Lower Bounds for QCDCL via Formula Gauge

Abstract

QCDCL is one of the main algorithmic paradigms for solving quantified Boolean formulas (QBF). We design a new technique to show lower bounds for the running time in QCDCL algorithms. For this we model QCDCL by concisely defined proof systems and identify a new width measure for formulas, which we call gauge. We show that for a large class of QBFs, large (e.g. linear) gauge implies exponential lower bounds for QCDCL proof size. We illustrate our technique by computing the gauge for a number of sample QBFs, thereby providing new exponential lower bounds for QCDCL. Our technique is the first bespoke lower bound technique for QCDCL.

1 Introduction

The satisfiability problem for propositional formulas (SAT) is one of the central problems of computer science. Traditionally perceived as a hard problem due to its NP completeness, SAT is nowadays very efficiently tackled by SAT solvers, building on the paradigm of conflict-driven clause learning (CDCL) [29], which solve problems in even millions of variables on many industrial problems.

The success of SAT solving has been transferred to computationally even more challenging settings, with quantified Boolean formulas (QBF) receiving key attention during the last decade [12]. One of the main approaches to QBF solving lifts CDCL to the quantified level, resulting in QCDCL [36]. In addition to QCDCL there are a number of further competing approaches to QBF solving [22, 26, 30]. Due to its PSPACE completeness, QBFs allow to encode many problems more succinctly, thus allowing to tackle even further applications [33].

Understanding which formulas are hard for (Q)CDCL is one of the most fascinating questions, both from a theoretical and a practical point of view. The main approach to this problem is through interpreting runs of SAT and QBF solvers on unsatisfiable formulas as formal proofs of their unsatisfiability. Since learned clauses in CDCL are derivable in resolution, it was noted early on that each run of a CDCL solver on an unsatisfiable formula can be efficiently translated into a resolution refutation [3]. Somewhat surprisingly, the converse holds as well, and when allowing arbitrary non-deterministic decision schemes, CDCL and propositional resolution are equivalent [31]. However, practical CDCL using decision schemes such as VSIDS [35] is exponentially weaker than the full resolution system [34].

Nevertheless, practical CDCL schemes are simulated by resolution and thus proof size lower bounds for resolution translate into lower bounds for CDCL running time. To obtain such lower bounds we can utilise the vast proof complexity machinery of resolution lower bound techniques [24] to show a plethora of lower bounds for combinatorial, random, and further formulas. Indeed, resolution is arguably the best-understood proof system, intensively studied long before the advent of SAT solving.

The situation is somewhat more intricate regarding the relation between QCDCL and Q-resolution, the latter being the simplest and most-studied analogue of propositional resolution for QBF [23]. The first result regarding their relative strength is due to Janota [21], who proved that practical QCDCL does not simulate Q-resolution. This can be interpreted as the QBF analogue of Vinyals result for practical CDCL vs resolution [34] (though [21] actually predates [34]). In contrast, the celebrated result on the equivalence of non-deterministic CDCL and resolution [31] does not lift to QBF as very recently shown in [7]: (non-deterministic) QCDCL and Q-resolution are incomparable, i.e., there exist formulas exponentially hard for Q-resolution, but easy for QCDCL, and vice versa.

This leaves us with the conundrum of how to show lower bounds for QCDCL. Though we understand Q-resolution fairly well and have a number of dedicated techniques for lower bounds in that system [5, 6, 8, 10, 13,14,15], unlike in the SAT case, these do not automatically apply to QCDCL.

The existing information on QCDCL lower bounds can be summarized as follows. In addition to the above-mentioned lower bound of [21] for practical QCDCL, we showed in [7] that under certain conditions, lower bounds from Q-resolution can be lifted to QCDCL. Also, while QCDCL runs on false QBFs cannot be efficiently transformed into Q-resolution proofs, they can be translated into long-distance Q-resolution proofs, an exponentially stronger proof system designed to model clause learning in QCDCL [2, 18]. However, we only have very few examples of hard formulas for long-distance Q-resolution [1, 10, 14], which again are lifted from Q-resolution hardness.

In summary, it is fair to say that QCDCL is rather poorly understood from a theoretical point of view and in particular lower bound techniques that would allow to show exponential lower bounds for QCDCL are lacking.

Our Contributions. We devise the first dedicated lower bound technique for QCDCL (with arbitrary clause learning mechanisms including those used in practise). In contrast to previous lower bounds for QCDCL, our technique does not import Q-resolution hardness and thus applies to different formulas, regardless of whether they are hard for Q-resolution or not (note that although Q-resolution was shown to be incomparable to QCDCL [7], basically all QCDCL lower bounds were imported from long-distance Q-resolution and hence are lower bounds also for Q-resolution). We already mention at this point though, that our technique is not completely general, but is restricted to $\Sigma _3^b$-formulas that meet a certain XT-condition, considered already in [7].

Technically, our approach rests on interpreting QCDCL runs in a formal framework of proof systems, already used in [7]. Further, we define a property of long-distance Q-resolution proofs, which we call quasi level-ordered. This is inspired by the notion of level-ordered proofs, introduced in [22], where the order of resolution steps in proofs must follow the quantification order in the prefix. Quasi level-order proofs relax that condition (Definition 4).

Our lower bound technique then rests on two steps: (1) We show that for $\Sigma _3^b$-formulas with the XT-condition, QCDCL proofs can be efficiently translated into quasi level-ordered Q-resolution proofs. (2) We define a new measure called the gauge of a QBF and show that large (i.e. linear) gauge implies exponential size in quasi level-ordered Q-resolution. Together, (1) and (2) imply that formulas with the XT-property and large gauge are hard for QCDCL (our main Theorem 6).

We illustrate our technique on a couple of examples on which computing the gauge is fairly straightforward. Thus, though showing (1) and (2) above is rather technical, the lower bound technique itself is quite easily applicable.

It is also interesting to mention that our new notion of gauge is some kind of width measure on clauses. Showing proof size lower bounds via width lower bounds is a very well-explored theme in proof complexity, both propositionally [4] and in QBF [6, 9]. We show, however, that gauge and proof width are not related in general.

Organisation. The remainder of this article is organised as follows. We start in Sect. 2 by reviewing notions from QBF, including Q-resolution and long-distance Q-resolution. In Sect. 3 we sketch QCDCL and explain how to model it as a formal proof system QCDCL. In Sect. 4 we introduce a new notion of quasi level-ordered proofs and give an algorithm to translate QCDCL proofs into quasi-level ordered Q-resolution. Section 5 introduces our lower bound method for quasi-level ordered proofs via the gauge measure, which we apply in Sect. 6 to a number of old and new QBF families. We conclude in Sect. 7 with some open questions.

2 Preliminaries

Propositional and Quantified Formulas. Variables and negated variables are called literals, i.e., for a variable x we can form two literals: x and its negation ${\bar{x}}$. We denote the corresponding variable as $\text {var}(x):=\text {var}({\bar{x}}):=x$.

A clause is a disjunction of literals, sometimes also viewed as a set of literals. The empty clause is the clause consisting of zero literals, denoted $(\bot )$. Terms are conjunctions of literals. Again, terms can be considered as sets of literals. A CNF (conjunctive normal form) is a conjunction of clauses. For $C=\ell _1\vee \ldots \vee \ell _m$ we define $\text {var}(C):=\{ \text {var}(\ell _1),\ldots ,\text {var}(\ell _m) \}$. For a CNF $\phi =C_1\wedge \ldots \wedge C_n$ we define $\text {var}(\phi ):=\bigcup _{i=1}^n\text {var}(C_i)$. A clause C is called tautological, if there is a variable x with $x,{\bar{x}}\in C$.

An assignment $\sigma$ of a set of variables X is a non-tautological set of literals, such that for all $x\in X$ there is $\ell \in \sigma$ with $\text {var}(\ell )=x$. The restriction of a clause C by an assignment $\sigma$ is defined as $C\vert _{\sigma }:=\top$ (true) if $C\cap \sigma \ne \emptyset$, and $\bigvee _{\ell \in C, \ell \not \in \sigma } \ell$ otherwise. One can interpret $\sigma$ as an operator that sets all literals from $\sigma$ to the Boolean constant 1. We denote the set of assignments of X by $\langle X\rangle$.

A QBF (quantified Boolean formula) $\Phi ={\mathcal {Q}}\cdot \phi$ is a propositional formula $\phi$ (also called matrix) together with a prefix ${\mathcal {Q}}$. A prefix $Q_1x_1Q_2x_2\ldots Q_kx_k$ consists of variables $x_1,\ldots ,x_k$ and quantifiers $Q_1,\ldots ,Q_k\in \{\exists ,\forall \}$. We obtain an equivalent formula if we unite adjacent quantifiers of the same type. Therefore we can always assume that our prefix is in the form of ${\mathcal {Q}}=Q'_1 X_1 Q'_2 X_2 \ldots Q'_s X_s$ with non-empty sets of variables $X_1,\ldots ,X_s$ and quantifiers $Q'_1,\ldots ,Q'_s\in \{\exists ,\forall \}$ such that $Q'_{i}\ne Q'_{i+1}$ for $i\in [s-1]$. For a variable x in ${\mathcal {Q}}$ we denote the quantifier level with respect to ${\mathcal {Q}}$ by $\text {lv}(x)=\text {lv}_{\Phi }(x)=i$, if $x\in X_i$. Variables from $\Phi$ are called existential, if the corresponding quantifier is $\exists$, and universal if the quantifier is $\forall$.

