Bounded-depth Frege complexity of Tseitin formulas for all graphs

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Abstract

We prove that there is a constant K such that Tseitin formulas for a connected graph G requires proofs of size 2tw(G)Ω(1/d)in depth-d Frege systems for d<Klognloglogn, where tw(G)is the treewidth of G. This extends Håstad's recent lower bound from grid graphs to any graph. Furthermore, we prove tightness of our bound up to a multiplicative constant in the top exponent. Namely, we show that if a Tseitin formula for a graph G has size s, then for all large enough d, it has a depth-d Frege proof of size 2tw(G)O(1/d)poly(s). Through this result we settle the question posed by M. Alekhnovich and A. Razborov of showing that the class of Tseitin formulas is quasi-automatizable for resolution.

MSC

03F20
03B05
05C83
68Q27

Keywords

Proof complexity
Bounded-depth Frege
Resolution
Tseitin formulas
Treewidth

1. Introduction

Propositional proof complexity is motivated by the result of Cook and Reckhow [15] saying that if there is a propositional proof system in which any unsatisfiable formula F has a short proof of unsatisfiability (of size polynomial in the size of F), then NP=coNP. In the last 30 years the complexity of proofs was investigated for several proof systems with the aim of finding concrete evidence, and eventually a proof, that for all proof systems there is a propositional formula which is not efficiently provable, i.e. requires super-polynomial proof size. The approach followed to prove such lower bounds was essentially borrowed from circuit complexity. Lines in a proof are Boolean formulas and we can define different proof system according to the circuit complexity of such formulas. For example resolution, a well-known refutational system for CNFs, corresponds to a system where formulas are of depth 1. In circuit complexity we keep on trying to strengthen lower bounds to computationally more powerful class of circuits. In proof complexity we follow the analogous approach: to strengthen lower bounds to systems working on formulas computationally more powerful. The hope is that techniques used to prove lower bounds for classes of Boolean circuits could be lifted to work with proof systems operating with formulas in the same circuit class. At present however we are far from such ideal situation and in fact, in terms of circuit classes, lower bounds for proof systems are well below those for Boolean circuits.

The complexity of proofs in resolution is largely studied. The first lower bound for (a restriction of) resolution was given by Tseitin in [43]. To obtain his result Tseitin introduced a class of formulas (nowadays known as Tseitin formulas) encoding a generalisation of the principle that the sum of the degrees of all vertices in a graph is an even number. A Tseitin formula T(G,f)is defined for every undirected graph G(V,E)and a charging function f:V{0,1}. We introduce a propositional variable for every edge of G so that T(G,f)is a CNF representation of a linear system over the field GF(2)such that each equation corresponds to a vertex vVans states that the sum of all edges incident to v equals f(v). Together with the unsatisfiable formulas PHPnencoding the Pigeonhole principle, Tseitin formulas are among the most commonly used examples to prove hardness results for proof systems. Haken [25], Beame and Pitassi [6] and Ben-Sasson Wigderson [10] proved exponential resolution lower bounds for the CNF formula encoding the negation of PHPn, results which were later generalized and improved in several other works [11], [16], [30], [37], [39], [40]. For unrestricted resolution it was Urquhart in [44] and later Ben-Sasson and Wigderson [10] who proved exponential lower bounds for Tseitin formulas defined over constant-degree expander graphs.

Bounded-depth Frege extends resolution since the formulas in the proof-lines are computable by AC0-circuits, i.e. constant-depth circuits with unbounded fan-in gates. The importance of understanding the complexity of proofs in bounded-depth-Frege systems was due at least to two reasons: (1) for general Frege systems, where formulas have no restrictions, i.e. are of depth O(logn), Buss in [12] proved that the Pigeonhole principle can be proved in polynomial size, hence obtaining an exponential separation with resolution. (2) Lower bounds for AC0-circuits were known [20], [27] and hence we could hope for applying lower bound techniques for AC0to lower bounds to bounded-depth Frege. Studying the complexity of proofs in bounded-depth Frege is of the utmost importance since it is a frontier proof system, i.e. is one of the strongest propositional proof systems with known significant lower bounds at the moment. Any advance is then a step towards proving lower bounds for AC0[2]-Frege, i.e. a bounded-depth Frege admitting also formulas with parity gates, which are unknown at the moment, though we know since a long time exponential lower bounds for AC0[2]circuits [32], [38], [42]. In this work we contribute to the complexity of proofs in bounded-depth Frege proving new lower bounds for Tseitin formulas.

Ajtai in [1] was the first to prove a lower bound in bounded-depth Frege. He showed that a refutation of PHPnmust have a super-polynomial size. His result was later followed by several results simplifying his technique [7] and improving the lower bound [33], [34] showing that any polynomial-size Frege refutation of PHPnmust have depth Ω(loglogn). The proof complexity of Tseitin formulas in bounded-depth Frege was first considered by Urquhart and Fu in [45], a work where they simplified and adapted the lower bound for the PHPnto the case of Tseitin formulas over a complete graph. Ben-Sasson in [9], proved exponential lower bounds for the Tseitin formulas over constant-degree expander graphs using a new reduction from the pigeonhole principle [45]. All these lower bounds are adaptation of the technique of [33], [34], hence vanish when the depth of formulas in the proof is more than loglogn. In a very recent major breakthrough [35] showed that Tseitin formulas over a 3-expander graph on n nodes requires super-polynomial bounded-depth Frege refutations at depth O(logn). This result was later improved to depth up to Clognloglognby Håstad in [28] but for Tseitin formulas defined only on the 2-dimensional grid, where C is a positive constant.

Proofs of T(G,f)were studied in terms of the treewidth of G, tw(G), for resolution [3], [22] and for OBDD proof systems [23]. We use Håstad result to prove tight bounds on the size of refutations in bounded-depth Frege of T(G,f)over any graph G in terms of the treewidth of G. Our main result is the following theorem:

Theorem 1

There is a constant K such that for any connected graph G over n nodes and for all dKlognloglogn, every depth-d Frege proof of ¬T(G,f)has size at least 2tw(G)Ω(1/d). Furthermore, for all large enough d there exist depth-d Frege proofs of ¬T(G,f)of size 2tw(G)O(1/d)poly(|T(G,f)|).

1.1. Overview of the proof technique

In Theorem 18 we prove the lower bound from Theorem 1. The proof is based on the improvement of the Excluded Grid Theorem by Robertson and Seymour recently obtained by Chuzhoy [13]: an arbitrary graph G contains as a minor a r×rgrid, where r=Ω(tw(G)λ)and λ is a constant; the latest improvement [14] establishes the theorem for λ<1/10, however, it is known that λ can not be greater than 1/2. For several classes of graphs it is possible to improve the value of λ, for example, for planar graphs λ=1[24], [41]. More precisely we use the corollary of this result (see Corollary 9) stating that any graph G has a wall of size r as a topological minor (i.e. can be obtained from G by several removing of vertices, edges and suppressions, see Fig. 1 and Fig. 3). Our proof consists of two parts: at first, we show that if H is a topological minor of G, then any bounded-depth Frege proof of ¬T(G,f)can be transformed to a proof of ¬T(H,f), with constant increase in depth and polynomial increase in size. And then we prove a lower bound on the size of depth-d Frege proof of ¬T(H,f)based on walls. In this proof we use the lower bound for grid graphs proved by Håstad [28].

Fig. 1
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Fig. 1. Suppression of v from G.

In Theorem 19 we prove the upper bound from Theorem 1. We consider the compact representation of linear functions F2nF2on variables x1,x2,,xnby propositional formulas of depth d and of size 2nO(1/d). We show that for linear functions f and g if the equations f(x)=aand g(x)=bare given in our representation, then there is a derivation of (f+g)(x)=a+bof depth d and of size 2nO(1/d). We also show that if a linear equation is represented in CNF, then it is possible to infer its compact representation with depth d and size 2nO(1/d). Since a Tseitin formula is an unsatisfiable system of linear equations written in CNF, hence it is possible to refute a Tseitin formula in size 2mO(1/d)and depth d, where m is the number of edges in G. However we wish to have the treewidth of G instead of m. We consider a tree-partition of a graph G, the vertices of G are split into bags and there exists a tree such that bags are nodes of this tree and if two vertices of G are connected, then they are either in one bag or in adjacent bags. It is known that there is a tree partition where the size of bags are at most O(tw(G)Δ(G))[46]. Since the number of edges touching a given bag is O(tw(G)Δ(G)2)we can use the compact representation to take care of the equations involving the parity of sum of adjacent bags with proofs growing in terms of the treewidth of G.

