# Help proving a 3CNF related prob. is in P

I need help proving that this problem is decidable in polynomial-time:

Input: a 3CNF formula with more than one clause.

Question: can the formula be divided into two satisfiable 3CNF formulas ?

Example: given the formula:

(x1 or not(x2) or x3) and (x1 or x2 or x3) and (not(x1) or not(x2) or not(x3))

the answer is "yes", as we can divide into these two satisfiable 3CNF formulas:

1. (x1 or not(x2) or x3) and (x1 or x2 or x3)
2. (not(x1) or not(x2) or not(x3))
• Crossposted from SO. Jul 13 '12 at 14:21
• Is this a homework? Jul 13 '12 at 19:20
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Take a formula $\phi$ in 3-CNF over variables $V=\{v_{i}\}$ and clauses $C=\{c_{j}\}$, and let $A:V\rightarrow \{TRUE,FALSE\}$ be an arbitrarily chosen assignment.

The clauses of $\phi$ can be divided into two categories, those that are satisfied by $A$ and those that aren't. All those that are satisfied have at least one literal that evaluates to $TRUE$, however those that aren't have no such literals. Let $C_{T} = \{c^{T}_{j}\}$ be set of clauses that evaluate to $TRUE$ under $A$, and $C_{F} = \{c^{F}_{j}\}$ be the rest.

Assuming that neither $C_{F}$ nor $C_{T}$ are empty, then $\bigwedge_{j} c^{T}_{j}$ is satisfiable under $A$, and $\bigwedge_{j} c^{F}_{j}$ is satisfiable under $\bar{A}$, the inverse assignment of $A$.

In the trivial cases (thanks to Marzio! q.v. the comments) where either $C_{F} = \emptyset$ or $C_{T} = \emptyset$ any split is valid, with $A$ being a satisfying assignment for both subformula in the first case, and $\bar{A}$ being a satisfying assignment in the second.

These two formula are in 3-CNF, and partition the clauses of $\phi$.

• thank you for your answer, i would just like to clarify some points to better my understanding: Jul 13 '12 at 14:12
• (sorry pressed enter by mistake...) does this mean that we can always find such a partition to satisfiable formulas (and so the answer is always "yes") ? Jul 13 '12 at 14:23
• Yep, whatever the truth assignment, you can split the formula. Jul 13 '12 at 15:42
• @LukeMathieson: the only trivial cases that must be handled differently is when $A$ is a satisfying assignment for $\phi$ (i.e. all clauses are true) or $A$ doesn't make true any clause at all. In the first case every split is valid, in the second case every split is valid using $\bar{A}$. Jul 13 '12 at 16:16
• Hehe, the stupid thing is I thought of this, just failed to actually write it in... (I'll add it for smooth reading). Jul 15 '12 at 4:28