# To prove 4-SAT CNF is NP-complete [closed]

I'm looking to prove that 4-SAT, (which will be momentarily defined) is NP-complete.

4-SAT: Given a formula in Conjunctive Normal Form, where each clause contains exactly 4 literals, does it have a satisfying truth assignment?

I was trying to prove that 4-SAT is in NP-complete. And was trying to take inspiration from the proof that CNF-SAT is in NP-complete, but I think there should be a direct reduction to prove 4-SAT is in NP-complete.

Could I get some direction on how to construct the explicit reduction?

• Why not reduce 3-SAT to 4-SAT? Jun 9, 2018 at 4:17
• so reducing 3-SAT to 4-SAT is to prove that 4-SAT is in NP-hard? Jun 9, 2018 at 4:27
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– D.W.
Jun 10, 2018 at 6:59
• I don't understand your question. The thing you describe as "the question below" is just a definition of the 4-SAT problem. You then ask about answering "that question" but you haven't told us what the question is. The only question I can imagine is "Is 4-SAT NP-complete?" but you can't be asking if the way to prove 4-SAT is NP-complete is to prove that 4-SAT is NP-complete because that's a tautology. Could you edit your question to clarify? Jun 10, 2018 at 11:43

In order to prove that 4-SAT is NP-complete, you need to prove that it is in NP and that it is NP-hard.

## Prove 4-SAT $\in$ NP

Given an instance of 4-SAT and an answer that evaluates to TRUE, it's pretty quick to verify.

## 4-SAT is NP-hard

Like the comments suggested, reduce 3-SAT, which is a known NP-complete problem, to 4-SAT:

Assume that you are given an instance of 3-SAT. The goal is to convert it in polynomial time into an instance of 4-SAT in such a way that the answer is preserved.

Take each clause $(x \lor y \lor z)$ and turn it into $(x \lor y \lor z \lor a) \land (x \lor y \lor z \lor \neg a)$, where $a$ is arbitrarily set.

• If any $(x \lor y \lor z)$ clause is satisfied, then $(x \lor y \lor z \lor a) \land (x \lor y \lor z \lor \neg a)$ will also be satisfied. If the instance of 3-SAT is satisfiable, then the new instance of 4-SAT is satisfiable.

• If $(x \lor y \lor z \lor a) \land (x \lor y \lor z \lor \neg a)$ is satisfied, then $(x \lor y \lor z)$ must also be true because $a$ is opposite of $\neg a$. If the new instance of 4-SAT is satisfiable, then the instance of 3-SAT is satisfiable.