Should Decisions in QCDCL Follow Prefix Order?

Abstract

Quantified conflict-driven clause learning (QCDCL) is one of the main solving approaches for quantified Boolean formulas (QBF). One of the differences between QCDCL and propositional CDCL is that QCDCL typically follows the prefix order of the QBF for making decisions. We investigate an alternative model for QCDCL solving where decisions can be made in arbitrary order. The resulting system $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ is still sound and terminating, but does not necessarily allow to always learn asserting clauses or cubes. To address this potential drawback, we additionally introduce two subsystems that guarantee to always learn asserting clauses ($\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$) and asserting cubes ($\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$), respectively. We model all four approaches by formal proof systems and show that $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ is exponentially better than $\mathsf{{QCDCL}}$ on false formulas, whereas $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ is exponentially better than $\mathsf{{QCDCL}}$ on true QBFs. Technically, this involves constructing specific QBF families and showing lower and upper bounds in the respective proof systems. We complement our theoretical study with some initial experiments that confirm our theoretical findings.

1 Introduction

SAT solving was revolutionised in the late 1990 s by the advent of conflict-driven clause learning (CDCL), which has since been the dominating paradigm in propositional SAT solving [24, 25, 37]. A few years later, the CDCL approach was lifted to the computationally even harder setting of quantified Boolean formulas (QBF) in the form of quantified CDCL (QCDCL) [38]. Though a number of competing approaches to QBF solving exist (cf. [7] for a recent overview), QCDCL is one of most competitive. State-of-the-art implementations include DepQBF [21] and Qute [29, 33].

In comparison to the propositional case, QCDCL poses additional technical challenges, stemming from partitioning the variables into existential and universal (SAT can be viewed as using only existential variables) and the dependencies between the variables imposed by the quantifier prefix. The presence of universal variables entails additional rules for unit propagation (universal reductions), while the variable dependencies imposed by the prefix are typically observed by decision heuristics in the sense that QCDCL follows the prefix order in decision making. The latter is arguably the most severe restriction when transitioning from CDCL to QCDCL. Another difference between CDCL and QCDCL arises from the fact that unlike in SAT, a satisfying assignment to the QBF matrix does not imply that the QBF is true. Instead, this is witnessed by additionally learning cubes (i.e., conjunctions of literals, also called terms) and producing a cube certificate for true QBFs.

Though CDCL and QCDCL are very efficient in practice and in particular on industrial instances (cf. [32] for an overview of QBF solving applications and [16, 22] for experimental studies of solver performance), their success and their inherent limitations are not at all well-understood from a theoretical perspective. The main theoretical approach is through proof complexity [10]. For SAT it is known that CDCL—viewed as a non-deterministic procedure—is equivalent to propositional resolution [1, 3, 30]. In particular, resolution refutations can be efficiently extracted from CDCL runs, whereby lower bounds for resolution proof size imply lower bounds for CDCL running time. However, when using CDCL with practical decision heuristics such as VSIDS [26], the model becomes exponentially weaker than resolution [36].

The situation is even more intricate in QBF. Again, from QCDCL runs, proofs can be efficiently extracted in the format of long-distance Q-Resolution [2, 38].¹ However, QCDCL—even as a non-deterministic procedure—is exponentially weaker than long-distance Q-Resolution and incomparable to the simpler system of Q-Resolution [5]. Thus it is very interesting, both from a theoretical and practical perspective, to gauge the precise power of QCDCL.

In this paper we introduce and investigate QCDCL models that drop the requirement of making variable decisions along the prefix order. Though it has been recently shown that following the prefix order in QCDCL is not needed for correctness,² existing prefix-relaxing techniques do not exploit this as much as they could. Dependency schemes [23, 28, 31, 34] work with the assumption that the prefix has to be observed, but notice that certain parts (often called spurious dependencies) can be relaxed in preprocessing. With dependency learning [29], a more recent, orthogonal technique, instead of calculating dependencies upfront the solver assumes independence until it runs into a problem, from which it learns a dependency on the fly (dependency learning can be combined with schemes [27]). These strategies are executed differently: with dependency schemes the solver can fully rely on the relaxed prefix and use it for decisions, propagation, and clause/cube learning alike; with dependency learning the solver can only use the relaxed prefix for decisions and propagation and must learn clauses and cubes with the original prefix in order to detect dependencies. However, both approaches share the restriction that once dependencies are found, decisions must respect them.

Our contributions. We propose a new QCDCL model where decisions can ignore quantification entirely; only propagation and clause/cube learning use the prefix information.

