Resolution proof system Presenter Valeriy Balabanov NTU, GIEE, AlCom lab

Outline Basic definitions Key-facts about resolution proofs Intractability of resolution Heuristics for proof minimization Resolution in QBF Conclusion and future work References

Basic definitions

Resolution is a deductive rule in a form: where a, b, c are some distinct logical facts “a” is called pivot (b or c) is called resolvent A Resolution refutation proof for F is a sequence of clauses R = (C1, ..,Ct), where Ct = ∅; Ci ∈ F or Ci is derived from two previous clauses by the resolution rule

The length of the proof = # of clauses in the derivation Resolution proof can also be seen as DAG, where the nodes represent clauses, and edges represent resolution steps; the single sink node is an empty clause Tree-like resolution is a resolution, with special property – each parent node has exactly one child (in other words each clause in a proof is resolved only once) Note: Tree-like resolution can be derived from DAG resolution by splitting multiply used nodes into separate nodes

Key-facts about resolution proofs For 2SAT it is possible to find the shortest resolution proof in polynomial time (2SAT∈ P) For HornSAT polynomial resolution proof exists (HornSAT ∈ P-complete), but finding the shortest proof is NP-hard Generally, finding the shortest resolution proof is NP-hard (generally, as we will see the shortest proof can be exponential in number of clauses)

Intractability of resolution Resolution is complete and sound Proof: Soundness: every clause, resolved from the formula is implied by that formula, thus, if resolved clause is empty – formula is UNSAT Completeness: elimination of variable “a” from CNF, is a procedure, when we make all possible resolutions using “a” as a pivot, and then eliminating all the clauses containing “a” from the original formula

Completeness(continued): Let F be UNSAT CNF with m-variables a1,a2…am Let Si be the set of clauses, which are left after elimination of i variables from F; S0 is the original formula F; Sm has at most the empty clause. Let’s prove by induction on i, that every truth assignment to variables in F will make some clause in Si to be false For i=0 S0 is UNSAT, and thus has false clause for every assignment Assume for Sk it is also true, and for some assignment V, the false clause is θ, then if θ doesn’t contain variable ak+1, then θ also will be present after elimination of ak+1;

Completeness(continued): now, if θ has variable ak+1, let W be the truth assignment, same as V, but with different assignment to variable ak+1; let β be the clause which is false for W; if β doesn’t contain variable ak+1, then β will be in Sk+1; if it does – then the resolvent of β and θ will be present in Sk+1 and obviously will be false for V(also W); thus for every truth assignment, Si must contain a clause which will be false under it Thus, Sm should contain the empty clause, and by the construction of Sm it was derived by resolution

Pigeonhole principle: Let A be a sequence of n=sr + 1 distinct numbers. Then either A has: an increasing subsequence of s + 1 terms or a decreasing subsequence of r + 1 terms (or both). Consequence: Suppose we have n=s+1 pigeons (r=1) If we put them in at most s holes, then there definitely will be at least 2 pigeons in the same hole In other words it is impossible to put every pigeon to it’s own hole

Proof: xi is the longest increasing subsequence ending at ai Every number in sequence ai has score (xi, yi). xi is the longest increasing subsequence ending at ai yi is the longest decreasing subsequence starting at ai (xi, yi) ≠ (xj, yj) whenever i ≠ j. Assume i < j, then: if ai < aj → xi < xj if ai > aj → yi > yj Thus we have rs+1 points on a plane, and there is ai with coordinate (xi, yi) outside the rs-square. So, for that ai we will have xi ≥ s+1 or yi ≥ r+1

Formalizing PHP to CNF formula xi,j - pigeon i sits in hole j (type 1): xi,1 ∨ xi,2 ∨ .. ∨ xi,n−1 for i = 1..n (every pigeon sits in at least one hole) (type 2): (¬xi,k ∨ ¬xj,k) for 1 ≤ i ≠ j ≤ n ; 1 ≤ k ≤ n − 1 (no two pigeons sit in the same hole) From pigeonhole principle conjunction of above clauses is UNSAT Example: Note: deleting any clause will lead to SAT

