I am studying for NP problems.
To prove k-CNF-SAT is NP-hard, there must exists something that can be reduced to k-CNF-SAT. So what I thought is to reduce 3-CNF-SAT to k-CNF-SAT and reduce k-CNF-SAT to 3-CNF-SAT both proves that it is NP-hard.
I know that 3-CNF-SAT is NP-Complete, because of its number of literals, but this property seems dedicate no effect to proof.
Thanks for any suggestion.
Note: k-CNF-SAT is a conjunctive normal form which k > 3.
And my question is same as the title, I have no idea about this question, since I just only know 3-SAT is a special case of k-SAT.