A QBF with CNF matrix is called a QCNF. We require that all clauses from a matrix of a QCNF are non-tautological, otherwise we just delete these clauses. We further require that all variables in the matrix appear in the prefix. Since we will only discuss refutational proof systems, we only consider false QCNFs.

A QBF can be interpreted as a game between two players $\exists$ and $\forall$. These players have to assign the respective variables one by one along the quantifier order from left to right. The $\forall$-player wins the game if and only if the matrix of the QBF gets falsified by this assignment. It is well known that for every false QBF $\Phi ={\mathcal {Q}}\cdot \phi$ there exists a winning strategy for the $\forall$-player.

Q-Resolution and Long-Distance Q-Resolution. Let $C_1$ and $C_2$ be two clauses. Let also $\ell$ be an existential literal with $\text {var}(\ell )\not \in \text {var}(C_1)\cup \text {var}(C_2)$. Then the resolvent of $C_1\vee \ell$ and $C_2\vee {\bar{\ell }}$ over $\ell$ is defined as

$$\begin{aligned} (C_1\vee \ell ){\mathop {\bowtie }\limits ^{\ell }}(C_2\vee {\bar{\ell }}) :=C_1\vee C_2 \text {.} \end{aligned}$$

Let $C:=u_1\vee \ldots \vee u_m\vee x_1\vee \ldots \vee x_n \vee v_1\vee \ldots \vee v_s$ be a clause from $\Phi$, where $u_1,\ldots ,u_m,v_1,\ldots ,v_s$ are universal literals, $x_1,\ldots ,x_n$ are existential literals and $v_1,\ldots ,v_s$ are exactly those literals $v\in C$ such that v is universal and $\text {lv}(v)>\text {lv}(x_i)$ for all $i\in [n]$. Then we can perform a reduction step and obtain

$$\begin{aligned} \text {red}(C):=(u_1\vee \ldots \vee u_m\vee x_1\vee \ldots \vee x_n)\text {.} \end{aligned}$$

For a CNF $\phi =\{ C_1,\ldots ,C_k \}$ we define $\text {red}(\phi ):=\{\text {red}(C_1),\ldots ,\text {red}(C_k) \}$.

Q-resolution [23] is a refutational proof system for false QCNFs. A Q-resolution proof $\pi$ of a clause C from a QCNF $\Phi ={\mathcal {Q}}\cdot \phi$ is a sequence of clauses $\pi =C_1,\ldots ,C_m$ with $C_m=C$. Each $C_i$ has to be derived by one of the following three rules:

• Axiom: $C_i\in \phi$;

• Resolution: $C_i= C_j{\mathop {\bowtie }\limits ^{x}}C_k$ for some $j,k<i$ and $x\in \text {var}_\exists (\Phi )$, and $C_i$ is non-tautological;

• Reduction: $C_i=\text {red}(C_j)$ for some $j<i$.

Note that none of our axioms are tautological by definition. A refutation of a QCNF $\Phi$ is a proof of the empty clause $(\bot )$.

To model clause learning in QCDCL, the proof system long-distance Q-resolution was introduced in [2, 36]. This extension of Q-resolution allows to derive universal tautologies under specific conditions such that the resulting system is still sound (this would not be the case for arbitrary tautologies). As in Q-resolution, there are three rules by which a clause $C_i$ can be derived. The axiom and reduction rules are identical to Q-resolution, but the resolution rule is changed to

• Resolution (long-distance): $C_i= C_j{\mathop {\bowtie }\limits ^{x}}C_k$ for some $j,k<i$ and $x\in \text {var}_\exists (\Phi )$. The resolvent $C_i$ is allowed to contain a tautology $u\vee {\bar{u}}$ if u is a universal variable. If $u\in \text {var}(C_j)\cap \text {var}(C_k)$, then we additionally require $\text {lv}(u)>\text {lv}(x)$.

Note that a long-distance Q-resolution proof without tautologies is just a Q-resolution proof.

If $\pi =C_1,\ldots ,C_m$ is a proof, we define a path in $\pi$ as a subsequence $C_{i_1},\ldots ,C_{i_s}$ of $\pi$, such that each $C_{i_j}$ is a parental clause of $C_{i_{j+1}}$ for each $j\in [s-1]$, either by resolution or reduction.

3 QCDCL as a Formal Proof System

In this section we review quantified conflict-driven clause learning (QCDCL) and its formalisation as a proof system from [7]. This provides the formal framework for our subsequent proof complexity analysis.

QCDCL is the quantified version of the well-known CDCL algorithm (see [29, 35] for further details on CDCL, and [19, 28, 36] for QCDCL). Let $\Phi ={\mathcal {Q}}\cdot \phi$ be a false QCNF. Roughly speaking, QCDCL consists of two interleaved processes: propagation and learning.

In the propagation process we generate assignments with the goal to either find a satisfying assignment or to obtain a conflict. We start with clauses from $\phi$ that force us to assign literals such that we do not falsify these clauses (called unit clauses). The underlying idea of this process is unit propagation. One can think of a clause $x_1\vee \ldots \vee x_n$ as an implication $({\bar{x}}_1\wedge \ldots \wedge {\bar{x}}_{n-1})\rightarrow x_n$. That is, if we already assigned the literals ${\bar{x}}_1,\ldots ,{\bar{x}}_{n-1}$, then we are forced to assign $x_n$ in order to satisfy this clause. In QBF, we also insert reduction steps into this process, i.e., we are interested in clauses that become unit after reduction. For example, the clause $({\bar{x}}_1\wedge \ldots \wedge {\bar{x}}_{n-1})\rightarrow (x_n\vee u)$ for an existential literal $x_n$ and a universal literal u with $\text {lv}(x_n)<\text {lv}(u)$ can also be used as a ground clause for propagating $x_n$.

Performing unit propagation, the goal is to prevent a conflict for as long as possible. However, it is not guaranteed that we can even perform any unit propagations by just starting with the formula. Therefore we will make decisions, i.e., we assign literals without any solid reason. With the aid of these decisions (one can also think of assumptions) we can provoke further unit propagations. Since decision making is one of the non-deterministic components of the algorithm, we only make decisions if there are no more unit propagations available. In QCDCL these decisions follow the quantification order, i.e., we always decide a variable from the leftmost quantifier block.

After obtaining a conflict, i.e., falsifying a clause, we start the clause learning process. Here the underlying idea is to use Q-resolution resp. long-distance Q-resolution. We start with the clause that caused the conflict and resolve it with clauses that implied previous literals in the assignment in the reverse propagation order. At the end we get a clause such that is derived from existing clauses by long-distance Q-resolution. We add the learned clause to $\phi$, backtrack to a state before we assigned all literals of this clause and restart the propagation process. The algorithm ends when we learn the empty clause $(\bot )$ and therefore obtain a refutation of $\Phi$.

QCDCL has to handle both refutations of false formulas as well as prove the validity of true formulas. Therefore one would additionally need to implement cube learning (or term learning) for satisfying assignments. Since we are only interested in refutations (otherwise we could not compare with Q-resolution), we will omit this aspect of QCDCL.

To prove rigorous lower bounds on the running time of QCDCL we cast QCDCL as a formal proof system. We recall the relevant details from [7], where we fully formalised all components of QCDCL. Each QCDCL run consists of backtracking steps and restarts. Between them we create trails, in which we store all information on decisions and unit propagations.

Definition 1

(trails, repeated from [7]) Let $\Phi ={\mathcal {Q}}\cdot \phi$ be a QCNF in n variables. A trail ${\mathcal {T}}$ for $\Phi$ is a sequence of literals (or $\bot$) of variables from $\Phi$ with some specific properties. We distinguish two types of literals in ${\mathcal {T}}$: decision literals, that can be both existential and universal, and propagated literals, that are either existential or $\bot$. We write a trail ${\mathcal {T}}$ as

$$\begin{aligned} {\mathcal {T}}=(p_{(0,1)},\ldots , p_{(0,g_0)};\mathbf {d_{1}},p_{(1,1)},\ldots ,p_{(1,g_{1})};\ldots ; \mathbf {d_r},p_{(r,1)},\ldots ,p_{(r,g_r)} )\text {,} \end{aligned}$$

where we denote decision literals by $d_i$ and propagated literals by $p_{(i,j)}$. We are not allowed to make a new decision unless there are no more propagations possible. Also, decision literals have to be level-ordered, i.e., we have to choose a leftmost quantified variable (still unassigned) as the next decision.

There are some further requirements on ${\mathcal {T}}$, for which we refer to [7]. However, as they are not crucial for our lower bounds, we can safely ignore them for now.

For unit propagation we need the notion of unit clauses that allow us to assign a variable without making a decision. We call a clause C a unit clause if $\text {red}(C)=(x)$ for an existential literal x or $x=\bot$.

The next definition presents the main framework for the analysis of QCDCL as a proof system. After having defined trails in a general way, we want to specify the way a trail can be generated during a QCDCL run.

Definition 2

(QCDCL proof systems [7]) Let $\Phi ={\mathcal {Q}}\cdot \phi$ be a QCNF. We call a triple of sequences

$$\begin{aligned} \iota =(({\mathcal {T}}_1,\ldots ,{\mathcal {T}}_m),(C_1,\ldots ,C_m),(\pi _1,\ldots ,\pi _m)) \end{aligned}$$

a $\mathsf{{QCDCL}}$ proof from $\Phi$ of a clause C, if for all $i\in [m]$ the trail ${\mathcal {T}}_i$ uses the QCNF ${\mathcal {Q}}\cdot (\phi \cup \{ C_1,\ldots ,C_{i-1} \})$, where $C_j$ is a clause learnable from ${\mathcal {T}}_j$ and $C_m=C$. Each $\pi _i$ is the long-distance Q-resolution derivation of the clause $C_i$ from ${\mathcal {Q}}\cdot (\phi \cup \{ C_1,\ldots ,C_{i-1} \})$ that we learned from the trail ${\mathcal {T}}_i$.

