ where A is DFA fixed beforehand decidable?
I know I've to find a reduction, and as a hint they told me that it is related to Acceptance Problem or Post Correspondence Problem, or probably with Non-empty Intersection problem ( $\{\langle G1, G2\rangle \in \text{CFG} \times \text{CFG} \mid L(G1) \cap L(G2) \not= \emptyset\}$), which are undecidable.
I've been reading about these problems for hours, and reading about reductions from these problems to others trying to find any idea, but I'm stuck.
I'd really appreciate any help, whether some hints or the answer.