1.2. The complexity of Tseitin formulas in resolution

A recent line of research [3], [22], [31] investigates the exact relation between the treewidth of G and the proof complexity of T(G,f)in resolution and its refinements. Since resolution is a special case of bounded-depth Frege, Theorem 1 is directly applicable to resolution. In Section 5 we contribute this line of research partially answering to some questions left open before. First we improve Theorem 1 for resolution. In Theorem 34 we show 2Ω(twλ(G))lower bounds for resolution refutations of T(G,f), where λ is the constant from the grid-minor theorem in [13]. A stronger lower bound 2Ω(tw(G)/logn)was recently proved for regular resolution in [31] and it is open whether this holds for unrestricted resolution as well.

A conjecture of Urquhart asserts that the shortest resolution refutations of T(G,f)are in fact regular. A consequence of our results, stated in Corollary 40, shows that the size of refutations of T(G,f)in proof systems between tree-like resolution and bounded-depth Frege are quasi-polynomially correlated, i.e. if T(G,f)has a refutation of size S in bounded-depth Frege, then it has a refutation of size at most 2poly(logS)in treelike resolution and vice versa. Since regular resolution is in between treelike resolution and bounded-depth Frege, this provides evidence (showing it for a quasi-polynomial size) to the conjecture of Urquhart.

Finally we consider the question posed in [3] of (quasi)-automatization of T(G,f)formulas in resolution, i.e. whether there exists a deterministic algorithm that returns a resolution refutation of T(G,f)in time which is (quasi-)polynomial in |T(G,f)|+|τT(G,f)|, where |τT(G,f)|is the size of the shortest refutation of T(G,f)in resolution. From [3] it is known that T(G,f)are (quasi)-automatizable in resolution only when G has a specific covering property known as bounded cyclicity. In Corollary 39 we extend their result proving the quasi-automatization of T(G,f)for any graph G.

Organization. The paper is divided into four sections. After the Preliminary section, we have Section 3 for the lower bound (Theorem 18), and Section 4 for the upper bound (Theorem 19). Section 5 is devoted to the results on the proof complexity of T(G,f)in resolution.

2. Preliminaries

Formulas and restrictions. We consider propositional formulas over binary ∨ and ∧, unary ¬ and Boolean constants 0,1. We represent formulas as rooted trees such that internal vertices are labelled with connectives and leaves are labelled with propositional variables or Boolean constants. The depth of a formula is the maximal number of alternations of types of connectives over all the paths from the root to a leaf plus one.

We assume that disjunctions with unbounded fanin are represented via binary disjunctions. By default, we mean that i=1nxiis right-associative; i.e. it denotes ((x1x2))xn1)xn; we also assume the same for ⋀.

We denote by vars(F)the set of variables of a formula F. A partial assignment α for a formula F is mapping from vars(F){0,1,}, where α(x)=if x is unassigned. We denote by dom(α)=α1({0,1})the set of variables in F which α assigns a Boolean value.

Pudlák-Buss games. We use the game interpretation of Frege proofs introduced by Pudlák and Buss [36]. Let us define a game with two players Pavel and Sam. The game starts with initial conditions of the form φ1=a1,,φk=ak, where φ1,φ2,,φkare propositional formulas and a1,a2,,ak{0,1}such that i=1k(φi=ai)is identically false. Sam claims that he knows an assignment of variables that satisfies i=1k(φi=ai), the goal of Pavel is to convict Sam. On each Pavel's turn he asks Sam the value of a propositional formula and Sam gives an answer. The game stops when Pavel convicts Sam, namely Pavel finds an immediate contradiction among initial conditions and Sam's answers. An immediate contradiction with a Boolean connective ∘ of arity t is a set of (t+1)formulas α1,,αtand (α1,,αt)with claimed values a1,,atand b such that (a1,,ak)b. In particular, 0 with claimed value 1 is an immediate contradiction.

A strategy of Pavel is a function that maps initial conditions and the history of a game to a propositional formula (request). A winning strategy is a strategy that allows Pavel to convict Sam for any behaviour of Sam. A winning strategy of Pavel can be represented as a binary tree whose nodes are labelled with Pavel's requests and edges correspond to Sam's answers. A leaf of the tree corresponds to an immediate contradiction among initial conditions and the equalities corresponding to the path from the root to this leaf.

A Pudlák-Buss game derivation of a formula ψ from formulas φ1,φ2,,φsis a tree of a Pavel's winning strategy in a game with initial conditions φ1=1,φ2=1,,φs=1,ψ=0. In that follows by derivations we always mean Pudlák-Buss game derivations. We are interested in the two complexity parameters of derivations: 1) the size of a derivation S that equals the total size of formula ψ and all formulas that are used as labels of nodes; 2) the depth of a derivation d is the maximum depth of ψ and formulas that are used as labels of nodes. We use the notation φ1,,φsdψfor a derivation of ψ from φ1,φ2,,φsof depth at most d. A derivation of φ is a derivation of φ from the empty set of formulas.

Lemma 2

Assume that there is a derivation φ1,,φkd1ψ1of size S1and also there is a derivation φ1,,φk,ψ1d2ψ2of size S2, then there is a derivation φ1,,φkmax{d1,d2}ψ2of size S1+S2.

Proof

Let us create the new tree with the root labelled with ψ1such that edge form the root labelled with 0 goes to the root of the first derivation and edge labelled with 1 goes to the root of the second derivation. □

Lemma 3

1. If a formula φ has a Frege derivation of size S and depth d, then φ has a Pudlák-Buss game derivation of size O(S2)and depth d. 2. If φ has a Pudlák-Buss game derivation of size S and depth d, then φ has a Frege derivation of size O(S3)and depth d+O(1).

Proof

For proof see Appendix A. □

Lemma 4

Let ψ1and ψ2be two formulas of depth at most d such that |vars(ψ1)vars(ψ2)|=kand ψ1semantically implies ψ2. Then there exists a derivation ψ1dψ2of size at most 2k(|ψ1|2+|ψ2|2).

Proof

Consider a game with initial conditions ψ1=1,ψ2=0.

Let |vars(ψ1)vars(ψ2)|={x1,,xk}. In the first k rounds of the game Pavel asks Sam x1,,xk. There are 2kpossible combinations of Sam's answers. For each combination Pavel asks Sam all subformulas of ψ1(the value of ψ1is 1 from initial conditions) starting with the deeper ones and then asks all subformulas of ψ2starting with the deeper ones (the value of ψ2is 0 from initial conditions). Since all variables are known, one of possible answers to each of these questions yields an immediate contradiction. The total size of all subformulas of ψ1and ψ2is at most |ψ1|2+|ψ2|2. Consider the branch of the game tree, where we still no get an immediate contradiction. In that case Sam ψ1=1and ψ2=0are consistent with the values of x1,,xk, but it is impossible since ψ1semantically implies ψ2. Hence, there should be an immediate contradiction. □

Corollary 5

let ψ1and ψ2be two formulas of depth d1such that vars(ψ1)vars(ψ2)={x1,x2,,xk}and ψ2is semantically implied by ψ1. Let φ1,,φkbe formulas of depth d2. Let ψi[x1φ1,,xkφk]for i{1,2}be the formula obtained by substitution φ1,,φkinstead of the variables of x1,x2,,xk. Then there exists a derivationψ1[x1φ1,,xkφk]d1+d2ψ2[x1φ1,,xkφk]of size at most 2k(|ψ1|2+|ψ2|2)(i=1k|φi|)

Proof

Use Lemma 4 for ψ1and ψ2and then substitute φ1,,φkinstead of the variables x1,,xk. □

A shortcut contradiction for the disjunction is a situation where Pavel asks Sam formulas i=1kαiand αjfor j[k]and gets the answers 0 and 1 respectively. Similarly a shortcut contradiction for the conjunction is a situation where Pavel asks Sam formulas i=1kαiand αjfor j[k]and gets the answers 1 and 0. We may consider shortcut games (and shortcut derivations) that has one difference from the already defined games: Pavel can convict Sam using shortcut contradictions instead of immediate contradictions. An ordinary derivation is a derivation which does not use shortcut contradictions.

Lemma 6

Consider a shortcut derivation of size S and of depth d. Then there is an ordinary derivation of size at most S3and of depth d.

Proof

We transform a tree with shortcut contradictions to a tree without them. Let us consider a leaf with a shortcut contradiction for the disjunction (case of conjunction is similar), which means that on the path from root to that leaf Pavel has asked Sam formulas i=1kαiand αjfor j[k]and got the answers 0 and 1 respectively. i=1kαiis represented as i=1qαii=q+1kαifor some index q, where ∨ is binary disjunction. Pavel asks Sam formulas i=1qαiand i=q+1kαi.

  • 1.

    If one of Sam's answers is 1, we get an immediate contradiction.

  • 2.

    If both Sam's answers are 0, then if jq, we continue this process with formula i=1qαiotherwise with formula i=q+1kαi.