Haken’s super-polynomial lower bound Original proof shows the bound for n>200 We present modified proof: Ω(2√n/32) Definition: A critical assignment is a one-to-one mapping of n − 1 pigeons to n − 1 holes, with one pigeon unset. Having i-th pigeon unset defines a i-critical assignment. Presenting the assignments of the xi,j as a matrix, the critical assignments would look like this: Example of 9-critical assignment for PHP with n=9

Let R be the proof of unsatisfiability of PHPn Replace xi,j’ in all clauses C by: Definition: The resulting sequence of positive clauses R+ = (C1+ , ..,Ct+ ) is a positive pseudo-proof of PHPn Lemma: C+(α) = C(α) for any critical α Proof: Suppose ∃C+(α) ≠ C(α) ⇒ ∃xi,j’ ∈ C s.t. Ci,j(α) ≠ xi,j’(α) ⇔(x1,j ∨ .. ∨ xn,j)(α) ≠ xi,j’(α). This is impossible, since α is critical, therefore has exactly one 1 in the column j.

We will show now, that t ≥ 2n/32. For a contradiction, assume t < 2n/32, t is the number of clauses in R+. Definition: A long clause has at least n2/8 variables. (more than 1/8 of all possible n(n − 1) variables). l is the number of long clauses in R. l ≤ t < 2n/32 By the pigeonhole principle, there exists a variable xi,j, which occurs in at least l/8 of the long clauses. Set the special variable xi,j to 1. Set all xi,j’, xi’,j for j’≠j, i’≠i to 0. Clauses containing xi,j are set to 1 and therefore disappear from the proof. The variables set to 0 disappear from all clauses.

We are left with a pseudo-proof of PHPn−1 with at most l(1 − 1/8) long clauses. Doing this d = 8log(l) times, we will eliminate all long clauses, since We are left now with a pseudo-proof of PHPm with no long clauses (of length more than n2/8). Since m = n – d, and from assumption l < 2n/32, we can obtain

Proof: let R’ be a positive pseudo-proof of PHPm Lemma: Any positive pseudo-proof of PHPm must have a clause with at least 2m2/9 variables. Proof: let R’ be a positive pseudo-proof of PHPm Definition: ∀C∈R’, W is a witness of C if W is a set of clauses from PHPm, whose conjunction implies C for critical assignments. (∀ critical α: α satisfies all ω∈W → α satisfies C). The weight of C = # clauses in minimal witness. Note: for any C there exist witness W Clauses of (type 2) are not the part of a minimal witness Clauses of (type 1) have weight 1 The weight of the final clause is m The weight of a clause is at most the sum of the two clauses its been derived from There exists a clause C∈R’ of weight s, m/3 ≤ s ≤ 2m/3.

Let Also let S is a set of indices of witness clauses for C W = {Ci|i ∈ S}, |S| = s, Ci = xi,1 ∨ xi,2.. ∨ xi,m−1; Ci ∈ PHPm ∧Ci → C Also let i ∈ S α is i-critical assignment with C(α)=0 j ∉ S; α’ is j-critical α’ is obtained from α, by swapping rowi and rowj: If α maps pigeonj to holek, then α’ maps pigeoni to holek

Since j ∉ S α’ satisfies all Ci ∈ W, so C(α’)=1 From the construction α differs from α’ only in xi,k, xj,k This implies xi,k ∈ C We can run this argument for current i-critical assignment under all (m − s) different choices for j ∉ S Thus C contains the variables xi,k1, xi,k2, .., xi,km−s And by repeating this for all i ∈ S, we conclude that C contains at least (m-s)s different variables Since m/3 ≤ s ≤ 2m/3, we have (m-s)s ≥ 2m2/9, concluding the proof for lemma We reached a contradiction to our assumption that t ≤ 2n/32

Extended resolution can find polynomial proofs for pigeonhole formulas Thus we conclude, that pigeonhole family of clauses requires super-polynomial minimal proofs for large n People have also found many exponentially hard examples for resolution using graph theory Definition: extended resolution, is a regular resolution, but with additional property: any definition can be added to original formula, if it doesn’t change its satisfiability Example: if x is not in original formula, we can add Extended resolution can find polynomial proofs for pigeonhole formulas Extended resolution is one of the strongest known proof systems

