I've tried to prove it for several days but I can't make sure if it is equivalent to max-3-SAT problem? This problem seems similar to the proof of SAT ∝ 3-SAT except 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.
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 last clause into 3-SAT is exactly the same with 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.
is there any general proof that could solve 3-SAT, ≤3SAT, n-SAT, ≤nSAT simultaneously?