Between two trails ${\mathcal {T}}_i$ and ${\mathcal {T}}_{i+1}$ we backtrack to some point which we can choose freely. Backtracking to the start (before any variable was assigned) is called restarting. If $C=(\bot )$ we call $\iota$ a refutation.

By sticking together $\pi _1,\ldots ,\pi _m$, we obtain a long-distance Q-resolution derivation $\pi$ of C from $\Phi$. We identify $\mathsf{{QCDCL}}$ proofs with this exact $\pi$.

We require that all trails are naturally created, which means that we are not allowed to skip unit propagations if they are possible, as we explained before. A more detailed description of this condition is given in [7].

We remark that though QCDCL proofs are basically long-distance Q-resolution derivations (i.e., QCDCL is simulated by long-distance Q-resolution), these system are not equal as QCDCL imposes a particular structure on long-distance Q-resolution proofs. Indeed, long-distance Q-resolution is exponentially stronger than QCDCL (cf. [7]).

4 Quasi Level-Ordered Proofs

For the remainder of this article we will entirely focus on $\Sigma _3^b$ formulas and throughout fix the prefix $\exists X\forall U \exists T$, where X, U, and T are pairwise disjoint and non-empty sets of variables.

Our ultimate aim will be to develop a lower bound technique for such formulas for QCDCL. Conceptually, our technique is inspired by an approach for level-ordered proofs, which is why we recall that notion from [22].

Definition 3

( [22]) A long-distance Q-resolution proof $\pi$ from a QCNF $\Phi$ of a clause C is called level-ordered if for each path P in $\pi$ and two resolution steps in P over variables $\ell _1$ and $\ell _2$ the following holds: if the resolution over $\ell _1$ is closer to the root C than the resolution over $\ell _2$, then $\text {lv}(\ell _1)\le \text {lv}(\ell _2)$.

For level-ordered proofs one can devise lower bounds as follows. A level-ordered long-distance Q-resolution refutation $\pi$ of a $\Sigma _3^b$-formula $\Phi =\exists X\forall U \exists T\cdot \phi$ always starts with T-resolutions and ends with X-resolutions. We then count the clauses consisting only of X-literals at the transitions from a T-resolution to some X-resolution. For each $\tau \in \langle X\rangle$ we can find such a clause $C_\tau$ that is falsified by $\tau$. Note that $C_\tau$ does not necessarily need to contain literals from all X-variables. Hence, the $C_\tau$ clauses do not need to be pairwise distinct. However, we will show that for a particular class of formulas, the $C_\tau$ clauses cannot be too small. Therefore each $C_\tau$ can only cover few assignments $\tau '\in \langle X\rangle$ and the number of these clauses is still exponential.

We will use this idea in a more general setting by introducing the notion of quasi level-ordered proofs where only the existence of these $C_\tau$ is required.

Definition 4

A long-distance Q-resolution refutation $\pi$ of a $\Sigma _3^b$ formula with prefix $\exists X\forall U \exists T$ is called quasi level-ordered, if for each assignment $\tau \in \langle X\rangle$ there exists an X-clause $C_\tau$ which is falsified by $\tau$ and the subproof $\pi _{C_\tau }\subseteq \pi$ of $C_\tau$ is level-ordered.

Clearly, level-ordered proofs are quasi level-ordered, but the converse does not hold in general.

In Sect. 5 we will devise a lower bound technique for quasi level-ordered proofs. To get the connection to QCDCL, we show that each QCDCL refutation of $\Sigma _3^b$ formulas with a special property can be efficiently transformed into a quasi level-ordered Q-resolution refutation. The property needed is the XT-property, which we recall from [7].

Definition 5

[7] Let $\Phi$ be a QCNF of the form $\exists X \forall U \exists T \cdot \phi$. We call a clause C in the variables of $\Phi$

• X-clause, if $\text {var}(C)\cap U=\emptyset$ and $\text {var}(C)\cap T=\emptyset$,

• T-clause, if $\text {var}(C)\cap X=\emptyset$, $\text {var}(C)\cap U=\emptyset$ and $\text {var}(C)\cap T\ne \emptyset$,

• XT-clause, if $\text {var}(C)\cap X\ne \emptyset$, $\text {var}(C)\cap U=\emptyset$ and $\text {var}(C)\cap T\ne \emptyset$,

• XUT-clause, if $\text {var}(C)\cap X\ne \emptyset$, $\text {var}(C)\cap U\ne \emptyset$ and $\text {var}(C)\cap T\ne \emptyset$.

We say that $\Phi$ fulfils the XT-property if $\phi$ contains no XT-clauses as well as no unit T-clauses and there do not exist two T-clauses $C_1,C_2\in \phi$ that are resolvable.

Intuitively, this says that there is no direct connection between the X- and T-variables, i.e., $\Phi$ does not contain clauses with X- and T-variables, but no U-variables. This XT-property allows us to prove several properties regarding QCDCL refutations.

Lemma 1

[7] Let $\Phi$ be a QCNF that fulfils the XT-property. Then the following holds:

• 1. It is not possible to derive XT-clauses by long-distance Q-resolution.

• 2. It is not possible to resolve two XUT-clauses over an X-literal in a QCDCL proof.

• 3. Each QCDCL refutation of $\Phi$ is a Q-resolution refutation (not just a long-distance Q-resolution refutation).

For the following results, we will always assume that all three properties from Lemma 1 are fulfilled.

Now we will work towards the transformation of QCDCL proofs into quasi level-ordered Q-resolution refutations. This transformation is described as an algorithm in the following theorem.

Intuitively, the algorithm takes as input a long-distance Q-resolution refutation $\pi$ that was extracted from a QCDCL refutation of a QCNF that fulfils the XT-property (which is a Q-resolution proof by Lemma 1) and adds a polynomial number of clauses (and resolution steps), such that the obtained proof is quasi level-ordered (i.e., it contains $C_\tau$ for each $\tau \in \langle X \rangle$, c.f. Definition 4). The idea is that all the $C_\tau$ from the definition of quasi level-ordered proofs are already somehow contained in $\pi$, but they might be hidden in XUT-clauses. The algorithm detects these XUT-clauses and eliminates the T-literals by changing the order of resolutions over X-variables and T-variables. Note that the algorithm will only add parts to $\pi$ and never delete anything (which is fine as quasi level-ordered proofs only require the existence of these $C_\tau$, but they do not need to contribute to the refutation).

Initially, the algorithm checks if the proof of the empty clause is already level-ordered. If this is the case, then there is nothing to do as the technique for level-ordered proofs can be applied to find the $C_\tau$ as described above. Otherwise, the algorithm tries to find the last step in the subproof of the currently considered clause (which is the empty clause at the beginning, but might also be non-empty later) that violated the level order, which is just the last transition from an X-resolution to a T-resolution. When this transition is found, we have detected a path of clauses that starts with one X-resolution step, followed by an at most linear-sized sequence of T-resolution steps that ends with the currently considered clause (c.f. Figure 2). The algorithm then adds a path of at most the same size that now starts with all the T-resolutions and reductions (we do not need the X-resolution at the and as we are only interested in parent clauses, c.f. Figure 4). Constructing this path can be done easily because we know (by Lemma 1) that only one of the two clauses from the X-resolution step can contain T-literals (if were both did, we would need to create two new paths, and therefore potentially increase the proof size exponentially in further loops). The clause at the end of this added path as well as the X-clause from the previous X-resolution step will become an observed clause for the next loop (in Figure 4, these would be clauses $C_{a'_{k+1}}$ and $C_e$).

Theorem 2

Let $\Phi$ be a $\Sigma _3^b$ QCNF that fulfils the XT-property. Then, using Algorithm 1, each QCDCL refutation $\pi$ (or more formally, the extracted Q-resolution refutation $\pi$) of $\Phi$ can be efficiently transformed into a quasi level-ordered Q-resolution refutation $\pi '$ of $\Phi$ with $\vert \pi '\vert \in {\mathcal {O}}(\vert \pi \vert ^4)$.

Proof

First, because of the XT-property each refutation extracted from a QCDCL run is also a Q-resolution refutation (cf. Lemma 1). That means we will only consider $\pi$ as the Q-resolution proof that was extracted from the QCDCL run.

Let $\pi =C_1,\ldots ,C_m=\bot$. Note that clauses could occur more than once in a proof since we cannot simply shorten a proof in QCDCL. Hence we will use indices to identify clauses in a proof. Each index not only determines the clause itself, but also its position in the proof. This is the reason why we will only use indices in the algorithm in order to store information about a particular clause.

Technically, we define an order that will help us determine if a resolution $C_d{\mathop {\bowtie }\limits ^{}} C_e$ takes place before or after another resolution $C_{d'}{\mathop {\bowtie }\limits ^{}} C_{e'}$ in a given proof.

For this we define a total order $\preccurlyeq$ on $\{ \{ d,e \}:\, d,e\in \mathbb {N},\, d\ne e \}$ as follows:

$$\begin{aligned} A\preccurlyeq B \Leftrightarrow \max A<\max B \text { or } ( \max A=\max B \text { and } \min A\le \min B ). \end{aligned}$$

We use the notation $A\prec B$ for $A\preccurlyeq B$ and $A\ne B$.