In the end of that process we get an immediate contradiction. We repeat this for all leaves with shortcut contradictions, the size of the resulting tree is at most S3since there are at most S leaves and the size of added formulas is at most S2for each leaf. □

Tseitin Formulas. Let G(V,E)be an undirected graph and vV. We denote by E(v)the set of edges in E incident with v and by N(v)the set of neighbours uVof v, i.e. the u such that (u,v)E(v).

A vertex-charging for G(V,E)is a mapping f:V{0,1}. We say that f is an odd-charging of G if vVf(v)1mod2. The Tseitin formulas defined on G using variables xe,eEare the formulas: T(G,f):=vVPar(v), where Par(v)is the canonical CNF representation of eE(v)xe=f(v).

Lemma 7

[44]

T(G,f)is unsatisfiable if and only if there is a connected component U of G such that the restriction of f on U is an odd-charging.

In this work we will work with the tautological form of Tseitin formulas in the form of ¬T(G,f).

Grids, Walls, Minors, Topological Minors and Treewidth. We consider 4 structural operations on undirected graphs G=(V,E)possibly with parallel edges, but without loops. We follow [8], [19].

  • edge removal of eE. It produces the graph [Ge]=(V,E{e}).

  • vertex removal of vV. It produces the graph [Gv]=(V{v},EE(v)), where E(v)is the set of edges in E incident with vV.

  • edge contraction of e=(u,v)E. Is the replacement of u and v with a single vertex such that edges incident to the new vertex are the edges other than e that were incident with u or v. The resulting graph Gehas one edge less than G.

  • vertex suppression of a vertex v in G of degree 2. Let u and w be v's neighbours in G. The suppression of v is obtained by deleting v along with two edges (u,v)and (w,v)and adding a new edge (w,u)(possibly parallel to an existing one). The resulting graph [Gsv]has one vertex less than G. See Fig. 1.

A graph H is a minor of G if H can be obtained from G by a sequences of edge and vertex removals and edge contractions. A graph H is a topological minor of G if H can be obtained from G by a sequence of edge removals, vertex removals and by vertex suppressions [8], [19].

The grid Hm,nis the graph of the cellular rectangle m×n; it has (m+1)(n+1)vertices and n(m+1)+m(n+1)edges, among them n(m+1)horizontal and m(n+1)vertical edges. See Fig. 2.

Fig. 2
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Fig. 2. The grid H5,5.

The wall Wnis a subgraph of Hn,nthat is obtained by the removing of several vertical edges. Vertical edges of Hn,nare in n rows and we enumerate them in every row from the left to the right. In the odd rows we remove all vertical edges with even numbers and in even rows we remove all vertical edges with odd numbers. See Fig. 3.

Fig. 3
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Fig. 3. The wall W5.

A tree decomposition of an undirected graph G(V,E)is a tree T=(VT,ET)such that every vertex uVTcorresponds to a set XuVand it satisfies the following properties: 1. The union of Xufor uVTequals V. 2. For every edge (a,b)Ethere exists uVTsuch that a,bXu. 3. If a vertex aVis in the sets Xuand Xvfor some u,vVT, then it is also in Xwfor all w on the path between u and v in T.

The width of a tree decomposition is the maximum |Xu|for uVTminus one. A treewidth of a graph G is the minimal value of the width among all tree decompositions of the graph G.

Recall the following Theorem proved in [13].

Theorem 8

[14]

If G has a treewidth t, then it has the grid Hr,ras a minor, where r=Ω(tλ)and λ110is a constant.

The following Corollary was mentioned in [8].

Corollary 9

If G has a treewidth t, then it has the wall Wras a topological minor, where r=Ω(tλ)and λ is the constant from Theorem 8.

Proof

Since Wris a minor of Hr, Wris a minor of G. Consider a transformation T from G to Wrmade by (1) edge removals, (2) vertex removals and (3) edge contractions that has the minimal number of edge contractions. Since operations (1), (2) and (3) commute, we may assume that in T no removal follows an edge contraction (i.e. we remove everything at first, and then contract edges). Now it is easy to understand that all edge contractions should be actually suppressions. Indeed, assume that we contract (u,v)and both u and v has degrees at least 3, then we will get a vertex {u,v}with degree at least 4 and this is a contradiction since Wrdoes have degrees at most 3 and edge contractions can decrease degree only in one case when we contract (u,v)and v has degree 1, but in this case we may just remove v and it decreases the number of edge contractions. □

3. The lower bound

3.1. Topological minors and Tseitin formulas

Le φ be a formula and let α be a partial assignment to variables of φ. Define φ[α]to be the formula obtained from φ substituting each variable x in the domain of α, with the constant assigned to x by α. Notice that φ and φ[α]have the same size and depth.

Lemma 10

Let Φaand Φafor aAbe propositional formulas of depth at most d such that |vars(Φa)vars(Φa)|k. Assume that for all aA, Φais semantically equivalent to Φa. Then ¬aAΦad+O(1)¬aAΦaof size at most 2kpoly(aA(|Φa|+|Φa|)), where A={aA|Φais not identically true}.

Proof

At first Pavel asks aAΦa. The answer 1 leads to an immediate contradiction. Then Pavel proceeds to ask Φafor each aA. If all the answers are 1, Pavel asks all subformulas of the big conjunction aAΦastarting with the deeper ones and at each step one of Sam's answers leads to an immediate contradiction.

If Sam claim that Φa0=0for some a0A, we use Lemma 4 to construct a derivation ¬Φa0d¬Φa0of size at most 2k(|Φa0|2+|Φa0|2). If a0A, then Φa0is identically true, thus, by Lemma 4, there is a derivation 1dΦa0of size at most 2k(|Φa0|2+1), using it we get an immediate contradiction.

If a0A, then Pavel asks aAΦa. If Sam's answer is 0, we get a shortcut contradiction with Φa0and proceed with Lemma 6, otherwise we get an immediate contradiction with the initial condition ¬aAΦa=0. □

Lemma 11

[23]

Let G(V,E)be a connected graph and H(V,E)be a connected subgraph of G with Ethat is obtained from G by the deletion of some vertices and edges. For every unsatisfiable Tseitin formula T(G,f)there exists a partial assignment α to variables xefor eEEsuch that α does not falsify any clause of T(G,f).

Lemma 12

Let G(V,E)be a connected graph and H(V,E)be a connected subgraph of G. Assume that there is a derivation d¬T(G,f)of size S. Then for some fthere is a derivation d+O(1)¬T(H,f)of size S+poly(|T(G,f)|).

Proof

Let T be the game tree of d¬T(G,f). Let α be given by Lemma 11 that is defined on all variables xefor eEEand does not falsify any clause of T(G,f). T[α]be the tree obtained form T applying the substitution α to all the queried formulas. Size and depth do not change, hence T[α]defines a derivation d¬T(G,f)[α]of size S. ¬T(G,f)has the form ¬vVPar(v), where Par(v)is a parity condition of the vertex v. Hence, ¬T(G,f)[α]is of the form ¬iPar(v)[α]. If vV, then α assigns values to all variables from Par(v), since α does not falsify Par(v), α satisfies Par(v), hence Par(v)[α]is identically true. If vV, then Par(v)[α]is a parity statement depending on variables xe, where eEis incident to v. Hence, for vV, Par(v)[α]is semantically equivalent to a parity condition of a Tseitin formula T(H,φ)for some charging φ. Let Δ be the maximal degree of G. Then every parity condition of T(H,φ)or T(G,φ)depends on at most Δ variables. Notice that since we represent parities in CNF, |T(G,f)|2Δ. By Lemma 10, there is a derivation ¬T(G,f)[α]O(1)¬T(H,f)of size poly(|T(G,f)|). The claim follows using the size S, depth d derivation of ¬T(G,f)[α]together with Lemma 2. □

A 1-substitution for a formula φ is a partial function mapping variables of φ into literals of φ. After applying a 1-substitution σ to φ, the depth of the new formula φ[σ]can increase by one. However 1-substitutions are closed under composition: if σ1maps [y¬z]and σ2maps [x¬y], then σ=σ1σ2is the 1-substitution [xz,y¬z]. We use 1-substitutions to handle in T(G,f)the operation of vertex suppression on the graph G. Let G=(V,E)be a graph and vVbe a node and let T(G,f)be a Tseitin formula on G. Let v be a degree-2 vertex v in G with neighbours u and w. Consider the following 1-substitution σvand the charge function fvfor [Gsv]:σv={[x(v,w)x(w,u),x(v,u)x(w,u)]iff(v)=0[x(v,w)x(w,u),x(v,u)¬x(w,u)]iff(v)=1fv(z)={f(z)ifzV{u,v}f(u)+f(v)ifz=u

Let G(V,E)be a graph and f:V{0,1}be a charging. Let A be a finite set. We say that a formula Ψ is a pseudo Tseitin formula based on G and f with fake vertices in A, and we write Ψ is TA(G,f), if Ψ has the form vVAψv, where

  • 1.

    for all vV, ψvis a propositional formula depending on variables xefor all edges e incident to v. And ψvis semantically equivalent to the parity condition Par(v)of T(G,f).