When suggesting a new model for solving, there are at least two possible approaches: (1) to give a formal account of the model, prove its correctness, and theoretically quantify the gains on running time; or (2) provide an implementation and experimentally evaluate its practical performance. In this paper, our main focus is to contribute towards (1). While we also perform some initial proof-of-concept experiments, an extensive practical evaluation of the competitiveness of the approach is left for future work (cf. the conclusion).³

Specifically, our contributions are as follows:

1. Formal proof complexity models for QCDCL using arbitrary decisions. We provide a formal proof-complexity model for QCDCL with arbitrary decisions. This follows a recent line of research to formalise and rigorously analyse QCDCL from a proof complexity perspective [5, 9].

Our most general model $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ allows arbitrary decisions. Care has to be taken to ensure that we can always learn new clauses and cubes, as otherwise termination of proof search is no longer guaranteed. We ensure this by adding a simple new constraint condition (NCC), which forbids making decisions that immediately falsify a clause or satisfy a cube (which is already trivially impossible in prefix-observing QCDCL).

A potential further drawback of not following prefix order is that we can no longer guarantee to learn asserting clauses or cubes.⁴ In order to address this, we introduce two subsystems of $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$—termed $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$—that allow to always learn asserting clauses and cubes, respectively. We prove that all three systems are sound, complete, and terminating.

2. Exponential separations between the QCDCL models. The main contribution of this paper lies in proving that both $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ allow for exponentially shorter proofs than the prefix-following $\mathsf{{QCDCL}}$ model. The resulting simulation order is depicted in Fig. 1.

Fig. 1

Hasse diagram of the simulation order of QCDCL proof systems. Solid lines represent p-simulations and exponential separations. Waved lines represent p-simulations, for which separations are not known

To show this we construct two QBF families that exponentially separate the systems. Both employ general constructions—using a ‘twin’ and a ‘reverse’ construction—that could potentially be used for further formulas. Technically, we use the recently developed lower bound approach via the gauge of QBFs [9]. However, different from previous work [5, 9], which only considered clause learning, our lower bounds work against a more realistic QCDCL system that uses both clause and cube learning. Interestingly, the separation of $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ from $\mathsf{{QCDCL}}$ works on false QBFs, while the separation of $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ from $\mathsf{{QCDCL}}$ uses true QBFs. The latter is the first dedicated QBF proof-complexity lower bound on true formulas.⁵ In fact, we provide a general method how to transform hardness of false QBFs into hardness of true formulas.

3. Proof-of-concept experiments. Though this is not our main focus, we provide initial experiments that confirm our theoretical findings. These experiments are only meant to illustrate that our approach is in principle competitive with plain QCDCL, without considering the impact of other techniques like preprocessing or dependency learning, etc. (cf. the discussion of future work in the conclusion).

Organisation. The remainder of this paper is organised as follows. We start in Sect. 2 with reviewing QBF preliminaries. In Sects. 3 and 4 we introduce and formally model the new QCDCL versions. Their proof-complexity analysis and the separations are proven in Sect. 5. Section 6 describes our proof-of-concept experiments and Sect. 7 outlines further work.

2 Preliminaries

Propositional and quantified formulas. Variables x and negated variables ${\bar{x}}$ are called literals. We denote the corresponding variable as $\text {var}(x):=\text {var}(\bar{x}):=x$.

A clause is a disjunction of literals and a cube is a conjunction of literals. We will sometimes interpret clauses and cubes as sets of literals on which we can perform set-theoretic operations.

A unit clause $(\ell )$ is a clause that consists of only one literal. The empty clause consists of zero literals, denoted $(\bot )$. We sometimes paraphrase $(\bot )$ as a unit clause with the ‘empty literal’ $\bot$. A clause C is called tautological if $\{ \ell ,\bar{\ell }\}\subseteq C$ for some literal $\ell$.

We define a unit cube of a literal $\ell$, denoted by $\left[ \ell \right]$, and the empty cube $\left[ \top \right]$ with ‘empty literal’ $\top$. A cube D is contradictory if $\{\ell ,\bar{\ell }\}\subseteq D$ for some literal $\ell$. If C is a clause or a cube, we define $\text {var}(C):=\{ \text {var}(\ell ):\,\ell \in C \}$. The negation of a clause $C=\ell _1\vee \ldots \vee \ell _m$ is the cube $\lnot C:=\overline{C}:=\bar{\ell }_1\wedge \ldots \wedge \bar{\ell }_m$.

