How can i prove that the conversion from CNF to DNF is NP-Hard. I'm not asking for an answer, just some suggestions about how to go about proving it.
After converting a CNF formula to DNF I can solve the NP-complete problem SAT in linear-time by checking each DNF clause until I find one that does not contain a literal and its negation. This means that SAT is polynomial-time Turing reducible to CNF-to-DNF conversion. The existence of such a reduction is the definition of NP-hardness.