We sketch how the transformation (Algorithm 1) works: Throughout the whole process we work with two sets $M_X$ and $M_{XUT}$. The set $M_X$ contains indices of X-clauses, where initially we start with $M_X=\{m\}$ (remember that $C_m=(\bot )$). For each $c\in M_X$ we check whether the clause $C_c$ has a level-ordered subproof. If the subproof is not level-ordered, and if the last step before $C_c$ (i.e., the last step in the subproof $\pi _{C_c}$) was an X-resolution, we just add the indices both parent clauses of $C_c$ to $M_X$ and delete c from it. Otherwise, if the subproof is not level-ordered, but the last step before $C_c$ was no X-resolution, we search for the last transition that violates the level-order condition. This must be a transition from an X-resolution to a T-resolution. After this transition there will be only T-resolutions until we reach $C_c$. One of the parent clauses of this X-resolution, which we call $C_d$ and $C_e$, is an X-clause and the other one is an XUT-clause due to the XT-property (Lemma 1). The index of the X-clause (either d or e) is again stored in $M_X$, while we delete c from $M_X$. However, for the XUT-clauses, which are stored as triples (d, e, c) in $M_{XUT}$ (where $C_d$ is the XUT-clause), we have to add several clauses to the proof, including a new X-clause $C_{a'}$. This clause $C_{a'}$ is then added to $M_X$ as well, and the loop repeats until there are no more clauses in $M_X$ left. Note that these added clauses will be part of a dead end in the proof and therefore are not necessary for the refutation itself. However, we need these new clauses for a counting argument in our lower bound technique.

We will show that at the end we return a proof that is quasi level-ordered. More specifically, the X-clauses we detect during the run whose subproofs are level-ordered will be exactly the clauses $C_\tau$ from the definition of quasi level-ordered proofs. This holds because, starting from the empty clause, whenever we detect an X-resolution we can choose which parent clause we will consider next. Hence we can choose the polarity of the X-variable we resolve over in the current step. At the end, this last X-clause (whose subproof is level-ordered) only consists of variables with the right polarity as previously chosen. Figure 1 depicts how the algorithm transforms a proof. $\square$

Fig. 1

Sketch of the functionality of the algorithm. Below each clause $C_j$ we specify the type of clause (X- or XUT-clause). Newly added parts are coloured red. Triangles labeled with “l.-o.” are level-ordered subproofs, otherwise they are not level-ordered and we can find a transition from an X-resolution to a T-resolution. The corresponding clause $C_c$ is then one of the $C_\tau$ clauses for a particular $\tau$

Claim 1

Each step is well-defined and the algorithm terminates.

Proof

Let us consider the first inner while loop from line 7 to 32. For each $c\in M_X$ that we delete during the loop, we will add d and e (or sometimes only one of them) to $M_X$ such that $C_d$ and $C_e$ both have smaller depth than $C_c$. Therefore this loop will repeat only finitely often.

Note that for each Q-resolution proof that is not level-ordered, we can find at least one transition from an X-resolution to a T-resolution. Because of the XT-property, we do not have any XT-clauses and also no X-resolutions over two XUT-clauses. The only two remaining possibilities for X-resolutions are between two X-clauses or between an XUT-clause and an X-clause. Let $C_d{\mathop {\bowtie }\limits ^{x}}C_e=C_f$ and $C_f{\mathop {\bowtie }\limits ^{t}} C_g=C_j$ be the transition we detected in the algorithm and as sketched in Figure 2. The case where both $C_d$ and $C_e$ are X-clauses is impossible since the next step is a T-resolution. So we can assume that we find an XUT- and an X-clause. For each $(d,e,c)\in M_{XUT}$ we have that $C_d$ is the XUT-clause and $C_e$ is the X-clause.

Fig. 2

Last transition from an X-resolution $C_d{\mathop {\bowtie }\limits ^{x}}C_e$ to a T-resolution $C_f{\mathop {\bowtie }\limits ^{t}}C_g$ in the subproof of $C_c$ as it is detected in line 12 of Algorithm 1. The literals x (resp. t) are placeholders for some X- (resp. T-)literals. The T-literals do not need to be equal

Fig. 3

Suppose that after the detected $\{d,e\}$ there is another set $\{ d',e' \}$ which initializes a transition from an X-resolution to a T-resolution. However, this would contradict the maximality of $\{d,e\}$ since we would have $d'>\max \{d,e\}$ and therefore $\{d,e\}\prec \{d',e'\}$

There cannot be another transition from an X-resolution to a T-resolution on a path downwards starting with the above transition since this would contradict the maximality of $\{d,e\}$, cf. Figure 3. Hence the found transition is indeed the (or “a”) last one.

In the second inner while-loop from line 25 to 31 we will add only finitely many new clauses to the proof. Note that all added clauses are inserted after the original clauses. Since we have only added finitely many triples to $M_{XUT}$ until this point, we will repeat this loop only finitely often, as well.

Let us now concentrate on the outer loop from line 6 to 36. We will show that this loop will repeat only $\vert \pi \vert ^2$ times.

For each iteration i let

$$\begin{aligned} K_i:=\text {max}_\preccurlyeq \{ \{ d,e \}:\, (d,e,c)\in M_{XUT}^{(i)} \text { for some index} c\in \mathbb {N}\}. \end{aligned}$$

For each $(d_i,e_i,c_i)\in M_{XUT}^{(i)}$ let $c'_i$ be the index $a'_{k+1}$ of the clause we add to $\pi '$ corresponding to $(d_i,e_i,c_i)$ as described in the algorithm. If these $c'_i$ are contained in a triple in the next $M_{XUT}^{(i+1)}$, say $(d_{i+1},e_{i+1},c'_i)\in M_{XUT}^{(i+1)}$, then we have $\{d_{i+1},e_{i+1} \}\prec \{ d_i,e_i \}$. We cannot have $\{d_i,e_i\}=\{d_{i+1},e_{i+1} \}$ simply due to the fact that $c'_i$ has no path to the resolution $C_{d_i}{\mathop {\bowtie }\limits ^{x}}C_{e_i}$ since we skipped the resolution with $C_{e_i}$. We cannot get $\{ d_i,e_i \}\prec \{d_{i+1},e_{i+1} \}$ either because otherwise we would have chosen $\{d_{i+1},e_{i+1}\}$ instead of $\{ d_i,e_i \}$ when we considered $c_i$ in the iteration before.

We conclude that we have $K_{i+1}\prec K_i$ for each iteration i. These $K_i$ are sets consisting of indices from original clauses since the corresponding clauses $C_d$ and $C_e$ got resolved over an X-variable in $\pi$ and new clauses that were added during the run of the algorithm appear only in resolution steps over T-literals and reduction steps. Hence we can argue that we will repeat the outer while-loop at most $\vert \pi \vert ^2$ times. $\square$

Claim 2

At the end, we have $\vert \pi '\vert \in {\mathcal {O}}(\vert \pi \vert ^4)$.

Proof

We have to count the number of clauses we add to $\pi '$ in each iteration. A visualization of this part of the algorithm can be seen in Figure 4. Let $\pi '_{(q)}$ be the current proof $\pi '$ after the $q^\text {th}$ time we added a path to $\pi '$ in line 28. For each q we prove by induction that each possible path in $\pi '_{(q)}$ has at most length $\vert \pi \vert$. For $q=0$ this is trivial since $\pi '_{(0)}=\pi$. Let the statement be true for $\pi '_{(q)}$ and consider the case $\pi '_{(q+1)}$. Let $C_{j_1},\ldots ,C_{j_\ell }$ be a path in $\pi '_{(q+1)}$. If all of these clauses were already contained in $\pi '_{(q)}$, then the result follows immediately. Therefore let us suppose the path contains some clauses we have newly added, say that $C_{j_p}$ is the leftmost new clause compared to $\pi '_{(q)}$. But then all clauses $C_{j_p},C_{j_{p+1}},\ldots ,C_{j_\ell }$ are new clauses as well, since each new clause in inserted at the end of the proof. By the method we constructed the clauses $C_{j_p},C_{j_{p+1}},\ldots ,C_{j_\ell }$ in line 28, we conclude that these clauses are some of the $C_{a'_2},\ldots ,C_{a'_{k+1}}$, say $C_{a'_v},\ldots ,C_{a'_w}$. But then we can find another path $C_{j_1},\ldots ,C_{j_{p-1}},C_{a_v},\ldots ,C_{a_w}$ (we have to set $C_{a_w}:=C_c$ if $w=k+1$ and we have to insert $C_{a_1}$ after $C_{j_{p-1}}$ if $C_{j_{p-1}}=C_d$), which has the same (or even a greater) length as the original path and is completely contained in $\pi '_{(q)}$. Hence, the original path has length at most $\vert \pi \vert$.

Fig. 4

Visualization of lines 29 and 28 in Algorithm 1. Newly added clauses and resolutions are coloured in red

All in all, for each $(d,e,c)\in M_{XUT}$ we will add a path of length at most $\vert \pi \vert$. Each c can occur only once in the triples in $M_{XUT}$. After we added the path corresponding to (d, e, c), we can ignore potential future occurrences of (d, e, c) if this particular triple is detected more than once in the algorithm.