Heuristics for proof minimization Resolution proofs are useful for Extracting unsatisfiable cores Extracting interpolants Detecting useful clauses for incremental SAT-solving Run-till-fix and Trim-till-fix Use SAT-solver repeatedly to minimize UNSAT-core Use incremental SAT-solver to analyze the structure of the proof and restructure it Running time is usually large, since we need to rerun SAT-solver again and again

Recycling learned unit clauses If (x) is a unit clause that was learned by the SAT solver, it can be used for simplifying resolution inferences that used x as the pivot prior to learning this clause May lead to circular reasoning, so must be applied carefully Let P – is a resolution proof of the empty clause For a given node n in P: n.C - is the clause represented by n n.L and n.R are parents of n n.piv – is the variable used to resolve n.C from n.L.C and n.R.C

Example:

Example: It is easy to see, that recycling units will only make proof stronger The size of the proof also will be reduced The time complexity is quadratic in size of the proof, and no SAT-solving is used

Runtime of Recycling Pivots is linear in proof size Observation: along each path from root to sink in a proof graph there is no need for resolving on the same variable more than once Proof: Key point here is: why do we want to use resolution? We use current resolution step to eliminate variable “x” If in few steps variable “x” will reappear again – then what was the purpose of first resolution? The proof with above mentioned property is called Regular The shortest proof for a given problem must be regular The Reconstruct-Proof algorithm will be the same as that for Recycling Units Runtime of Recycling Pivots is linear in proof size

Example:

Experimental results Run-till-fix finds the smallest UNSAT core (# of roots), but it increases the proof-size Recycle Units and Pivots significantly simplify the proof, but cannot make UNSAT core small enough

Resolution in QBF SAT vs. QBF Universal reduction Q-resolution Example but Q-resolution combines resolution and universal reduction

Example: Red lines: universal reduction Green lines: exist. resolution

Q-resolution is both complete and sound Soundness: if the empty clause was generated, as in SAT, QBF obviously evaluates to 0 Completeness: Induction on number of quantifiers: For single ∃-variable it is just a usual resolution For single ∀-variable, falsity of formula->there is at least one non-tautological clause, which can be universally reduced Induction step for ∀-variable (a) will choose the value of a, which leads to UNSAT, and use the same resolution steps; For ∃-variable (a) both assignments to a lead to a conflict; we use Q-res steps for those assignments; if in one of them a (a’) was not present – we are done; if both present – we resolve resulting clauses on a, and thus get the conflict clause

As QBF is a general case of SAT, Q-resolution is also intractable More definitions: ∃-unit clause is clause with only one ∃-variable Q-unit resolution is a Q-resolution where one of the clauses is a positive ∃-unit clause Horn clause is a clause with only one positive literal Extended quantified Horn formula has every clause’s existential part to be a Horn clause

Q-resolution can’t simulate usual resolution can’t conclude x Theorem: Q-unit resolution is complete and sound for extended quantified Horn formulas Proof: look into [7] Theorem: For every t>0 there exists a quantified extended Horn formula of length 18t+1 which is FALSE, and the refutation to the empty clause requires at leas 2t Q-resolution steps Q-resolution can’t simulate usual resolution Example can’t conclude x

Conclusion and future work Resolution is simplest, but yet efficient proof system Resolution is intractable Existence of exponential lower bounds Resolution proofs are used in model checking Shorter proofs can be produced using some heuristics Q-resolution is an extension of resolution in QBF

Other proof systems Exchange of the nodes in the resolution graph Different heuristics for proof-length reduction Interpolants in first-order logic Q-resolution vs. QBF’s certificates

References “The relative efficiency for propositional proofs”, Stephen A. Cook and Robert A. Reckhov “Hard examples for Resolution”, Alasdair Urquhart “On the complexity of derivation in propositional calculus”, G.S. Tseitin “Optimal length tree-like resolution refutations for 2SAT formulas”, K. Subramani “The intractability of resolution”, Armin Haken “Reducing the size of resolution proofs in linear time”, O.B.Ilan, O. Fuhrmann, S. Hoory, O. Shacham, O.Strichman “Resolution for Quantified Boolean Formulas”, H.Buning, M. Karpinski, A. Flogel

Thank you very much!!!