  • 2.

    for all vA, ψvis a tautology.

Lemma 13

Let G(V,E)be a connected constant-degree graph over n vertices. Let [Gsv]be the graph obtained after the suppression of a degree-2 vertex v in G. If Ψ is TA(G,f), then Ψ[σv]is TA{v}([Gsv],fv).

Proof

Assume that v is linked to two vertices w and u in G. Let A be the set of fake vertices of Ψ so Ψ has the form xVAψx, hence Ψ[σv]is xVAψx[σv]. For xA, ψ(x)is a tautology, hence ψx[σv]is also a tautology. By the definition of σv, ψv[σv]is a tautology. It is not hard to verify that for xV{v}, ψx[σv]is equivalent to parity condition of T([Gsv],fv). Hence, Ψ[σv]is TA{v}([Gsv],fv) □

Lemma 14

Let G(V,E)be a graph and f:V{0,1}and W={v1,,vk}be degree 2 nodes in V suppressed in that order from G and [GsW]be the resulting graph. Let σibe the corresponding 1-substitutions and let σ=σkσ1. Then there is a charging fkof G such that if Ψ is TA(G,f), then Ψ[σ]is TAW([GsW],fk).

Proof

By induction on k0and using Lemma 13 we prove that Ψ[σ1][σk]is TAW([GsW],fk), for fkgiven by repeated applications of Lemma 13.

Since the composition of 1-substitutions removes double negations, then the only difference between Ψ[σ1][σk]and Ψ[σ]can be only in the number of negations in front of the same occurrence of a variable, though the parity of the number of negations is preserved. Hence for all vVthe semantical equivalence of the ψv's in Ψ[σ1][σk]and Ψ[σ]is preserved. Then Ψ[σ]too is TAW([GsW],fk). □

Lemma 15

Let G be a connected graph on n vertices and with the maximal degree at most 3. Let H be obtained from G by several suppressions. Assume that there is a derivation of ¬T(G,f)of size S and depth d. Then for some charging fkthere is a derivation of ¬T(H,fk)of size O(S)+poly(n)and depth d+O(1).

Proof

Assume that, in order, to get H from G we have to apply suppressions for vertices W={v1,,vk}. Let σibe the 1-substitutions corresponding to the suppression of vi, and let σ=σkσ1. T(G,f)is T(G,f). Let fkbe the charging given by Lemma 14 applied to T(G,f)and [GsW]=H. Then T(G,f)[σ]is TW(H,fk). We apply the 1-substitution σ to the given derivation of ¬T(G,f)and we get a derivation of ¬T(G,f)[σ]of size O(S)and depth at most d+1. By Lemma 10, applied on T(G,f)[σ]and T(H,fk), there is a derivation ¬T(G,f)[σ]d+O(1)¬T(H,fk)of size poly(n). Combining the two derivations together by Lemma 2 we obtain a derivation d+O(1)T(H,fk)of size O(s)+poly(n). □

3.2. From walls to grids

Lemma 16

If there exists a derivation d¬T(Wn,f)of size S, then there exists a derivation d+O(1)¬T(Mn,f)of size O(S)+poly(n), where Mnis a connected constant-degree graph that contains Hn,n12as a subgraph.

Proof

Consider a set I of all the horizontal edges of Wnthat belong to odd columns (on Fig. 4, Fig. 5 edges from I are wavy). I is a matching, i.e. no two edges from I are incident to the same vertex. If we contract all edges from I, we get the graph Mnthat for odd n coincides with Hn,n12and for even n coincides with a graph that is obtained from Hn,n2by the removal of several edges from the last vertical (see Fig. 4, Fig. 5). For every eIwe denote its left vertex by ueand the right vertex by ve. Let Euebe the set of edges of Wnincident to ueexcept e. Let τedenote a CNF formula encoding fEuexf=f(ue).

Fig. 4
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Fig. 4. W6is contracted to M6.

Fig. 5
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Fig. 5. W5is contracted to M5.

Consider a game tree T for the derivation of the Tseitin tautology ¬T(Wn,f)of size S and depth d. To every formula used in this tree we apply the substitution that replaces every occurrence of xewith τe. We denote the resulting tree by T.

Notice that Tis a correct game tree of a derivation d+O(1)¬F, where F is obtained from T(Wn,f)by the same substitution. The depth of this derivation is increased by at most a constant since in several leaves we hang a formula of constant depth; here we also use that I is a matching and thus we do not add new occurrences of variables corresponding edges from I. The size of τeis O(1), hence any formula from the derivation is increased in at most a constant factor, thus the size of the derivation defined by the tree Tis O(S).

We define a function fon vertices of Mnas follows. If a vertex w of the graph Mnis obtained by merging the vertices w,wof the graph Wn, then f(w)=(f(w)+f(w))mod2. If the vertex w of Hn,n/2is obtained from the vertex w of Wn, then f(w)=f(w).

Now we show how to derive ¬T(Mn,f)from ¬F. T(Wn,f)is a Tseitin formula and it has the following structure: vVψv, where V is the set of vertices of Wnand ψvis a CNF formula encoding a parity condition for the vertex v. F differs from T(Wn,f)only in conditions corresponding to vertices that are incident to an edge from I (if n is even, then there are vertices in Wnthat are not incident to any edge from I). Notice that F has the form vVψvwhere ψvis obtained by substitution from ψv. Let w=uefor some eI, then the formula ψwis identically true. If w=ve, then the condition ψwis equivalent to the parity condition of the merged vertex {ue,ve}in the Tseitin formula T(Mn,f), but ψwis not written in canonical form.

Since all degrees in Mnare at most 4, then by Lemma 10 there exists a derivation ¬Fd+O(1)¬T(Mn,f)of size poly(n). The claim follows by Lemma 2. □

3.3. Putting it all together

We use Håstad's Theorem from [28].

Theorem 17

[28]

There is a constant K>0such that for dKlognloglognany depth d derivation of ¬T(Hn,n,f))has size at least 2nΩ(1/d).

Theorem 18

There exist constants K>0and C>0such that for every connected graph G of treewidth t and every dKlognloglognC, any depth d derivation of ¬T(G,f))has size at least 2tΩ(1/d).

Proof

Suppose that ¬T(G,f)have a derivation of size S and depth d. By Corollary 9 we know that G contains the wall Wras a topological minor, where r=Ω(tλ). Consider a sequence of operations (edge/vertex removals and suppressions) that transform G to Wr. Assume that removals do not follow suppressions. And let Gbe a subgraph of G that is obtained from G by application of all removals (hence, Wrcan be obtained from Gby application of several suppressions).

By Lemma 12, for some fthere is a derivation of ¬T(G,f)of size poly(|T(G,f)|)+Sand depth d+O(1). Since Wrcan be obtained from Gby application of several suppressions, Gis connected. Suppressions can not increase the degrees, hence all degrees in Gare at most 3. By Lemma 15, for some fthere is a derivation of ¬T(Wr,f)of size poly(|T(G,f)|)+Sand depth d+O(1). By Lemma 16, for some fthere is a derivation of ¬T(Mr,f)of size poly(|T(G,f)|)+O(S)and depth d+O(1), where Mris connected constant-degree graph containing H(r1)/2as a subgraph. And finally by Lemma 12, for some fthere is a Frege derivation of T(H(r1)/2,f)of size poly(|T(G,f)|)+O(S)and depth d+O(1). Notice that S is the size of a derivation of ¬T(G,f), hence S|T(G,f)|. Thus, for some constants C and c there is a derivation of ¬T(H(r1)/2,f)of size Scand depth d+C.

By Theorem 17, there is a constant K such that if d+CKlognloglogn, then Sc2(r1)/2Ω(1/(d+C)). Hence S2rΩ(1/d)and, thus, S2tΩ(1/d). □

4. The upper bound

In this section we prove the following Theorem:

Theorem 19

Let G(V,E)be a connected undirected graph and T(G,f)be an unsatisfiable Tseitin formula. Then for all large enough d the formula ¬T(G,f)has a derivation of depth d and size 2tw(G)O(1/d)poly(|T(G,f)|).

Let us give the outline of the proof:

  • 1.

    First, we define the compact representation of Parity. The main idea behind this is using associativity of the operation: we split the input on chunks of suitable size and construct the formula recursively. The first part of the section describes the technical lemmas concerning the compact representation.

  • 2.