We want to show next that in each outer while-loop we will only add at most $\vert \pi \vert ^2$ new clauses (resp. at most $\vert \pi \vert$ paths) to $\pi '$. The number of added paths in the $i^\text {th}$ loop is determined by $\vert M_{XUT}^{(i)}\vert$. For $i=1$ this is obvious, as all c in $(d,e,c)\in M_{XUT}^{(1)}$ are indices from original formulas and each c appears only once. Let us now assume that $M_{XUT}^{(i')}\le \vert \pi \vert$ for all $i'<i$. Then we have to show that $M_{XUT}^{(i)}\le \vert \pi \vert$ holds as well (for $i>1$). For each $(d,e,c)\in M_{XUT}^{(i)}$ it holds that c is either some $a'_{k+1}$ from the loop before, or an original index (because d and e from the upper loop can only be original indices, the only newly added index that can be contained in a triple is the last index $a'_{k+1}$ of some added path from the loop before), from which we get $\vert M_{XUT}^{(i)}\vert \le \vert \pi \vert + \vert M_{XUT}^{(i-1)}\vert$. However, if the c from some $(d,e,c)\in M_{XUT}^{(i)}$ was original, it must hold $(d,e,c)\not \in M_{XUT}^{(i-1)}$. In fact, c cannot be included in any triple $(d',e',c)$ from any $M_{XUT}^{(i')}$ for $i'<i$. Therefore, we can restrict the above inequality even more: $\vert M_{XUT}^{(i)}\vert \le (\vert \pi \vert - \vert M_{XUT}^{(i-1)}\vert )+\vert M_{XUT}^{(i-1)}\vert =\vert \pi \vert$.

We conclude that in each outer while-loop we will add at most $\vert \pi \vert ^2$ new clauses to $\pi '$. Since we will repeat the outer loop at most $\vert \pi \vert ^2$ times, the new proof $\pi '$ will at the end consist of at most ${\mathcal {O}}(\vert \pi \vert ^4)$ clauses. $\square$

Claim 3

$\pi '$ is quasi level-ordered.

Proof

Let us now briefly explain why the obtained proof is quasi level-ordered. We just have to argue how one can find the clauses $C_\tau$ for each $\tau \in \langle X \rangle$. For this, we simply backtrace all steps of the algorithm. We start by defining the empty clause as the currently considered clause and check whether its derivation is already level-ordered. If this is the case, then we can set $C_\tau = (\bot )$ for all $\tau$. Otherwise, we look at the last X-resolution step in this derivation and find two clauses that got resolved over X. Depending on the choice of $\tau$, we choose that clause which contains the X-literal that got negated by $\tau$ (as we would do in the level-ordered setting). At most one of these clauses is an XUT-clause (by Lemma 1), the other one is always an X-clause. If the clause we have chosen is the X-clause, then this X-clause becomes the new currently observed clause. If it is the XUT-clause, then we go into the path that was newly added during the algorithm and its last clause becomes the next observed clause (which, by construction, consists of the same X-literals as the original XUT-clause, but lacks all T- and U-literals). That also means that all observed clauses are X-clauses.

This search ends as soon as we get to a clause that has a level-ordered derivation, which will happen when the considered clause is an axiom (in the worst case). Because we always chose the right X-literal, this clause will be falsified by $\tau$ and can therefore serve as $C_\tau$.

In more detail, we prove that the clauses we have added to L are exactly the clauses $C_\tau$ from the definition of quasi level-ordered proofs. Let us fix an assignment $\tau \in \langle X\rangle$. Starting from $C_m=(\bot )$, for each X-clause $C_c$ we check if the subproof $\pi '_{C_c}$ is level ordered. If it is not, we can find clauses $C_d,C_e\in \pi '_{C_c}$ as described in the algorithm that are resolved over an X-literal x. We pick the clause which contains x if $\tau (x)=0$ and the other clause otherwise. If we pick an XUT-clause, say $C_d$, then we have to jump to the corresponding X-clause $C_{a'_{k+1}}$ which we have added when we chose $(d,e,c)\in M_{XUT}$ in the second inner while-loop. Note that $C_{a'_{k+1}}$ is a subclause of $C_c\vee x$ (resp. $C_c\vee {\bar{x}}$) since we only omitted the resolution with $C_e$ over x. We continue by checking the subproof of $C_d$ (resp. $C_e$ or $C_{a'_{k+1}}$).

At the end, when the X-clause $C_c$ has finally a level-ordered subproof $\pi '_{C_c}$, we will stop there and we set $C_\tau :=C_c$ since we have $\tau (x)=0$ for each $x\in C_c$. Therefore $C_\tau$ is falsified by $\tau$. $\square$

Algorithm 1 can be easily modified to also transform long-distance Q-resolution refutations by adding more case distinctions to line 16. However, this might lead to an exponential blow up.

We give an example of a formula with a refutation which we transform into a quasi level-ordered refutation.

Example 1

Let $\Psi$ be the QCNF with prefix $\exists X\forall U \exists T$ with $X=\{x,y\}$, $U=\{u\}$, $T=\{s,t\}$ and the matrix

$$\begin{aligned}&(u\vee {\bar{s}})\wedge (x\vee u \vee s)\wedge ({\bar{u}}\vee {\bar{s}})\wedge (y\vee u\vee s)\wedge ({\bar{x}}\vee u\vee {\bar{s}})\wedge (x\vee {\bar{y}})\\ {}&\wedge (y\vee u\vee t)\wedge ({\bar{s}}\vee {\bar{t}}). \end{aligned}$$

Further, let $\pi$ be the Q-resolution refutation of $\Psi$ as represented in Figure 5. We want to transform this proof $\pi$ to a quasi level-ordered proof $\pi '$ by carrying out the instructions as described in the algorithm. Note that for the sake of simplicity this proof is exceptionally not necessarily a QCDCL proof since finding a QCDCL proof that is representative enough to serve as an example is not a trivial thing to do. However, $\pi$ fulfils at least the properties we need in order to get polynomially transformed, namely we never resolve two XUT-clauses over X. Also, $\pi$ is most likely not the shortest possible refutation of $\Psi$, as one can see that the clause $C_6=y\vee u\vee s$ is derived although $C_1=y\vee u\vee s$ is an axiom clause.

Fig. 5

Q-resolution refutation of $\Psi$

First, we have $M_X=\{16\}$ and $M_{XUT}=\emptyset$. The proof of $C_{16}$, which is just $\pi$ itself, is obviously not level ordered. The last transition from an X-resolution to a T-resolution is at $C_{11}{\mathop {\bowtie }\limits ^{y}}C_{12}=C_{13}$ to $C_{13}{\mathop {\bowtie }\limits ^{s}} C_{14}=C_{15}$. Since the last step in $\pi$ was no X-resolution, we have to add the triple (12, 11, 16) to $M_{XUT}$ (note that the first number of the triple has to be the index of the XUT-clause). Further, we add 11 to $M_X$ and delete 16 from it. The subproof $\pi _{C_{11}}$ of $C_{11}$ is not level-ordered either. The last X- to T-transition in $\pi _{C_{11}}$ is $C_4{\mathop {\bowtie }\limits ^{x}}C_5=C_6$ to $C_6{\mathop {\bowtie }\limits ^{s}}C_7=C_8$. Because the last step in $\pi _{C_{11}}$ was a reduction and no X-resolution, we have to add (5, 4, 11) to $M_{XUT}$ and replace 11 with 4 in $M_X$. Now, the subproof $\pi _{C_{4}}$ is level-ordered, so we can add 4 to L and delete it from $M_X$. Because $M_X$ is now empty, we can continue by adding new clauses to $\pi$.

Fig. 6

Adding the new path of clauses corresponding to $(12,11,16)\in M_{XUT}$

First, we add the clauses $C_{17}=C_{12}{\mathop {\bowtie }\limits ^{s}}C_{14}$ and $C_{18}=\text {red}(C_{17})$. This new path, which can be seen in Figure 6, corresponds to the triple (12, 11, 16), that can now be deleted from $M_{XUT}$. After this we have to add 18 to $M_X$ and continue with the next available triple from $M_{XUT}$, which is (5, 4, 11). We add the clauses $C_{19}=C_5{\mathop {\bowtie }\limits ^{s}}C_7$, $C_{20}=C_9{\mathop {\bowtie }\limits ^{t}}C_{19}$ and $C_{21}=\text {red}(C_{20})$ to $\pi$, delete (5, 4, 11) from $M_{XUT}$ and add 21 to $M_X$. After that, $M_{XUT}$ is empty and $M_X=\{18,21\}$.

In the next iteration, we have to consider the subproofs $\pi _{C_{18}}$ and $\pi _{C_{21}}$, which are luckily both level-ordered. That means we can immediately delete both 18 and 21 from $M_X$ and add them to L. Both $M_X$ and $M_{XUT}$ are now empty and hence our algorithm terminates. Our new proof $\pi '$ which is represented in Figure 7 is now quasi level-ordered. The clauses whose indexes are contained in L are exactly the clauses $C_\tau$ we need for a quasi level-ordered proof. More precisely, $L=\{4,18,21\}$ with $C_{x\mapsto 1,y\mapsto 1}=C_{x\mapsto 0,y\mapsto 1}=C_{18}=({\bar{y}})$, $C_{x\mapsto 1,y\mapsto 0}=C_4={\bar{x}}\vee y$ and $C_{x\mapsto 0,y\mapsto 0}=C_{21}=x\vee y$.

Fig. 7

Adding the new path of clauses corresponding to $(5,4,11)\in M_{XUT}$. This new proof $\pi '$ is now quasi level-ordered

5 A Lower Bound Technique via Gauge

Now that we have proven that QCDCL is simulated by quasi level-ordered proofs, we continue by introducing a measure for $\Sigma _3^b$ QCNFs that will provide an exponential lower bound for quasi level-ordered refutations of these formulas.

