When using resolution, if the empty set is derived from a formula like , does that mean the formula is unsatisfiable? If this is the case, why is ,, satisfiable, and ,, is not satisfiable?
72.8k 30 30 gold badges 180 180 silver badges 392 392 bronze badges asked Nov 16, 2013 at 21:19 469 1 1 gold badge 5 5 silver badges 15 15 bronze badges$\begingroup$ What's your question exactly? Which formula are you trying to test whether it is satisfiable? Sorry, I can't understand your question. I encourage you to edit it to explain your question more clearly -- right now I find it very difficult to follow what you are asking. $\endgroup$
Commented Nov 17, 2013 at 1:40 $\begingroup$ I have now edited. $\endgroup$ Commented Nov 17, 2013 at 12:55$\begingroup$ $\, \$, $\$ is satisfiable because the assignment $x=\mathrm$, $y=\mathrm$ (or vice-versa) satisfies it. $\, \, \mathrm$ is unsatisfiable because the first clause requires $x$ to be true and the second requires it to be false. $\endgroup$
Commented Nov 17, 2013 at 18:46 $\begingroup$ <Ø>is not the empty set. $\endgroup$Ø> Commented Jan 23, 2014 at 9:03To complete the Kyle's answer:
A CNF (in Conjunctive normal form) formula, $F$, can be considered as a set of clauses, eg : $F=\$ and each clause can be regarded as a set of literals, eg : $C_1=\, C_2=\<\lnot x, \lnot y\>$.
If $F=\varnothing$ i.e. $F$ is the empty set and does not contain any clause then $F$ is satisfiable (=tautology).
If $F=\$ i.e. $F$ contains the empty clause then $F$ is not satisfiable. The empty clause can be obtained by the resolution rule eg : the resolvent of $C_3=\$ and $C_4=\<\lnot x\>$ is the empty clause $\varnothing=\$.
answered Nov 17, 2013 at 14:00 Xavier Labouze Xavier Labouze 521 2 2 silver badges 11 11 bronze badges$\begingroup$ Thanks, that cleared it up a bit more, but why is the first clause satsfiable, but not the second? $\endgroup$
Commented Nov 17, 2013 at 14:04$\begingroup$ I think you are confusing clause and formula : your first formula is satisfiable because you can't infer the empty clause by the resolution rule. $\endgroup$
Commented Nov 17, 2013 at 14:09 $\begingroup$ both the clauses i suggested are not satisfiable check $\endgroup$ Commented Nov 17, 2013 at 14:10 $\begingroup$You seem to be trying to describe and use the resolution proof system. Two points:
To address why the first example is satisfiable but the second is not, it may be helpful to informally think about what's going on geometrically.
The set of all truth-assignments may be represented as the n-cube, or Boolean poset $B_n$. Every literal corresponds to a distinct (n-1)-face; each positive literal corresponds to the filter of a distinct rank 1 element and each negative literal corresponds to the ideal of a distinct rank n-1 element.
If a k-clause encodes the set of all truth-assignments that satisfy it, its dual encodes the (n-k)-face of all truth-assignments that do not satisfy it. (Equivalently, if an input in CNF represents the intersection of unions of maximal ideals and filters of $B_n$, its dual represents the union of closed intervals.) Resolution on the dual input essentially glues unsatisfiable faces together.
Thus, if you can produce the empty clause via resolution, you have shown that the set of unsatisfying faces can be glued together to produce the n-cube itself. Why? Because a 0-clause must be unsatisfied by an (n-0)-face of assignments, which is all of them.
In the dual formulation of your first example, you have three 0-faces of the 2-cube. Obviously three 0-faces cannot cover all four 0-faces of the 2-cube, so the collection must be satisfiable.
In the dual formulation of your second example, you have two 3-faces and one 0-face of the 4-cube. Since we can glue the two 3-faces together to form the 4-cube, the collection must be unsatisfiable; the 4-face will absorb all other subfaces since it entails them, rendering them redundant.