A (total) assignment $\sigma$ of a set of variables V is a non-tautological set of literals such that for all $x\in V$ there is some $\ell \in \sigma$ with $\text {var}(\ell )=x$. A partial assignment $\sigma$ of V is an assignment of a subset $W\subseteq V$. A clause C is satisfied by an assignment $\sigma$ if $C\cap \sigma \ne \emptyset$. A cube D is falsified by $\sigma$ if $\lnot D\cap \sigma \ne \emptyset$. A clause C that is not satisfied by $\sigma$ can be restricted by $\sigma$, defined as

$$\begin{aligned} C|_{\sigma }:=\bigvee _{\ell \in C, \bar{\ell }\not \in \sigma }\ell . \end{aligned}$$

Similarly we can restrict a non-falsified cube D as

$$\begin{aligned} D|_\sigma :=\bigwedge _{\ell \in D\backslash \sigma }\ell . \end{aligned}$$

Intuitively, an assignment sets all its literals to true.

A CNF (conjunctive normal form) is a conjunction of clauses and a DNF (disjunctive normal form) is a disjunction of cubes. We restrict a CNF (resp. DNF) $\phi$ by an assignment $\sigma$ as

$$\begin{aligned} \phi |_{\sigma }:=\bigwedge _{C\in \phi \text { non-satisfied}}C|_\sigma \,\,\,\left( \text {resp. } \phi |_\sigma :=\bigvee _{D\in \phi \text { non-falsified}}D|_\sigma \right) . \end{aligned}$$

For a CNF (DNF) $\phi$ and an assignment $\sigma$, if $\phi |_\sigma =\emptyset$, then $\phi$ is satisfied (falsified) by $\sigma$.

A QBF (quantified Boolean formula) $\Phi =\mathcal {Q}\cdot \phi$ consists of a propositional formula $\phi$, called the matrix, and a prefix $\mathcal {Q}$. A prefix $\mathcal {Q}=\mathcal {Q}'_1V_1\dots \mathcal {Q}'_sV_s$ consists of non-empty and pairwise disjoint sets of variables $V_1,\dots ,V_s$ and quantifiers $\mathcal {Q}'_1,\ldots ,\mathcal {Q}'_s\in \{\exists ,\forall \}$ with $\mathcal {Q}'_i\ne \mathcal {Q}'_{i+1}$ for $i\in [s-1]$. For a variable x in $\mathcal {Q}$, the quantifier level is $\text {lv}(x):=\text {lv}_\Phi (x):=i$, if $x\in V_i$. For $\text {lv}_\Phi (\ell _1)<\text {lv}_\Phi (\ell _2)$ we write $\ell _1<_\Phi \ell _2$, while $\ell _1\le _\Phi \ell _2$ means $\text {lv}_\Phi (\ell _1)\le \text {lv}_\Phi (\ell _2)$.

For a QBF $\Phi =\mathcal {Q}\cdot \phi$ with $\phi$ a CNF (DNF), we call $\Phi$ a QCNF (QDNF). We define $\mathfrak {C}(\Phi ):=\phi$ (resp. $\mathfrak {D}(\Phi ):=\phi$). $\Phi$ is an AQBF (augmented QBF), if $\phi =\psi \vee \chi$ with CNF $\psi$ and DNF $\chi$. Again we write $\mathfrak {C}(\Phi ):=\psi$ and $\mathfrak {D}(\Phi ):=\chi$.

We restrict a QCNF (QDNF) $\Phi =\mathcal {Q}\cdot \phi$ by an assignment $\sigma$ as $\Phi |_\sigma :=\mathcal {Q}|_\sigma \cdot \phi |_\sigma$, where $\mathcal {Q}|_\sigma$ is obtained by deleting all variables from $\mathcal {Q}$ that appear in $\sigma$. Analogously, we restrict an AQBF $\Phi =\mathcal {Q}\cdot (\psi \vee \chi )$ as $\Phi |_\sigma :=\mathcal {Q}|_\sigma \cdot (\psi |_\sigma \vee \chi |_\sigma )$.

If L is a set of literals (e.g., an assignment), we can get the negation of L, which we define as $\lnot L:= \overline{L}:=\{ \bar{\ell }|\, \ell \in L \}$.

(Long-distance) Q-resolution and Q-consensus. Let $C_1$ and $C_2$ be two clauses (cubes) from a QCNF (QDNF) or AQBF $\Phi$. Let $\ell$ be an existential (universal) literal with $\text {var}(\ell )\not \in \text {var}(C_1)\cup \text {var}(C_2)$. The resolvent of $C_1\vee \ell$ and $C_2\vee \bar{\ell }$ over $\ell$ is defined as

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

(resp. $(C_1\wedge \ell ){\mathop {\bowtie }\limits ^{\ell }}(C_2\wedge \bar{\ell }):=C_1\wedge C_2 )$.