Definition 6

For a $\Sigma _3^b$ QCNF $\Phi$ with prefix $\exists X \forall U \exists T$ let $W_\Phi$ be the set of all Q-resolution derivations $\pi$ from $\Phi$ of some X-clause such that $\pi$ only contains T-resolution and reduction steps. We define the gauge of $\Phi$ as

$$\begin{aligned} \text {gauge}(\Phi ):=\min \{\vert C\vert :\,C\text { is the root of some }\pi \in W_\Phi \}. \end{aligned}$$

Intuitively, $\text {gauge}(\Phi )$ is the minimal number of X-literals that are necessarily piled up in a level-ordered Q-resolution derivation in which we want to get rid of all T-literals (hence we consider proofs of X-clauses).

Before showing how gauge lower bounds imply proof size lower bounds let us consider an example for which we recall the formulas from [22].

Definition 7

( [22]) The QCNF consists of the quantifier prefix

$$\begin{aligned} \exists x_{(1,1)},\ldots ,x_{(1,n)},x_{(2,1)},\ldots ,x_{(2,n)}, \ldots ,x_{(n,1)},\ldots ,x_{(n,n)}\forall u \exists s_1,\ldots ,s_n,t_1,\ldots ,t_n \end{aligned}$$

and matrix clauses $(x_{(i,j)}\vee u\vee s_i)$, $({\bar{x}}_{(i,j)}\vee {\bar{u}}\vee t_j)$ for $i,j\in [n]$ as well as $\bigvee _{i\in [n]}{\bar{s}}_i$ and $\bigvee _{i\in [n]}{\bar{t}}_i$.

The formulas describe a ‘completion’ game on an $(n\times n)$-matrix (cf. [22]): The universal player has to set u to false iff for all i there exists a j such that $x_{(i,j)}$ is set to false. Otherwise, there exists an i such that for each j, the literal $x_{(i,j)}$ is set to true, hence the universal player has to set u to true.

It is readily checked that the formulas fulfil the XT-property. We can now compute their gauge. Note that according to our convention, the T-variables comprise of all variables $s_1,\ldots ,s_n, t_1,\dots , t_n$.

Lemma 3

We have .

Proof

Since there are no X-clauses as axioms, we necessarily need to resolve over T somehow. For this we need T-literals of negative polarity, hence each contains $\bigvee _{i\in [n]}{\bar{s}}_i$ or $\bigvee _{i\in [n]}{\bar{t}}_i$. In each every T-literal has to be resolved away. For this reason we need the corresponding clauses $x_{(i,j)}\vee u\vee s_i$ or ${\bar{x}}_{(i,j)}\vee {\bar{u}}\vee t_j$. Because we cannot resolve over X in , there are at least n X-literals that are piled up and therefore . $\square$

Towards our lower bound technique we now estimate the size of derivations of X-clauses in terms of gauge.

Lemma 4

Let $\Phi$ be a $\Sigma _3^b$ QCNF. Let $\pi$ be a level-ordered Q-resolution proof from $\Phi$ of a non-tautological X-clause D with $\vert D\vert =c$. Then $\vert \pi \vert \ge 2^{\text {gauge}(\Phi )-c}$.

Proof

Let $V:=X\backslash \text {var}(D)$. For each assignment $\tau \in \langle V\rangle$ we will find a path $P_\tau$ in $\pi$ by going backwards starting from D. For each resolution step over some $x\in V$ we choose the path whose literals are negated by $\tau$, hence we choose the clause that contains x if $\tau (x)=0$ and the other clause otherwise. If there are resolution steps over variables from $\text {var}(D)$, then we will always choose the literal from D. If we reach a reduction step, we will just expand the path by this one parental clause. If we detect a resolution step over a T-literal, we stop there.

Let $C_\tau$ be the clause at which we stop. Clearly, the subproof $\pi _{C_\tau }$ of $C_\tau$ is one of the derivations in $W_\Phi$, hence $\vert C_\tau \vert \ge \text {gauge}(\Phi )$. Then $C_\tau$ has to be a non-tautological X-clause with at least $\text {gauge}(\Phi )$ different X-literals. Then $C_\tau$ contains at least $\text {gauge}(\Phi )-c$ different X-literals whose variables are in V. These literals are negated by the assignment $\tau$.

Now let a be the number of these clauses $C_\tau$ by summing over all $\tau$. Since for each $C_\tau$ there are at most $\vert X\vert -\text {gauge}(\Phi )$ variables that are not contained as some literal in the clause, there are at most $2^{\vert X\vert -\text {gauge}(\Phi )}$ paths that can lead to each $C_\tau$. Multiplying with the number of $C_\tau$ gives us at least the number of assignments $\tau \in \langle V\rangle$, hence

$$\begin{aligned}&2^{\vert X\vert -\text {gauge}(\Phi )}\cdot a\ge 2^{\vert X\vert -c}\\ \Leftrightarrow \;&a\ge 2^{\vert X\vert -c}/2^{\vert X\vert -\text {gauge}(\Phi )}=2^{\text {gauge}(\Phi )-c}\text {.} \end{aligned}$$

Since each $C_\tau$ is a clause from $\pi$, we get $\vert \pi \vert \ge a\ge 2^{\text {gauge}(\Phi )-c}$. $\square$

Note that the bound from Lemma 4 is an exact lower bound (no asymptotics involved). We will now use Lemma 4 to get a lower bound for quasi level-ordered Q-resolution refutations. We will do this with a similar counting argument as in Lemma 4 by counting the number of clauses $C_\tau$ in quasi level-ordered proofs.

Proposition 5

Each quasi level-ordered Q-resolution refutation of a $\Sigma _3^b$ QCNF $\Phi$ has size $2^{\Omega (\text {gauge}(\Phi ))}$.

Proof

Let $\pi$ be the shortest quasi level-ordered refutation of $\Phi$. By the definition of quasi level-ordered proofs we can find clauses $C_\tau$ for each $\tau \in \langle X\rangle$.

Let $h:=\min _{\tau \in \langle X \rangle }{\vert C_\tau \vert }$. By Lemma 4 we get $\vert \pi \vert \ge 2^{\text {gauge}(\Phi )-h}$, hence $h\ge \text {gauge}(\Phi )- \log \vert \pi \vert$. Each clause $C_\tau$ can have at most $2^{\vert X\vert -h}$ assignments $\alpha \in \langle X\rangle$ such that $C_\alpha =C_\tau$. Let $a:=\vert \{ C_\tau :\, \tau \in \langle X\rangle \}\vert$, then $a\cdot 2^{\vert X\vert -h}\ge 2^{\vert X\vert }$ and thus

$$\begin{aligned} \vert \pi \vert \ge a\ge 2^h\ge 2^{\text {gauge}(\Phi )-\log \vert \pi \vert }=\frac{2^{\text {gauge}(\Phi )}}{\vert \pi \vert }. \end{aligned}$$

We conclude that $\vert \pi \vert ^2\in 2^{\Omega (\text {gauge}(\Phi ))}$. $\square$

We combine Theorem 2 and Proposition 5 above and obtain a lower bound for QCDCL on formulas with the XT-property.

Theorem 6

Each QCDCL refutation of a $\Sigma _3^b$ QCNF $\Phi$ that fulfils the XT-property has size $2^{\Omega (\text {gauge}(\Phi ))}$.

6 Applications of the Lower Bound Technique

We now apply our new lower bound technique via gauge to show exponential lower bounds for QCDCL proof size (and thereby for QCDCL running time) for a number of QBF families. First, by combining Lemma 3 with Theorem 6 we obtain hardness for the formulas from [22].

Corollary 7

The formulas require exponential-size proofs in QCDCL.

With this result we gain an improved separation between Q-resolution and QCDCL. It was already shown in [7] that Q-resolution and QCDCL are incomparable. This involves constructing QBFs that are easy for QCDCL, but hard for Q-resolution, and vice versa. One direction is shown via the formulas (Definition 9 below), which are hard for Q-resolution [14], but easy in QCDCL [7]. For the other direction, [7] used the Trapdoor [7] and Lonsing formulas [28], both of which are easy for Q-resolution, but hard for QCDCL. However, both QBF families incorporate the propositional pigeonhole principle (PHP) and the hardness of these formulas for QCDCL rests entirely on the hardness of PHP for propositional resolution [20]. This is somewhat unsatisfactory, as the hardness results do not refer to quantification and in particular do not hold in the presence of NP oracles (cf. [11, 25] for a detailed formal account on how to equip QBF proofs with NP oracles or equivalently QBF solving with SAT calls).

Our improved separation is shown in Corollary 7 above, as these formulas are hard in QCDCL, but easy in Q-resolution [22]. Unlike the separations from [7], this hardness result does not make any reference to propositional hardness but also holds under NP oracles in the framework of [11] (at least if each axiom of our formula contains at least one T-literal, otherwise the formula might be trivial to refute with just X-resolutions and without reductions). This is due to the fact that one could simply replace $\vert \pi \vert$ with the number of reduction steps in $\pi$ in Lemma 4 and Proposition 5 (note that we can assume that there is a reduction step before each $C_\tau$) and hardness in the presence of NP oracles basically corresponds to counting reduction steps (cf. [11]).

We also note that Janota [21] already proved hardness of the QBFs for QCDCL with UIP learning. Corollary 7 improves on that result as well as our hardness result holds for arbitrary learning schemes in QCDCL.

As our second example we introduce the following formulas.