    The next part of the section describes how to emulate summation of linear equations modulo 2 using the compact representation of Parity.

  • 3.

    In the last part of the section, we use the notion of tree-partition width of the graph. Using the small cuts of tree-partition, we construct a proof which tries to detect local contradictions of a number of variables, proportional to tree-partition width.

4.1. A compact representation of parity

Let t1,t2,,tdbe natural numbers, where d is a non-negative integer. Let U0,U1,,Udbe partitions of a finite set F; it is possible that some elements of Uifor i{0,1,,d}are empty. We say that a list of partitions U=(U0,U1,,Ud)is a (t1,,td)-refinement of F if the following conditions hold:

  • 1.

    U0consists of the only element U0,1=F.

  • 2.

    For every i, Ui+1is a subpartition of Uisuch that every element of Uiis split into ti+1parts. Hence, Uisplit F into miparts: Ui,1,Ui,2,,Ui,mi, where mi=j=1itj.

  • 3.

    All elements of Udhave cardinality at most 1.

We refer to elements of Uifor i{0,1,,d}as blocks of the refinement.

Let U be a (t1,,td)-refinement of a set F and let Ui,jbe one of the blocks of this refinement. Then U induces on each of the blocks Uija (ti+1,,td)-refinement Uwhich is obtained by restricting Ui,,Udto the set Uij. Uis called a sub-refinement of Uijin U.

Lemma 20

Let F be a set of size n and d0be an integer. Let t1,,tdbe integers such that t1t2tdn. Then there exists a (t1,,td)-refinement U of F.

Proof

We define Udas n singletons each containing one element of F and t1tdnempty sets. For each i<dwe construct Uifrom Ui+1as follows. We split the ti+1tdblocks of the partition Ui+1into groups of size ti+1and define Ui,1as the union of the elements of the first group, Ui,2as the union of the elements of the second group and etc. □

For a{0,1}and natural number n we define a Boolean function PARITYna:{0,1}n{0,1}nsuch that PARITYna(x1,,xn)=1iff i=1nxi=afor all x1,,xn{0,1}.

Lemma 21

Let n and d be positive integers and U be a (t1,t2,,td)-refinement of [n]. Then there exists a formula representing PARITYnbof depth at most 3d+1and of size i=1d2ti+1ti.

Proof

Let us construct a formula of the needed size and depth for each of the chunks in the refinement, starting from the smallest ones. More formally, we do that by backward induction on i from d to 0. We prove that for every j[k=1itk], there is a formula representing kUi,jxkof depth 3(di)and of size q=i+1d2tq+1tq.

If i=d, then |Ud,j|1, hence kUi,jxkis either 0 or a variable xkand thus has size 1 and depth 0.

Assume that i<d. Let 1,2,,ti+1be such that Ui,j=Ui+1,1Ui+1,2Ui+1,ti+1. Let for r[ti+1], βrbe a representation of kUi+1,rxkof size q=i+2d2tq+1tqand depth 3(di1)that exists by the induction hypothesis. Consider a CNF-representation of β1βti+1: kUi,jxk=S{1,,ti+1}|S|mod2=0(sS¬βssSβs). After the substitution of the representations of β1,,βti+1we obtain a formula of size at most 2ti+1ti+1q=i+2d2tq+1tq+2ti+1ti+1q=i+1q2tq+1tqand of depth 3(di1)+3=3(di).

Therefore we have constructed a representation of PARITYn1of the needed size and depth. The representation of PARITYn0could be constructed as ¬φ where φ is the obtained representation of PARITYn1. □

We call the representation of PARITYnaobtained by Lemma 21 the compact representation of PARITYnawith respect to a (t1,,td)-refinement U.

Let us define for S[n]and for a{0,1}, PARITYn,Sa(x1,,xn)=(¬a)iSxi. We define a compact representation of PARITYn,Sawith respect to a (t1,,td)-refinement U as the result of substitutions xj:=0for all jSto the compact representation of PARITYnawith respect to U. We denote the compact representation of PARITYn,Sa(x1,x2,,xn)w.r.t. U by Φa(S,U).

Lemma 22

Let U be a (t1,,td)-refinement of [n]and Ube a sub-refinement of Uijin U. Then for every SUijthere exists a derivation Φa(S,U)3d+O(1)Φa(S,U)of size at most 4|Φa(S,U)|3.

Proof

Let ψ1:=xand ψ2be the formula obtained from Φa(S,U)by replacement of all subformulas Φa(U)with x. Clearly ψ1and ψ2are semantically equivalent. Then let us apply Corollary 5 to ψ1, ψ2, k=1and φ1=Φa(S,U). Then there exists a derivation of ψ1[xΦa(S,U)]3d+O(1)ψ2[xΦa(S,U)]of size 21(|ψ1|2+|ψ2|2)|φ1|=2(1+|Φa(S,U)|2)|Φa(S,U)|4|Φa(S,U)|3. □

4.2. Summation of linear equations

Let STbe the symmetric difference of sets S and T i.e. ST=(ST)(ST).

Proposition 23

Let U be a (t1,,td)-refinement of [n]. For any S[n]there exists a derivation ¬Φa(S,U)3d+O(1)Φ1a(S,U)of size at most 2(|Φ1(S,U)|+|Φ0(S,U)|).

Proof

If a=1then the premise equals the conclusion, so an empty derivation is sufficient. If a=0we need to remove double negation so Pavel asks Φ0(S,U). If the answer is 1 then it is an immediate contradiction with the premise and if the answer is 0 it is an immediate contradiction with the conclusion. □

Lemma 24

Let U be a (t1,,td)-refinement of [n]. Then there exist a constant c and a derivationΦa(S,U),Φb(T,U)3d+O(1)Φab(ST,U)of size at most c|Φ1(,U)|6. We writehere to stress that the size of Φa(S,U)does not depend on S.

Proof

Let us prove this Lemma by induction on d. The base case: d=0. Since U0is the last layer of refinement, it contains only singletons. On the other hand by the definition U0={[n]}, thus, n=1and Φa(S,U),Φb(T,U)and Φab(ST,U)all depend on one variable x1. Thus by Lemma 4, there exists a derivation Φa(S,U)Φb(T,U)2Φab(ST,U)of constant size. Clearly Φa(S,U)Φb(T,U)is derivable from Φa(S,U)Φb(T,U)by a derivation of constant depth and size.

If d>0, let V1,,Vt1be the sub-refinements of U1,1,,U1,t1in U respectively where [n]=U1,1U1,t1. Recall that by definitionΦ1(X,U)=R[t1]|R|mod2=0(iR¬Φ1(XU1,i,Vi)iRΦ1(XU1,i,Vi)).We view this representation as CNF in variables Φ1(XU1,i,Vi)for i[t1]. Recall that by definition Φ0(X,U)=¬Φ1(S,U).

Let us describe a strategy of Pavel in the game with initial conditions Φa(S,U)=1, Φb(T,U)=1and Φab(ST,U)=0.

Pavel asks Sam the values of the following 3t1formulas: Φ1(SU1,i,Vi), Φ1(TU1,i,Vi), Φ1((ST)U1,i,Vi)for i[t1]. There are 23t1possible combinations of Sam's answers. The total size of the formulas is at most 223t1|Φ1(,V1)|. Let us show how Pavel wins in each of them. Let αi,βi,γidenote Sam's answers to Φ1(SU1,i,Vi), Φ1(TU1,i,Vi), Φ1((ST)U1,i,Vi)respectively for i[t1].

If i[t1]αia, then we can get a contradiction with Φa(S,U). Indeed consider a formula ψ:=i=1t1Φαi(SU1,i,Vi)Φa(S,U)and let L be the size of ψ if we consider it as a formula in variables Φ1(SU1,i,Vi). Clearly L2t1+1t1. By Corollary 5, there is a derivation of 0 from ψ of size at most 2t1L2(i=1t1|Φ1(,Vi)|)23t1+4t13|Φ1(,V1)|. Pavel asks Sam the value of ψ and in both cases gets a contradiction. If the answer is 0 then it is a contradiction with the conjuncts so Pavel proceeds to ask all subformulas of ψ starting with the deeper ones until he gets an immediate contradiction. If the answer is 1 we get a contradiction by Corollary 5. The total size of all such derivation for all possible combinations of Sam's answers is at most 23t1max{|ψ|2,23t1+4t13|Φ1(,V1)|}26t1+4t13|Φ1(,V1)|. Adding the size of the 3t1preliminary questions we get 26t1+4t13|Φ1(,V1)|+223t1|Φ1(,V1)|<2026(t1+1)t16|Φ1(,V1)|620|Φ1(,U)|6. The cases when i[t1]βibor i[t1]γiabare similar.