Let $C:=\ell _1\vee \ldots \vee \ell _m$ be a clause from a QCNF or AQBF $\Phi$ such that $\ell _i\le _\Phi \ell _j$ for all $i<j$, $i,j\in \{1,\ldots ,m\}$. Let k be minimal such that $\ell _k,\ldots ,\ell _m$ are universal. Then we can perform a universal reduction step and obtain

$$\begin{aligned} \text {red}_\Phi ^\forall (C):=\ell _1\vee \ldots \vee \ell _{k-1}. \end{aligned}$$

Analogously, we perform existential reduction on cubes. Let $D:=\ell _1\wedge \ldots \wedge \ell _m$ be a cube of a QDNF or AQBF $\Phi$ with $\ell _i\le _\Phi \ell _j$ for all $i<j$, $i,j\in \{1,\ldots ,m\}$. Let k be minimal such that $\ell _k,\ldots ,\ell _m$ are existential. Then $\text {red}_\Phi ^\exists (D):=\ell _1\wedge \ldots \wedge \ell _{k-1}$.

If it is clear that C is a clause or a cube, we can just write $\text {red}_\Phi (C)$ or even $\text {red}(C)$, if the QBF $\Phi$ is also obvious. We will write $\text {red}(\Phi )=\text {red}_\Phi (\Phi )$, if we reduce all clauses and cubes of the AQBF $\Phi$ according to its prefix.

As defined by Kleine Büning et al. [20], a Q-resolution (Q-consensus) proof $\pi$ from a QCNF (QDNF) or AQBF $\Phi$ of a clause (cube) C is a sequence of clauses (cubes) $\pi =(C_i)_{i=1}^m$, such that $C_m=C$ and for each $C_i$ one of the following holds:

• Axiom: $C_i\in \mathfrak {C}(\Phi )$ (resp. $C_i\in \mathfrak {D}(\Phi ))$;

• Resolution: $C_i=C_j{\mathop {\bowtie }\limits ^{x}} C_k$ with x existential (universal), $j,k<i$, and $C_i$ non-tautological (non-contradictory);

• Reduction: $C_i=\text {red}_\Phi ^\forall (C_j)$ (resp. $C_i=\text {red}_\Phi ^\exists (C_j)$) for some $j<i$.

We call C the root of $\pi$. In [2], an extension of Q-resolution (Q-consensus) proofs to long-distance Q-resolution (long-distance Q-consensus) proofs was introduced by replacing the resolution rule by

• Resolution (long-distance): $C_i=C_j{\mathop {\bowtie }\limits ^{x}}C_k$ with x existential (universal) and $j,k<i$. The resolvent $C_i$ is allowed to contain tautologies such as $u\vee \bar{u}$ (resp. $u\wedge \bar{u}$), if u is universal (existential). If there is such a universal (existential) $u\in \text {var}(C_j)\cap \text {var}(C_k)$, then we require $x<_\Phi u$.

Furthermore, a Q-resolution (Q-consensus) or long-distance Q-resolution (long-distance Q-consensus) proof $\pi$ from $\Phi$ of the empty clause $(\bot )$ (the empty cube $\left[ \top \right]$) is called a refutation (certificate) of $\Phi$. In that case, $\Phi$ is called false (true). We will sometimes interpret $\pi$ as a set of clauses (or cubes).

A proof system S p-simulates a system $S'$, if every $S'$ proof can be transformed in polynomial time into an S proof of the same formula.

3 Our QCDCL Models

To analyse the complexity of QCDCL procedures, we need to fully formalise them as proof systems. This approach was initiated in [5] and [9], and we follow that framework. We will only sketch this formalization here.

We store all relevant information of a QCDCL run in trails. Since QCDCL uses several runs and potentially also restarts, a QCDCL proof will typically consist of many trails.

Definition 1

(trails) A trail $\mathcal {T}$ for a QCNF or AQBF $\Phi$ is a (finite) sequence of pairwise distinct literals from $\Phi$, including the empty literals $\bot$ and $\top$. In general, a trail has the form

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

(1)

where the $d_i$ are decision literals and $p_{(i,j)}$ are propagated literals. A trail $\mathcal {T}$ has run into a conflict if $\bot \in \mathcal {T}$ or $\top \in \mathcal {T}$.

Decision literals are written in boldface. We use a semicolon before each decision to mark the end of a decision level. If one of the empty literals $\bot$ or $\top$ is contained in $\mathcal {T}$, then it has to be the last literal $p_{(r,g_r)}$. In this case, we say that $\mathcal {T}$ has run into a conflict.