Definition 8

Let with the matrix $\psi _n$ containing the clauses:

$$\begin{aligned}{} & {} x_1\vee u_1\vee t_1,\, {\bar{x}}_1\vee {\bar{u}}_1\vee t_1, \\{} & {} x_i\vee u_i \vee {\bar{t}}_{i-1}\vee t_{i},\,{\bar{x}}_i\vee {\bar{u}}_i \vee {\bar{t}}_{i-1}\vee t_{i}, \qquad for \qquad i=2,\ldots ,n\\{} & {} x_{n+1}\vee u_{n+1}\vee {\bar{t}}_n,\,{\bar{x}}_{n+1}\vee {\bar{u}}_{n+1}\vee {\bar{t}}_n. \end{aligned}$$

It is easy to see that fulfils the XT-property. can be interpreted as a variant of (cf. [5] or the corresponding definition later), in which the universal player wins by setting all $u_i$ equal to $x_i$. The only difference is that instead of having a long T-clause, the T-literals are now connected chain-like by only using XUT-clauses. Next we will show an exponential lower bound for in QCDCL.

Lemma 8

We have .

Proof

Let . Define the sets of clauses

$$\begin{aligned} Z_1&:=\{ x_1\vee u_1\vee t_1,\ {\bar{x}}_1\vee {\bar{u}}_1\vee t_1 \}\\ Z_i&:=\{ x_i\vee u_i\vee {\bar{t}}_{i-1}\vee t_i,\ {\bar{x}}_i\vee {\bar{u}}_i\vee {\bar{t}}_{i-1}\vee t_i \} \qquad \text {for} \quad i=2,\ldots ,n\\ Z_{n+1}&:=\{ x_{n+1}\vee u_{n+1}\vee {\bar{t}}_n,\ {\bar{x}}_{n+1}\vee {\bar{u}}_{n+1}\vee {\bar{t}}_n \}. \end{aligned}$$

Let C be an axiom clause in $\pi$. Then C has to be contained in some set $Z_i$ as above.

Case 1: $C\in Z_1$.

Then we have to get rid of $t_1\in C$, hence we need a clause from $Z_2$. But then we have to get rid of $t_2$ and so on:

$Z_1\rightsquigarrow Z_2\rightsquigarrow \ldots \rightsquigarrow Z_n \rightsquigarrow Z_{n+1}$.

We conclude that $\pi$ has to contain at least one clause from each $Z_j$, $j\in [n+1]$. Therefore we have to pile up $n+1$ X-literals.

Case 2: $C\in Z_i$ for some $i\in \{ 2,\ldots ,n \}$.

Then we have to get rid of ${\bar{t}}_{i-1}$ and $t_i\in C$, hence we need a clause from $Z_{i-1}$ and $Z_{i+1}$. After this we have to resolve over ${\bar{t}}_{i-2}$ and $t_{i+1}$ and so on, leading to a chain of resolutions

.

Again, we conclude that $\pi$ has to contain at least one clause from each $Z_j$, $j\in [n+1]$. Therefore we have to pile up $n+1$ X-literals.

Case 3: $C\in Z_{n+1}$.

This works similarly to Case 1, except that we start at $Z_{n+1}$ and go backwards: . $\square$

Corollary 9

The QBFs require exponential-size proofs in QCDCL.

The gauge of a formula is obviously some width measure and it seems natural to wonder how it relates to the notion of the existential proof width¹ of long-distance Q-resolution refutations of a formula as studied in [6, 9, 17]. However, it turns out that these two measures are not directly related. On the one hand, it is easy to see that has long-distance Q-resolution refutations of constant existential clause width. Hence these formulas have small (constant) existential proof width, but linear gauge.

On the other hand, there are also formulas with constant gauge and linear proof width. For this we revisit the parity formula from [14].

Definition 9

[14] consists of the prefix $\exists x_1 \ldots x_n\forall u\exists t_2 \ldots t_n$ and the matrix

$$\begin{aligned}{} & {} x_1\vee x_2\vee {\bar{t}}_2,\ x_1\vee {\bar{x}}_2\vee t_2,\ {\bar{x}}_1\vee x_2 \vee t_2,\ {\bar{x}}_1\vee {\bar{x}}_2\vee {\bar{t}}_2,\\{} & {} x_i\vee t_{i-1}\vee {\bar{t}}_i,\ x_i\vee {\bar{t}}_{i-1}\vee t_i,\ {\bar{x}}_i\vee t_{i-1}\vee t_i,\ {\bar{x}}_i\vee {\bar{t}}_{i-1}\vee {\bar{t}}_i \quad \text { for} i\in \{ 3,\ldots ,n \}\\{} & {} u\vee t_n,\ {\bar{u}}\vee {\bar{t}}_n. \end{aligned}$$

The formulas are built on the parity function. In more detail, the universal player only wins by setting u equal to $x_1\oplus \ldots \oplus x_n$. Each $t_i$ encodes the partial sum $x_1\oplus \ldots \oplus x_i$.

It was shown in [6, 9] that requires linear proof width. Here we modify this formula such that proof width remains unaffected, but gauge is small. Let be the modified variant of this formula that consists of the prefix $\exists x_1,\ldots ,x_n,y\forall u\exists t_2,\ldots ,t_n$ and the matrix . Obviously, because of the unit clause $({\bar{y}})$, we have , but still linear proof width (since we can simply consider the proof width of the proof restricted to the assignment $y\mapsto 0$, which is exactly a refutation of ).

We will see later that we can also use the formulas to show that large gauge alone is not sufficient to guarantee QCDCL hardness, but some further assumption such as the XT-condition is needed.

We continue with the equality formula from [5] as a further example of hard formulas for QCDCL. In [7] QCDCL hardness of was already proven by lifting Q-resolution hardness of these formulas to QCDCL. However, with our new lower bound technique it is possible to prove QCDCL hardness directly without importing Q-resolution lower bounds.

Definition 10

[5] The formula is defined as the QCNF

$$\begin{aligned} \exists x_1\ldots x_n\forall u_1\ldots u_n \exists t_1\ldots t_n \cdot ({\bar{t}}_1\vee \ldots \vee {\bar{t}}_n)\wedge \bigwedge _{i=1}^n(({\bar{x}}_i\vee {\bar{u}}_i\vee t_i)\wedge (x_i\vee u_i\vee t_i))\text {.} \end{aligned}$$

Proposition 10

We have . Consequently the formulas are exponentially hard for QCDCL.

Proof

Let . Since none of the axioms are X-clauses, we have to resolve over T somehow. For this we need the clause ${\bar{t}}_1\vee \ldots \vee {\bar{t}}_n$. But that means we have to resolve over each $t_i$ at least once in $\pi$, and therefore we will pile up all n X-variables.

$\square$

Our next example illustrates that some further condition (such as the XT-property) is indeed required for our lower bound method to work. For this we will take another look at the parity formula . These formulas are known to be hard for Q-resolution [14], but easy for QCDCL [7]. Nevertheless, we show that has large gauge. Hence this measure alone is not sufficient to imply QCDCL hardness.

Proposition 11

We have .

Proof

We define the following sets of clauses:

$$\begin{aligned} Z_1&:=\{x_1\vee x_2\vee {\bar{t}}_2,\ x_1\vee {\bar{x}}_2\vee t_2,\ {\bar{x}}_1\vee x_2 \vee t_2,\ {\bar{x}}_1\vee {\bar{x}}_2\vee {\bar{t}}_2\}\\ Z_i&:=\{x_{i+1}\vee t_{i}\vee {\bar{t}}_{i+1},\ x_{i+1}\vee {\bar{t}}_{i}\vee t_{i+1},\ {\bar{x}}_{i+1}\vee t_{i}\vee t_{i+1},\ {\bar{x}}_{i+1}\vee {\bar{t}}_{i}\vee {\bar{t}}_{i+1}\}\\ Z_n&:=\{u\vee t_n,\ {\bar{u}}\vee {\bar{t}}_n\} \end{aligned}$$

for $i=2,\ldots ,n-1$.

We show that each needs at least one clause from each $Z_j$ as an axiom, hence $\pi \cap Z_j\ne \emptyset$ for every $j\in [n]$.

Assume that there is a proof of an X-clause C and $j\in [n]$ with $\pi \cap Z_j=\emptyset$. Let S be the set of all symmetries $\sigma$ on such that $\sigma (x_k)\in \{ x_k,{\bar{x}}_k \}$ for each $k\in [n]$ and $\sigma (t_k)$ is chosen such that (one has to make sure that $\sigma (t_k)=\sigma (x_1)\oplus \ldots \oplus \sigma (x_k)$).

Then for each such $\sigma \in S$ we can derive $\sigma (C)$ via $\sigma (\pi )$ and still have $\sigma (\pi )\cap Z_j=\emptyset$. But then we could easily construct a refutation by just using $\{\sigma (C):\, \sigma \in S\}$. Then without the clauses from $Z_j$ would still be a false QCNF. However, this is not possible since we can construct a winning strategy A for :

$$\begin{aligned} A(x_k)&:=0 \text { for all} k\in [n]\\ A(t_\ell )&:=0 \text { for all} \ell \in \{2,\ldots ,j\}\\ A(t_{\ell '})&:=1\oplus u \text { for all} \ell '\in \{ j+1,\ldots ,n \} \end{aligned}$$

Therefore our assumption is false and we get $\pi \cap Z_j\ne \emptyset$. Using one clause from each $Z_j$ results in piling up all variables $x_1,\ldots ,x_n$ in some polarity, hence . $\square$

The next example is a formula that follows the same approach as from [5], where the universal player had to fulfil the task of assigning the U-variables in the same way as the existential X-variables. However, we can replace this task with another, more complex one. In our case, the universal player has to detect palindromes in the word that was input by the existential player.