Let us consider the case where i[t1]αi=a, i[t1]βi=band i[t1]γi=1ab. In this case there exists i0[t1]such that αi0+βi0γi0. We derive a contradiction using Sam's answers Φ1(SU1,i0,Vi0)=αi, Φ1(TU1,i0,Vi0)=βiand Φ1((ST)U1,i0,Vi0)=γi. By Proposition 23 we may assume that Sam's answers are Φαi0(SU1,i0,Vi0)=1, Φβi0(TU1,i0,Vi0)=1and Φγi0((ST)U1,i0,Vi0)=1. By the induction hypothesis there exists a derivationΦαi0(SU1,i0,Vi0),Φβi0(TU1,i0,Vi0)3d+O(1)Φ1γi0((ST)U1,i0,Vi0)of size at most c|Φ1(,V1)|so we get an immediate contradiction with Φγi0((ST)U1,i0,Vi0). The total size of the derivation is at most

Since |Φ1(,U)|=2t1+1t1|Φ1(,V1)|, for a large enough c the size of derivation does not exceed c|Φ1(,U)|6. □

Lemma 25

Let U be a (t1,,td)-refinement of [n]. Let S1,S2,,Sk[n]and a1,,ak{0,1}. Then there exists a constant c such that:

  • 1.

    There exists a derivation Φa1(S1,U),Φa2(S2,U),,Φak(Sk,U)3d+O(1)Φa1ak(S1Sk,U)of size at most ck|Φ1(,U)|6.

  • 2.

    If i[k](jSixj=ai)is unsatisfiable then there exists a derivation

    Φa1(S1,U),Φa2(S2,U),,Φak(Sk,U)3d+O(1)0of size at most ck|Φ1(,U)|6.

Proof

The proof of (1) is simply (k1)consecutive applications of Lemma 24 using Lemma 2. The proof of (2) is based on a well-known fact: a system of linear equations is unsatisfiable if and only if the contradiction 0=1can be obtained as a linear combination of the equations. In our case it means that there is a set R:={r1,,rm}[k]such that Sr1Sr2Srm=and iRai=1.

Φar1arm(Sr1Srm)=Φ1(,U). Φ1(,U)is identically false, hence by Lemma 4 there is a derivation Φ1(,U)3d+10of size at most 4|Φ1(,U)|2.

Let us construct Pavel's strategy in the game with initial conditions Φar1(Sr1,U)=1,,Φarm(Srm,U)=1and Φ1(,U)=0. Pavel asks Sam the value of Φ1(). If Sam answers 0 Pavel uses the strategy corresponding to Φar1(Sr1),Φar2(Sr2),,Φark(Srk)3d+O(1)Φar1ar2ark(Sr1Sr2Srk)from (1). If Sam answers 1 Pavel uses the strategy corresponding to the derivation Φ1(,U)3d+10. □

4.3. Tree-partition-width

Let G(V,E)be an undirected graph and S1,,Smbe a partition of V. S1,,Smis a tree-partition of G if there exists a tree T([m],ET)such that every edge e of G connects either two vertices from the same part Sior connects a vertex from Siand a vertex from Sj, where i and j are adjacent in T, i.e. (i,j)ET. A width of a tree-partition S1,S2,,Smis the size of the largest set Sifor i[m]. A tree-partition width of a graph G is the smallest width among all tree-partitions of G. We denote the tree-partition width of G by tpw(G).

If we add a new vertex in the middle of every edge (i,j)of the tree T and put the set SiSjon it, we will get a tree decomposition of G, hence tw(G)2tpw(G)1.

The following theorem shows an inequality in the other direction.

Theorem 26

[46]

If tw(G)1, then tpw(G)10Δ(G)tw(G), where Δ(G)is the maximum degree of G.

So, tw(G)and tpw(G)coincide up to a multiplicative constants for constant degree graphs.

Theorem 27

Let G(V,E)be a connected graph and let a Tseitin formula T(G,f)be unsatisfiable. Then there exists a derivation 3d+O(1)T(G,f)of size at most poly(|T(G,f)|)2(tpw(G)Δ(G))O(1/d), where Δ(G)is the maximum degree of G.

Proof

Let S1,,Smbe a tree-partition of G with width tpw(G)and let T([m],ET)be the corresponding tree. Without loss of generality we assume that T is a rooted tree with the root m; for all i[m1], p(i)denotes its parent and for all i[m], s(i)denotes the set of direct successors of i. Without loss of generality we assume that p(i)>ifor all i[m1].

Since T(G,f)is unsatisfiable and G is connected, vVf(v)=i[m]vSif(v)=1. We consider the sum i[m]eE(Si,VSi)xe. Since each xeoccurs in the sum exactly twice, the sum (modulo 2) is 0 for all values of xe. Then for each assignment to {xe}eEthere exists i0such that vSi0f(v)eE(Si0,VSi0)xe. The first part of Pavel's strategy is to find such i0.

Pavel will request parity of the sum of all edges between Siand Sjfor all (i,j)ET. In order to represent these formulas in a compact way we now define m different (t1,,td)-refinements W1,,Wm; for every i, Wiis a refinement of the set E(Si,js(i)Sj)of all edges connecting a vertex from Siwith a vertex from js(i)Sj. We construct appropriate refinements Wilater.

Pavel asks Sam the values of eE(Si,Sp(i))xerepresented as Φ1(E(Si,Sp(i)),Wp(i))for i[m1]in the increasing order until he finds i0such that eE(Si0,VSi0)xevSi0f(v).

At the moment when Sam has answered the value of Φ1(E(Si,Sp(i)),Wp(i))the values of eE(Si,Sj)xefor each j such that (i,j)ETare all determined, thus, the value of eE(Si,VSi)xeis determined. If eE(Si,VSi)xevSif(v)Pavel proceeds to the next part of his strategy. Otherwise he continues to ask Sam similar questions corresponding to the vertices with larger indices.

Now we describe the strategy of Pavel in case if he finds i0. We are going to describe this case in terms of derivation using Lemma 2 multiple times. Consider a linear system that consists of the equation eE(Si0,VSi0)xe=1vSi0f(v)and all parity conditions of T(G,f)of the vertices from Si0. This linear system is unsatisfiable. We are going to use Lemma 25. In order to do it we need to derive the representations of these linear equations w.r.t. some refinement Q of a superset of E(Si0,V).

Let for i[m], Uibe a (t1,t2,,td)-refinement of the set E(Si)of all edges connecting two vertices from Si(we construct these refinements in the end of the proof together with the refinements Wi). Let us define a (3,t1,,td)-refinement Q as a union of (t1,,td)-refinements Wi0,Wp(i0)and Ui0such that Q1={E(Si0,js(i0)Sj),E(Sp(i0),js(p(i0))Sj),E(Si0)}and for every j{2,3,,d+1}, Qjis the union of Wj1i0,Wj1p(i0)and Uj1i0.

Let ajbe Sam's answer to the question eE(Si0,Sj)xefor each j that is a neighbour of i0in T, hence we may assume that Φaj(E(Si0,Sj),Wi0)for js(i0)and Φap(i0)(E(Si0,Ep(i0)),Wp(i0))are already derived. By Lemma 22, we derive Φaj(E(Si0,Sj),Q)from Φaj(E(Si0,Sj),Wi0)for js(i0)and Φap(i0)(E(Si0,Sp(i0)),Q)from Φap(i0)(E(Si0,Sp(i0)),Wp(i0)), where ajare Sam's answers to the corresponding questions.

By the first part of Lemma 25 we derive Φ1(vSi0f(v))(E(Si0,VSi0),Q)from the set of formulas {Φaj(E(Si0,Sj),Q)|(i0,j)ET}. We assume that the parity conditions of the vertices of G in T(G,f)represented as CNF are asked at the beginning of the game i.e. for each vVwe know that the CNF representation of u:(u,v)Exeis true (if any clause of T(G,f)is false Pavel queries all subformulas of T(G,f)except subformulas of the clauses and gets an immediate contradiction, if any of the parity conditions is false it yields an immediate contradiction with the corresponding subset of clauses). Thus, by Lemma 4 we derive the representations of parity conditions of the vertices from Si0w.r.t. Q. Since the corresponding linear system is unsatisfiable, using the second part of Lemma 25 we get a contradiction.

Claim 28

The size of the described game tree is at most m23Δ(G)Δ2(G)tpw(G)2O(i=1dti).

Proof

The game tree consists of m subtrees corresponding to each possible value of i0, m1nodes corresponding to the questions Φ1(E(Si,Sp(i)),Wp(i))for i[m1]and O(|V|)nodes corresponding to the initial questions about parity conditions of the vertices of G in CNF. The total size of formulas in these O(|V|)nodes is poly(|T(G,f)|). Each of the m1formulas corresponding to the questions Φ1(E(Si,Sp(i)),Wp(i))has size i=1d2ti+1ti. Let us estimate the size of each of the m subtrees. It is the sum of the following values:

  • The derivation of the formulas Φ1(E(Si0,Sj),Q)for j{p(i0)}s(i0). By Lemma 22 the size of each of these derivations is O(i=1d23titi3).