Trails can be interpreted as non-tautological sets of literals, and therefore as (partial) assignments. We write $x<_\mathcal {T}y$ if $x,y\in \mathcal {T}$ and x is left of y in $\mathcal {T}$. Furthermore, we write $x\le _\mathcal {T}y$ if $x<_\mathcal {T}y$ or $x=y$.

As trails are produced gradually from left to right in an algorithm, we define $\mathcal {T}[i,j]$ for $(i,j)\in (\{0,\ldots ,r\}\times \{0,\ldots ,g_i\})\backslash \{(0,0)\}$ as the subtrail that contains all literals from $\mathcal {T}$ up to (and excluding) $p_{(i,j)}$ (resp. $d_i$, if $j=0$) in the same order. Intuitively, $\mathcal {T}[i,j]$ is the state of the trail before we assigned the literal at the point [i, j] (which is $p_{(i,j)}$ or $d_i$).

Each propagated literal $p_{(i,j)}\in \mathcal {T}$ belongs to an antecedent clause (if $p_{(i,j)}$ is existential) or an antecedent cube (if $p_{(i,j)}$ is universal) from $\Phi$, which we call $\text {ante}_{\mathcal {T}}(p_{(i,j)})$. At the point where $p_{(i,j)}$ was propagated in $\mathcal {T}$, we need that $\text {ante}_{\mathcal {T}}(p_{(i,j)})$ had become unit, hence $\text {red}_{\Phi }(\text {ante}_{\mathcal {T}}(p_{(i,j)})|_{\mathcal {T}[i,j]})=(p_{(i,j)})$ if $p_{(i,j)}$ is existential, and $\text {red}_{\Phi }(\text {ante}_{\mathcal {T}}(p_{(i,j)})|_{\mathcal {T}[i,j]})=[\bar{p}_{(i,j)}]$, if $p_{(i,j)}$ is universal.

Trails are not generated arbitrarily, as they follow some further rules in practice, such as propagations should not be skippable. We denote trails with these conditions as natural trails.

Definition 2

(natural trails) We call $\mathcal {T}$ a natural trail for the formula $\Phi$, if for each $i\in \{1,\ldots ,r\}$ the formula $\text {red}(\Phi |_{\mathcal {T}[i,0]})$ does not contain unit or empty constraints. Furthermore, the formula $\Phi |_{\mathcal {T}[i,j]}$ must not contain empty constraints for each $i\in \{1,\ldots ,r\}$, $j\in \{1,\ldots ,g_i\}$, except $[i,j]=[r,g_r]$. Intuitively, we require that decisions are only made if and only if there are no more propagations on the same decision level left. Also, conflicts must be detected immediately if there are any.

We state some general facts about trails and antecedent clauses/cubes one should keep in mind.

Remark 1

Let $\mathcal {T}$ be a trail, $\ell \in \mathcal {T}$ a propagated literal and $A:=\text {ante}_{\mathcal {T}}(\ell )$.

• If $\ell$ is existential, then $\ell \in A$ and for each literal $x\in A$ with $x\ne \ell$ we need $\bar{x}<_\mathcal {T}\ell$.

• If $\ell$ is universal, then $\bar{\ell }\in A$ and for each literal $u\in A$ with $u\ne \bar{\ell }$ we need $u<_\mathcal {T}\ell$.

An essential element of QCDCL is clause and cube learning. This guarantees to make ‘progress’ after each trail (at least under some conditions that we will specify later).

Definition 3

(learnable constraints) Let $\mathcal {T}$ be a trail for $\Phi$ of the form (1) with $p_{(r,g_r)}\in \{\bot ,\top \}$. Starting with $\text {ante}_{\mathcal {T}}(\bot )$ (resp. $\text {ante}_\mathcal {T}(\top )$) we reversely resolve over the antecedent clauses (cubes) that were used to propagate the existential (universal) variables, until we stop at some arbitrarily chosen point. The clause (cube) we so derive is a learnable constraint. Note that clause (cube) learning will skip propagations caused by cubes (clauses) and interpret the corresponding literal in the trail as a decision. Universal (existential) reduction will be performed whenever applicable (basically before and after each resolution step in the learning process). We denote the set of learnable constraints by $\mathfrak {L}(\mathcal {T})$.

We can also learn cubes from trails that did not run into conflict. If $\mathcal {T}$ is a total assignment of the variables from $\Phi$, then we define the set of learnable constraints as the set of cubes $\mathfrak {L}(\mathcal {T}):= \{ \text {red}_\Phi ^\exists (D)|\, D\subseteq \mathcal {T}\text { and }D\text { satisfies }\mathfrak {C}(\Phi ) \}$.

