Question

How to prove that the satisfiability problem with each clause containing at most 3 literals, denoted by <= 3SAT, is NP-complete.

Answer 1

I've tried to prove it for several minutes but I can't make sure if it is equivalent to the max-3-SAT problem? This problem seems similar to the proof of SAT ∝ 3-SAT except for the case where there are more than 3 literals in each clause. so, we could prove that a clause with more than 3 literals could be broken into several 3-SAT.

for example,

x1
& x2 ∨ x3
& x4 ∨ x5 ∨ x6
& x7 ∨ x8 ∨ x9 ∨ x10 ∨ x11

This is a typical SAT problem. If we could break the last clause into several clauses and prove they are equal to the original, then we prove SAT ∝ ≤3SAT.

Fortunately, the way to break the last clause into 3-SAT is exactly the same as how it is done in SAT ∝ 3SAT.

That is, we add y1 and y2 and make the last clause become

x7 ∨ x8 ∨ y1
x9 ∨ -y1 ∨ y2
x10 ∨ x11 ∨ -y2

suppose they are denoted S' and the original clause is S. and, if S is satisfiable, then S' is satisfiable and if S' is satisfiable, then S is satisfiable

Since SAT, which is NP-complete, is polynomially reduced to ≤3SAT, ≤3SAT is also NP-complete.

However, The above proof is quite trivial if we already understand the proof of SAT ∝ 3-SAT.