  • The derivation of the representations of parity conditions with respect to Q. By Lemma 4 the size of each of these derivations is O(2Δ(G)(2Δ(G)Δ(G)+i=1d2titi)2)

  • The derivation of Φ1(vSi0f(v))(E(Si0,VSi0),Q). By the first part of Lemma 25 the size of this derivation is Δ(G)tpw(G)2O(i=1dti).

  • The derivation of contradiction by the second part of Lemma 25 has size (Δ(G)+1)tpw(G)2O(i=1dti).

Each of these four summands is 23Δ(G)Δ2(G)tpw(G)2O(i=1dti). □

Let us choose ti=(Δ(G)tpw(G))2/dfor all i[d]. Since |Si|tpw(G), |E(Si)|+|E(Si,js(i)Sj)|Δ(G)tpw(G). Hence, the condition i=1dtiΔ(G)tpw(G)|E(Si)|+|E(Si,js(i)Sj)|holds and, thus, for all i[m]the refinements Ui,Wiexist by Lemma 20. If we substitute choosen values in the bound from Clam 28, we get the upper bound m2O(3Δ(G)+d(Δ(G)tpw(G))2/d)=poly(|T(G,f)|)2(Δ(G)tpw(G))O(1/d). □

Now we are ready to prove Theorem 19.

Proof of Theorem 19

Theorem 27 and Theorem 26 imply that there exists a constant c and a derivation 3d+O(1)¬T(G,f)of size at most poly(|T(G,f)|)2(10Δ2(G)tw(G))c/d. If tw(G)>Δ(G)then we can rewrite our upper bound on the size as poly(|T(G,f)|)2(10tw(G))3c/d. If tw(G)>1then it is poly(|T(G,f)|)2(tw(G))O(1/d). If tw(G)=1then it is simply poly(|T(G,f)|). Otherwise if tw(G)Δ(G)we can rewrite the upper bound as poly(|T(G,f)|)2(10Δ(G))3c/d=poly(|T(G,f)|)if 3c/d1. Thus, for d3cthe upper bound is poly(|T(G,f)|)2tw(G)3c/d. Therefore, for the both cases we have the needed upper bound. □

5. The complexity of Tseitin formulas in resolution

Resolution is a sound and complete propositional proof system suited for refuting unsatisfiable formulas in conjunctive normal form [17], [18]. A resolution refutation of a unsatisfiable CNF F=C1Cmis a sequence of clauses D1,,Dtsuch that Dtis the empty clause □ and each clause Diin the sequence is either a clause Cjof F or is obtained by the resolution rule applied to clauses Djand Dkpreceding Diin the sequence. The resolution rule is the following rule, where x is the resolved variable.AxB¬xABResolution rule

The size of a resolution refutation is the number of clauses in the sequence. With S(F)we denote the minimal size to refute F in resolution. The width of a clause is the number of literals in it; the width of a resolution refutation is the maximal width of a clause appearing in it. With w(F)we denote the minimal width to refute F in resolution.

Resolution refutations can be arranged as directed acyclic graphs with leaves corresponding to clauses of F and a root corresponding to the empty clause. An important refinement of resolution is the system known as regular (or read-once) resolution. In this system refutations are constrained in such a way that in the refutation-graph in each path from the empty clause to a leaf the sequence of resolved variables is read-once, i.e. each variable appears at most once. Tree-like resolution is the restriction obtained by imposing that the refutation-graph must be a tree.

Since their introduction [43] Tseitin formulas play an important role in understanding the complexity of refutations in resolution and its restrictions [10], [44]. Since resolution is a special case of bounded-depth Frege, Theorem 18 is applicable to it as well. The lower bound that can be obtained from this theorem is 2Ω(tw(G)γ)with γ<158λ, where λ is the constant from the grid minor theorem, and 158comes from Håstad's lower bound (Theorem 17).

Recently in [31] Itsykson et al. proved that regular resolution refutation of T(G,f)has size at least 2Ω(tw(G)logn)for every connected graph G and left open whether a similar result can be proved for unrestricted resolution. Theorem 34 improves the lower bound for general resolution comparing to the direct corollary from Theorem 18. In particular, Theorem 34 implies a lower bound 2Ω(tw(G))for planar graphs.

5.1. Resolution lower bounds for T(G,f)in terms of treewidth

Lemma 29

[31] Let T(G,f)and T(G,f)be two unsatisfiable Tseitin formulas based on a connected graph G. Then the sizes of the shortest resolution refutations of this formulas coincide.

Let φ be a CNF and α be a partial assignment to the variables of φ. Let φ|αbe a CNF which is obtained from φ[α]by removing all satisfied clauses and removing the satisfied literals from the rest of the clauses.

Lemma 30

Let G(V,E)and G(V,E)be connected graphs, and let Gis a topological minor of G, then for two unsatisfiable formulas T(G,f)and T(G,f), S(T(G,f))S(T(G,f)).

Proof

Gcan be obtained from G by a sequence of vertex and edge deletions and edge suppressions. It is easy to see that these operations commute, so we may assume that we first delete edges and vertices and then suppress edges. Let G(V,E)be the graph that is obtained from G by all edge and vertex deletions. Observe that Gis connected since Gis obtained from it by vertex suppressions which do not affect connectivity. By Lemma 11 there is a partial assignment α with support {xe|eEE}that does not falsify any clause of T(G,f).

Let us show that T(G,f)|α=T(G,f)where f(v)=(f(v)+eE;e is incident to vα(e))mod2. Indeed, T(G,f)|α=vVPar(v)|αwhere Par(v)is the parity condition of the vertex v encoded in CNF. Observe that for vVPar(v)|αis identically true and for vVit encodes in CNF the parity condition of v in T(G,f).

Consider a resolution refutation C1,,Ckof T(G,f). Observe that since α does not falsify axioms of T(G,f), C1|α,Ck|αcan be transformed into a resolution refutation of T(G,f)|αof size at most k. Thus S(T(G,f))S(T(G,f)).

Now we need to show that for a graph G(V,E)and for some fthat is obtained from Gby a sequence of vertex suppressions, S(T(G,f))S(T(G,f)). By induction it is sufficient to show this for just one suppression.

Assume that Gis the result of suppression of a vertex v neighbouring the vertices u and w, i.e. G=Gsv. Consider a vertex-charging of the set of vertices of G,f¯()={f() if wf(w)+f(v) if =w.f¯is an odd-charging since it has the same parity as f, which is an odd-charging since T(G,f)is unsatisfiable and Gis connected.

Consider the following procedure of obtaining T(G,f¯)from T(G,f). First substitute xuv:=xuw;xvw:=xuwif f(v)=0and xuv:=¬xuw;xvw:=xuwif f(v)=1. Such substitution satisfies the clauses of Par(v)in T(G,f). Observe that application of the same substitution to each clause of the resolution refutation of T(G,f)results in a resolution refutation of T(G,f¯)of the same or smaller length. By Lemma 29 there exists a refutation of T(G,f)of the same size, which completes the proof. □

The cyclicity of a graph G is the minimal for which the set of edges of G can be covered by cycles of length at most such that every edge is covered at most times.

Lemma 31

Let W¯nbe the result of removal all degree-1 vertices from Wn. Then W¯nhas cyclicity at most 6.

Proof

We can cover all the edges by cycles corresponding to “bricks” of the wall (i.e. two cells of the grid with the edge between them removed). The length of each such cycle is 6 and since the graph Wnis planar, each edge is covered by at most two cycles. □

Theorem 32

[3] S(T(G,f))=2Ω(w(T(G,f)))/(G)O(1)where (G)is the cyclicity of G.

Theorem 33

[22] For a constant-degree graph T(G,f), w(T(G,f))=Ω(tw(G)).

Theorem 34

S(T(G,f))=2Ω(twλ(G))where λ is a constant from the grid-minor theorem.

Proof

Since n2×ngrid is a minor of W¯n, tw(W¯n)=Ω(n). Then by Lemma 31 and Theorem 32, S(T(W¯n,f))=2Ω(n). By Corollary 9 W¯kis a topological minor of G where k=Ω(twλ(G)). By Lemma 30 S(T(G,f))S(T(W¯k,f))which yields the desired result. □

5.2. Searching for resolution refutations of T(G,f)

A key concept in the field of automatic generation of proofs is that of automatizability of a proof system. Resolution is automatizable if there is an algorithm that, given a unsatisfiable CNF formula F of size n as input, it outputs a resolution refutation of F in time polynomial in n+S(n), where S(n)is the length of a shortest refutation of F in resolution.