In QCDCL, our goal is to make ‘progress’ in each run/trail. Thus, we have to ensure that we can always learn new clauses or cubes from a constructed trail. Since we want to work with QCDCL models that do not necessarily follow the prefix order for decision making, it is not guaranteed that we can even learn new constraints from each trail. As we will show later, we need the following condition to prevent such a situation, which could easily lead to a loop in practical solving.

Definition 4

A trail $\mathcal {T}$ for a formula $\Phi$ fulfils the New Constraint Condition (NCC for short), if for each decision $d_i$ the formula $\text {red}(\Phi |_{\mathcal {T}[i,0]\cup \{d_i\}})$ does not contain the empty clause or cube.

Intuitively, this means that a decision must not lead to a conflict immediately. It will become clear later, why we can always find a decision that does not violate the NCC. In fact, classical QCDCL automatically fulfils this condition.

We will now formally define our four QCDCL proof systems, namely $\mathsf{{QCDCL}}$, $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$.

Definition 5

(QCDCL proof systems) Let S be one of $\mathsf{{QCDCL}}$, $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$. An S proof $\iota$ from a QCNF $\Phi =\mathcal {Q}\cdot \phi$ of a clause or cube C is a (finite) sequence of triples

$$\begin{aligned} \iota :=[(\mathcal {T}_i,C_i,\pi _i)]_{i=1}^m, \end{aligned}$$

where $C_m=C$, each $\mathcal {T}_i$ is a trail for $\Phi _{i}$ that fulfils the NCC, each $C_i\in \mathfrak {L}(\mathcal {T}_i)$ is one of the constraints we can learn from each trail and $\pi _i$ is the long-distance Q-resolution or long-distance Q-consensus proof from $\Phi _i$ of $C_i$ we obtain by performing the steps in Definition 3. If necessary, we set $\pi _i:=\emptyset$. We will denote the set of trails in $\iota$ as $\mathfrak {T}(\iota )$.

The AQBFs $\Phi _i$ are defined as follows:

$$\begin{aligned} \Phi _1:=\mathcal {Q}\cdot \left( \mathfrak {C}(\Phi )\vee \emptyset \right) \end{aligned}$$

and

$$\begin{aligned} \Phi _{j+1}:=\left\{ \begin{array}{ll} \hspace{-5.0pt}\mathcal {Q}\cdot \left( (\mathfrak {C}(\Phi _j)\wedge C_{j})\vee \mathfrak {D}(\Phi _j)\right) &{}\hspace{-5.0pt}\text { if}\,\, C_j \,\,\text { is a clause,} \\ \hspace{-5.0pt}\mathcal {Q}\cdot \left( \mathfrak {C}(\Phi _j)\vee (\mathfrak {D}(\Phi _j)\vee C_j)\right) &{}\hspace{-5.0pt}\text { if }\,\, C_j \,\, \text { is a cube,}\end{array}\right. \end{aligned}$$

for $j=1,\ldots ,m-1$.

The four systems differ from each other in the way decisions are made. We extend the definition of natural trails with decision rules that belong to the corresponding system S. A natural trail $\mathcal {T}$ for a formula $\Psi$ that fulfils the following rules for S is called a natural S trail:

• $\mathsf{{QCDCL}}$: For each decision $d_i$ we have that $\text {lv}_{\Psi |_{\mathcal {T}[i,0]}}(d_i)=1$. I.e., decisions are level-ordered.

• $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$: Decisions can be made arbitrarily as long as the NCC is fulfilled.

• $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$: An existential decision $d_i$ can only be made if all universal variables that are quantified left of $d_i$ were already assigned in $\mathcal {T}$. Universal decisions can be made in any order as long as the NCC is fulfilled.

• $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$: A universal decision $d_i$ can only be made if all existential variables that are quantified left of $d_i$ were already assigned in $\mathcal {T}$. Existential decisions can be made in any order as long as the NCC is fulfilled.

After each trail, we will backtrack to some arbitrary previous point in the trail and continue to decide or propagate from that point.

If $C=C_m=(\bot )$, then $\iota$ is called an S refutation of $\Phi$. If $C=C_m=\left[ \top \right]$, then $\iota$ is called an S certificate of $\Phi$. The proof ends once we have learned $(\bot )$ or $\left[ \top \right]$.