Example 2

Let be the QCNF with prefix

$$\begin{aligned} \exists X\forall U\exists T \end{aligned}$$

with

$$\begin{aligned} X&=\left\{ x_{{j}}:\, {j}\in \{ {1},\ldots ,{n} \} \right\} \\ U&=\left\{ u_{k,i}, v_k:\, k\in \{0,\ldots n-1\}, i\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\} \right\} \\ T&=\left\{ t_{k,i}, s_k:\, k\in \{0,\ldots n-1\}, i\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\} \right\} \\ \end{aligned}$$

where the indices from the X-variables are interpreted as integers modulo n. Let the matrix of the formula consist of the following clauses:

$$\begin{aligned}&x_{{i+k}}\vee x_{{n-i+1+k}}\vee {\bar{u}}_{k,i}\vee t_{k,i},\,{\bar{x}}_{{i+k}}\vee {\bar{x}}_{{n-i+1+k}}\vee {\bar{u}}_{k,i}\vee t_{k,i}\\&{x}_{{i+k}}\vee {\bar{x}}_{{n-i+1+k}}\vee {u}_{k,i}\vee {\bar{t}}_{k,i},\,{\bar{x}}_{{i+k}}\vee {x}_{{n-i+1+k}}\vee {u}_{k,i}\vee {\bar{t}}_{k,i}\\&{\bar{v}}_k\vee {\bar{t}}_{k,1}\vee \ldots \vee {\bar{t}}_{k,\lfloor \frac{n}{2}\rfloor }\vee s_k\\&v_k\vee t_{k,i}\vee s_k,\,{\bar{s}}_0\vee \ldots \vee {\bar{s}}_{n-1} \end{aligned}$$

for $k\in \{0,\ldots n-1\}, i\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\}$.

Intuitively, the X-variables represent the word in which palindromes have to be detected. We not only check the word $x_1\ldots x_n$ itself, but also all shifted variants $x_{1+k}\ldots x_{n+k}$. The $u_{k,i}$ encode whether or not, in the word $x_{1+k}\ldots x_{n+k}$, the $i^\text {th}$ letter from the left and the $i^\text {th}$ letter from the right are the same. If the universal player assigns the $u_{k,i}$ correctly, the existential player is forced to set $t_{k,i}$ equal to $u_{k,i}$ in their turn at the end. If and only if the word $x_{1+k}\ldots x_{n+k}$ was a palindrome, the universal player sets $v_k$ to true. Hence, if $x_{1+k}\ldots x_{n+k}$ was a palindrome, $v_k$ as well as $t_{k,i}$ for each i is set to true, hence the existential player has to set $s_k$ to true. Otherwise, if it was not a palindrome, then there was an i such that $x_{i+k}$ and $x_{n-i+1+k}$ were assigned differently, therefore $t_{k,i}$ and $v_k$ are both false and $s_k$ must be set to true, as well. However, setting all $s_k$ to true falsifies the matrix.

In a nutshell: Let $\tau$ be a total assignment of X. There is a winning strategy for the universal player by setting $u_{k,i}$ to 1 if and only if $\tau (x_{{i+k}})=\tau (x_{{n-i+k}})$ and $v_k$ to 1 if and only if the word $\tau (x_{{1+k}})\ldots \tau (x_{{n+k}})$ is a palindrome. Then the existential player has to set each $s_k$ to 1, negating the last clause in the matrix.

Remark 1

fulfils the XT-property.

Lemma 12

We have .

Proof

Let . Since we do not have any X-clauses as axioms, we need to resolve over T-variables at least once. We partition the matrix of into the following sets:

$$\begin{aligned} Z_{k,i}^{+}&:=\{x_{{i+k}}\vee x_{{n-i+1+k}}\vee {\bar{u}}_{k,i}\vee t_{k,i}, {\bar{x}}_{{i+k}}\vee {\bar{x}}_{{n-i+1+k}}\vee {\bar{u}}_{k,i}\vee t_{k,i}\}\\ Z_{k,i}^{-}&:=\{{x}_{{i+k}}\vee {\bar{x}}_{{n-i+1+k}}\vee {u}_{k,i}\vee {\bar{t}}_{k,i}, {\bar{x}}_{{i+k}}\vee {x}_{{n-i+1+k}}\vee {u}_{k,i}\vee {\bar{t}}_{k,i}\}\\ P_k&:=\{ {\bar{v}}_k\vee {\bar{t}}_{k,1}\vee \ldots \vee {\bar{t}}_{k,\lfloor \frac{n}{2}\rfloor }\vee s_k \}\\ N_{k,i}&:=\{ v_k\vee t_{k,i}\vee s_k \}\\ S&:=\{ {\bar{s}}_0\vee \ldots \vee {\bar{s}}_{n-1} \} \end{aligned}$$

Let $C\in \pi$ be an axiom. We claim that $S\subseteq \pi$, for which we will distinguish four cases.

Case 1: $C\in Z_{k,i}^+$ for some $k\in \{0,\ldots ,n-1\}$ and $i\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\}$.

Then we have to resolve away $t_{k,i}$, which can only be done with the clause in $P_k$ since the clauses from $Z_{k,i}^-$ are blocked because of the $u_{k,i}$. But now we have introduced $s_k$ which we can only resolve with the clause from S.

Case 2: $C\in Z_{k,i}^-$ for some $k\in \{0,\ldots ,n-1\}$ and $i\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\}$.

To get rid of ${\bar{t}}_{k,i}$, we have to use the clause from $N_{k,i}$ since $Z_{k,i}^+$ is blocked as before. But then we have introduces $s_k$ and we will need the clause from S.

Case 3: $C\in P_k$ or $C\in N_{k,i}$ for some $k\in \{0,\ldots ,n-1\}$ and $i\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\}$.

Then we have $s_k\in C$ and we need to use the clause from S in order to resolve it away.

Case 4: $C\in S$.

This case is trivial.

We have shown that in each case we have $S\subseteq \pi$. That means we have to resolve over each $s_k$ in $\pi$. For each $k\in \{0,\ldots ,n-1\}$ we can choose if we want to use $P_k$ or $N_{k,i}$ to get rid of the literal ${\bar{s}}_k$. If we choose $P_k$, then we have to resolve over each $t_{k,i}$ for $i\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\}$ by using the clauses from $Z_{k,i}^+$. However, if we choose $N_{k,i}$, it suffices to resolve over only one $t_{k,i}$ for some particular i. Hence, we only have to use one clause from $Z_{k,i}^-$ for only one i. In the worst case (that means in the case with the least resolutions over $t_{k,i}$), we will always pick $N_{k,i}$. More specific, for each $k\in \{0,\ldots ,n-1\}$ there exists an $i_k\in \left\{ 1,\ldots , \left\lfloor \frac{n}{2}\right\rfloor \right\}$ such that $\pi$ contains at least one clause from $Z_{k,i_k}^+$ or $Z_{k,i_k}^-$. That means we will pile up at least the X-variables $x_{{i_k+k}},x_{{n-i_k+1+k}}$ for each $k\in \{0,\ldots ,n-1\}$. If we can find a lower bound for the number of these X-variables, then this is also a lower bound for .

First of all, it its obvious that for each k we have $x_{{i_k+k}}\ne x_{{n-i_k+1+k}}$ since $i_k+k\not \equiv n-i_k+1+k$ (mod n). Next, we define the following sets of variables:

$$\begin{aligned} X_k:=\{ x_{{i_k+k}},x_{{n-i_k+1+k}} \} \end{aligned}$$

for $k\in \{0,\ldots ,n-1 \}$. As we have argued above, we already know that the gauge of is $\Omega (\vert \bigcup _{k}X_k\vert )$. Note that if a pair $\{ x_{{j}},x_{{\ell }} \}$ is equal to $X_k$, then their position in the word $x_{{1+k}}\ldots x_{{n+k}}$ is symmetric. If n is odd, then each pair $\{ x_{{j}},x_{{\ell }} \}$ can represent at most one $X_k$. For example, for $n=5$ the pair $\{x_{{3}},x_{{4}}\}$ can at most be $X_3$ since they are only symmetric in the word $x_{{4}}x_{{5}}x_{{1}}x_{{2}}x_{{3}}$. However, if n is even then each pair then each pair $\{ x_{{j}},x_{{\ell }} \}$ can represent up to two $X_k$. For example, for $n=4$ the pair $\{ x_{{1}},x_{{4}} \}$ is symmetric in both $x_{{1}}x_{{2}}x_{{3}}x_{{4}}$ and $x_{{3}}x_{{4}}x_{{1}}x_{{2}}$.

We conclude that for odd n we have $\vert \{ X_0,\ldots ,X_{n-1} \}\vert =n$ and for even n we have $\vert \{ X_0,\ldots ,X_{n-1} \}\vert \ge \frac{n}{2}$. Now, with m different variables we could create at most ${\mathcal {O}}(m^2)$ different pairs $X_k$. Hence we need at least $\Omega (\sqrt{n})$ different variables to create ${\mathcal {O}}(n)$ different pairs $X_k$, and therefore $\vert \bigcup _{k}X_k\vert \in \Omega (\sqrt{n})$ and also . $\square$

Corollary 13

The QCNF needs QCDCL refutations of size $2^{\Omega (\sqrt{n})}$.

7 Conclusion

We initiated the study of devising lower bound methods tailored to QCDCL. At the moment our techniques only apply to $\Sigma _3^b$-formulas. Though this is a quite relevant class of QBFs, also prominently represented in QBF benchmarks [27, 32], it would be very interesting to extend the method to QBFs of higher quantifier complexity.

In another direction, future research should explore further conditions (besides the XT-condition considered here) that allow to efficiently translate QCDCL into quasi level-ordered proofs and thus enable to show lower bounds via gauge.