Given the importance of resolution in applied fields, investigation on automatizability of resolution received particular attention. Improving a result of Alekhnovich and Razborov in [2], automatizating resolution was recently found to be NP-hard [4]. These hardness results motivate the search of algorithms to generate resolution refutations for specific family of formulas like T(G,f).

Consider the next definitions from [3].

Definition 35

A class of unsatisfiable CNF formulas Fis (quasi-)automatizable with respect to a proof system Π if there exists a deterministic automatizing algorithm that, given an unsatisfiable formula FF, returns its refutation in the proof system S in time which is (quasi-)polynomial in |F|+SΠ(F), where SΠ(F)denotes the size of the shortest Π-refutation of F.

Definition 36

A class of unsatisfiable CNF formulas Fis smooth (respectively quasi-smoooth) if for all FF, w(F)logS(F)(resp. there is a constant C such that w(F)logCS(F)).

Definition 37

A class Fof unsatisfiable CNFs is width-automatizable if there exists a deterministic algorithm that, given FF, returns its resolution refutation of width O(w(F))in time which is polynomial in |F|+2w(F)

Smoothness and quasi-smoothness are important properties for a class of formulas since, together with width-automatizability imply automatizability (quasi-automatizabilty) in resolution of that class of formulas.

Theorem 38

(Fact 2.7 [3]) If a class of unsatisfiable CNFs is both width-automatizable and (quasi-)smooth, then it is (quasi-)automatizable with respect to resolution.

The question whether Tseitin formulas T(G,f)are (quasi-)automatizable with respect to resolution using previous theorem was investigated in [3]. There they proved that T(G,f)are width-automatizable but obtained quasi-smoothess of T(G,f)only for a restricted class of graphs G, namely those with constant cyclicity.

As a consequence of Theorem 1, we show that for all graphs the class of Tseitin formulas T(G,f)is quasi-smooth, hence quasi-automatizable with respect to resolution answering to a question left in [3].

Corollary 39

For any graph G, T(G,f)are quasi-smoooth, i.e. there is a constant C such that w(T(G,f))logCS(T(G,f)). Hence the class of Tseitin formulas is quasi-automatizable.

Proof

In [3], [22] it was established that w(G)=max{Δ(G),tw(L(G))}1, where L(G)is the line graph (graph on edges) of G. It is known that max{Δ(G),12(tw(G)+1)}1tw(L(G))(tw(G)+1)Δ(G)1, (see, for example, [26]). By Theorem 1 and the fact that T(G,f)is a CNF, we have that S(G)max{2Δ(G),2tw(G)ϵ}, for some 0<ϵ<1given by Theorem 1. Then S(TG,f)2(tw(G)ϵ+Δ(G))/22(tw(G)Δ(G))ϵ/22w(G)ϵ/4. □

A conjecture of A. Urquhart asserts that the shortest resolution refutations of T(G,f)are in fact regular. If one believes the conjecture, then devising algorithms for automatically searching for regular resolution refutations of T(G,f)is important since produces the shortest resolution refutations. Alekhnovich and Razborov in [3] devised an algorithm called BWBATP which on input T(G,f), when G is of bounded cyclicity, produces a regular refutation of T(G,f)in time 2poly(logSreg(G)), that is polylogarithmic in the length of the shortest regular refutation of T(G,f).

Using Theorem 1 and other known results we can generalize Alekhnovich and Razborov result to any graph and we can provide positive evidence (i.e. showing it for quasi-polynomial size) to the Urquhart's conjecture. We prove the more general statement that the size of refutations of T(G,f)in all the proof systems between tree-like resolution and constant-depth Frege (hence in particular regular resolution) are quasi-polynomially correlated. i.e. if T(G,f)has a derivation of size S in one of them of size S then it has a refutation of size at most 2poly(logS)in the other.

Corollary 40

Let Pand Qbe two proof systems such that both polynomially simulate tree-like resolution and both are polynomially simulated by constant-depth Frege. If there are derivations of size S in P, then there are derivations of size at most 2poly(logS)in Q.

Proof

Constant-depth Frege polynomially simulates treelike resolution hence it is sufficient to show the corollary taking Pas constant-depth Frege and Qas treelike resolution. Assume that ¬T(G,f)have proofs of size S in constant-depth Frege, we have to prove that there are derivations of size at most 2poly(logS)in tree-like Resolution. By Theorem 1 Smax{2Δ(G),2tw(G)ϵ}. Hence tw(G)(logS)1ϵ. By a result in [5], [29] there are tree-like resolution refutations of T(G,f)of size nO(w(G))=2O(tw(G)Δ(G)logn), where n is the number of edges in G. Substituting the expression for the treewidth we obtain the claim. □

Declaration of Competing Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

Acknowledgements

The authors thank Navid Talebanfard for discussions on the lower bound. Nicola also thanks Paul Wollan for introducing him to the tree-cutwidth. Dmitry is a Young Russian Mathematics award winner and would like to thank sponsors and jury of the contest.

The research was supported by Russian Science Foundation (project 16-11-10123).

Appendix A. Frege vs Pudlák-Buss games

A set of formulas Γ semantically implies a formula ψ if every assignment that satisfies all formulas from Γ also satisfies ψ.

Every particular Frege system is defined by a finite set of derivation rules. A derivation rule has form φ1,,φkφ, where k0, φ1,,φkand φ are propositional formulas. It is required that every rule is sound, i.e. {φ1,,φk}semantically implies φ.

Let Γ be a list of propositional formulas. We say that φ can be derived from Γ if there exists a sequence of formulas ψ1,ψ2,,ψssuch that ψs=φand each ψiis either an element of Γ or can be derived from the formulas with smaller numbers by a derivation rule. It is allowed to substitute propositional formulas instead of variables into a derivation rule. A formula τ is derivable if it can be derived from an empty list of formulas.

A system of derivation rules is a Frege system if it is implicationally complete: if some set of formulas Γ semantically implies a formula φ, then φ must be derivable from Γ.

A depth of a derivation is the maximum depth of a formula that occurs in the derivation. A size of a derivation is the sum of sizes of all formulas from the derivation.

Lemma 3

1. If a formula φ has a Frege derivation of size S and depth d, then φ has a Pudlák-Buss game derivation of size O(S2)and depth d. 2. If φ has a Pudlák-Buss game derivation of size S and depth d, then φ has a Frege derivation of size O(S3)and depth d+O(1).

Proof

  • 1.

    Let φ1,φ2,,φsbe a Frege derivation of φ of size S. We construct a strategy for Pavel. Pavel asks Sam values of the formulas φ1,φ2,until he gets the answer 0. φs=φ, so at least for φsthe answer is 0. Assume that Sam has answered 1 for the formulas φ1,φ2,,φkand 0 for φk+1where 0ks1. φk+1is derived from φ1,φ2,,φkusing one of the rules in Frege system. Let us consider all subformulas of this rule (there is a constant number of such subformulas). Pavel asks Sam values of the formulas corresponding to such subformulas in the application of the rule where φk+1is derived. Due to correctness of the rules, we get an immediate contradiction. Number of vertices for constructed tree is O(S). In every vertex Pavel has asked only subformulas of formulas in the Frege derivation, so size of every asked formula does not exceed S and depth does not exceed d. The total size of the game is O(S2)

  • 2.

    Let T be a tree for Pudlák-Buss game with the initial condition φ=0. Every vertex in the tree corresponds to a sequence of questions and answers on the path from the root to this vertex. For the sequence φ1=a1,φ2=a2,,φj=ajwe write a formula ¬(¬φφ1a1φjaj), where ψ0denotes ¬ψ, and ψ1denotes ψ.

    We assume that Frege system includes the rule ¬(xy),¬(x¬y)¬x(it can be simulated with derivation of constant size and depth due to implicationally completeness of the system). With that rule, formulas corresponding to the inner vertices of the tree can be derived from the formulas corresponding to their descendants.

    We also need to derive formulas in the leaves. Let h be the depth of T. We construct the derivation for a leaf using O(h)formulas, because in every leaf we have an immediate contradiction. There exist only a constant number of ways to get an immediate contradiction, so we assume that all these axioms are present in our Frege system. In order to use these axioms we need to rearrange the formulas in the big conjunction and place the formulas needed for immediate contradiction first.

    For example, if the immediate contradiction is A=0,B=1,AB=1, we need to add an axiom ¬(¬AB(AB)X).

    If the tree T has N vertices then the constructed derivation has O(Nh)formulas, and every formula has size O(S+h). Given that NSand hS, we get a derivation with total size O(S3) □

References

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A preliminary version of this paper appeared at MFCS 2019 [21].

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