If C is a clause, we can stick together the long-distance Q-resolution derivations from $\{ \pi _1,\ldots ,\pi _m \}$ and obtain a long-distance Q-resolution proof from $\Phi$ of C, which we call $\mathfrak {R}(\iota )$. Similarly, if C is a cube, we can stick together the long-distance Q-consensus derivations and obtain a long-distance Q-consensus proof $\mathfrak {R}(\iota )$ from $\Phi$ of C.

The size of $\iota$ is defined as $|\iota |:=\sum _{i=1}^{m}|\mathcal {T}_i|$. Obviously, we have $|\mathfrak {R}(\iota )|\in \mathcal {O}(|\iota |)$.

Our formalisation above is based on [5, 9]. However, since in the present paper cube learning is always included, our plain model $\mathsf{{QCDCL}}$ now includes clause and cube learning (while in [5, 9], $\mathsf{{QCDCL}}$ denotes a system with just clause learning, but without learning cubes).

The concept behind the two models $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ was already introduced in [5] (albeit defined slightly differently, they were called $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ss}}-R-O\tiny {\MakeUppercase {rd}}}}_\textsf {{R\tiny {\MakeUppercase {ed}}}}$ and $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}-O\tiny {\MakeUppercase {rd}}}}_\textsf {{N\tiny {\MakeUppercase {o}}}-{R\tiny {\MakeUppercase {ed}}}}$ in those papers). However, since we include cube learning now, our models here match practical solving much better.

Remark 2

In $\mathsf{{QCDCL}}$, decision making can never violate the NCC if we create the trails ‘naturally’ (i.e., decisions are only made if and only if there are no more propagations on the same decision level left, and conflicts must be detected immediately if there are any).

Proof

If we made a level-ordered decision $d_i$ and get a conflict immediately afterwards on a clause (w.l.o.g.) $C=\text {ante}_{\mathcal {T}}(\bot )$, then $d_i$ must have been existential (otherwise we could have reduced $\bar{d}_i$ in order to get a conflict before) and we would need $\bar{d}_i\in C$. Furthermore, there must exist at least one universal literal $u\in C$ that was reduced while propagating $\bot$, otherwise C would have been a unit clause before we made the decision $d_i$. However, the reduction must have been blocked before deciding $d_i$, otherwise we could have used this reduction to propagate $\bar{d}_i$. That means u was quantified left of $d_i$, but this is a contradiction since the decision $d_i$ was level-ordered. Hence, deciding the leftmost unassigned literal according to the prefix order, we will never violate the NCC. $\square$

We still have to make sure to fulfil NCC when backtracking, though. We will explain later how this is achieved.

The next result states simulations between systems, cf. Fig. 1. They all follow by definition.

Proposition 1

Each $\mathsf{{QCDCL}}$ proof is also a $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ proof, and each $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ or $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ proof is also a $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ proof.

4 Learning Asserting Constraints

We recall the notion of an asserting clause (or cube). The concept originates from SAT solving [25], but directly lifts to QBF [13, 38]. Intuitively, asserting constraints are learnable constraints that become unit after backtracking. We give a more liberal definition as we do not refer to specific asserting constraints (such as UIP clauses).

Definition 6

(asserting constraints) Let $\mathcal {T}$ be a trail for a QCNF $\Phi$ that contains r decision literals. A clause (cube) $C\in \mathfrak {L}(\mathcal {T})$ is called asserting, if there exists some point [i, j] such that $\text {red}_\Phi ^\forall (C|_{\mathcal {T}[i,j]})$ is a unit clause (resp. $\text {red}_\Phi ^\exists (C|_{\mathcal {T}[i,j]})$ is a unit cube). Furthermore, we require that we backtrack by at least one decision level, i.e., $i<r$ or $j=0$.

Learning asserting clauses might be advantageous as it guarantees new unit propagations after backtracking to a suitable point. In addition, asserting clauses are always new.

Proposition 2

If $\mathcal {T}$ is a trail in a $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ (resp. $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$) proof of a formula $\Phi$, and if $\bot \in \mathcal {T}$ (resp. $\top \in \mathcal {T}$), then there exists a new asserting or empty clause (cube) $C\in \mathfrak {L}(\mathcal {T})$.

Furthermore, if C is non-empty, there exists a point [i, j] in the trail to which we can backtrack after learning C such that the NCC continues to hold.

Proof

We will show the case for conflicts on clauses for $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ proofs, the $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ case is completely dual.

Let the trail $\mathcal {T}$ look like

$$\begin{aligned} \mathcal {T}=( p_{(0,1)},\ldots ,p_{(0,g_0)};\mathbf {d_1},p_{(1,1)},\ldots ,p_{(1,g_1)};\ldots ;\mathbf {d_r},p_{(r,1)},\ldots ,p_{(r,g_r)} ). \end{aligned}$$

Then the sequence of learnable clauses is

$$\begin{aligned} \mathfrak {L}(\mathcal {T})=(C_{(r,g_r)},\ldots ,C_{(r,1)},\ldots ,C_{(1,g_1)},\ldots ,C_{(1,1)},C_{(0,g_0)},\ldots ,C_{(0,1)}). \end{aligned}$$

We can assume that there exists at least one existential decision literal $d_i$ such that $\bar{d}_i$ is contained in some $C\in \mathfrak {L}(\mathcal {T})$. Otherwise, the rightmost clause in $\mathfrak {L}(\mathcal {T})$ is empty since it contains negated decisions or universal literals only, which will be reduced to the empty clause $(\bot )$.

Let $k\in \{1,\ldots ,r\}$ be maximal such that an existential $\bar{d}_k$ is contained in some clause from $\mathfrak {L}(\mathcal {T})$. Let $p_{(\ell ,m)}\in \mathcal {T}$ be the propagated (non-empty) literal directly right of $d_k$ in $\mathcal {T}$ and set $D:=C_{(\ell ,m)}$. Note that $p_{(\ell ,m)}$ does not need to be on the same decision level as $d_k$. Such a $p_{(\ell ,m)}$ must exist by the NCC. We will show that D is asserting.

We consider the trail $\mathcal {T}$ at the point [k, 0], that means right before $d_k$ was decided. We will prove that $E:=\text {red}_{\Phi }^\forall (D|_{\mathcal {T}[k,0]})=(\bar{d}_k)$.

If there is an existential literal $\bar{d}_k\ne y\in E$, then $\bar{y}$ cannot have been assigned in $\mathcal {T}[k,0]$, hence we have $d_k<_\mathcal {T}\bar{y}$. But that means $\bar{y}$ had to be a decision, otherwise it would have been resolved away during clause learning. But this is a contradiction to the maximality of k. We conclude that such a y cannot exist.

Let us now assume there is a universal literal $u\in E$. Then we need $u<_\Phi d_k$ since it was not reduced during clause learning. But $\mathcal {T}$ was a trail in a $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ proof, hence $\text {lv}_{\Phi |_{\mathcal {T}[k,0]}}(d_k) =1$ and therefore $\bar{u}\in \mathcal {T}[k,0]$. Then we get ${u}\not \in E$, contradiction. Thus such a u cannot exist, and E is in fact a unit clause.

We can backtrack to the point [k, 0] (i.e., before we made the decision $d_k$) and will not hurt the NCC since the only new clause we have learned can only propagate the non-empty literal $\bar{d}_k$.

At the end, we have to show that D is new. In fact, if D was already known, we would get a conflict directly after deciding $d_k$, which would violate the NCC. Thus, D must be a new clause. $\square$

A similar result holds for the any-order model, albeit with the difference that we might not be able to learn asserting constraints. But at least we can guarantee to learn a new clause/cube.

Proposition 3

If $\mathcal {T}$ is a trail in a $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ proof for a formula $\Phi$, that has run into a conflict or in which we assigned all variables, then $\mathfrak {L}({\mathcal {T}})$ contains a new clause or cube C that is not contained in $\Phi$.

Further, if C is non-empty, there exists a point [i, j] in the trail to which we can backtrack after learning C such that the NCC continues to hold.

Proof

Case 1: $\mathcal {T}$ runs into a conflict.

Let the trail $\mathcal {T}$ look like

$$\begin{aligned} \mathcal {T}=( p_{(0,1)},\ldots ,p_{(0,g_0)};\mathbf {d_1},p_{(1,1)},\ldots ,p_{(1,g_1)};\ldots ;\mathbf {d_r},p_{(r,1)},\ldots ,p_{(r,g_r)} ). \end{aligned}$$

Then the sequence of learnable clauses is

$$\begin{aligned} \mathfrak {L}(\mathcal {T}_i)=(C_{(r,g_r)},\ldots ,C_{(r,1)},\ldots ,C_{(1,g_1)},\ldots ,C_{(1,1)},C_{(0,g_0)},\ldots ,C_{(0,1)}). \end{aligned}$$

By the NCC, we have that $g_r>1$. We will show that $C_{(r,1)}$ (which is the clause/cube we get after resolving over $p_{(r,g_r-1)},\ldots ,p_{(r,1)}$) is a new clause (cube).

Assume not. Consider the restricted clause (cube) $E:=C_{(r,1)}|_{\mathcal {T}_i[r,1]}$. Suppose that there is an existential (universal) literal $x\in E\subseteq C_{(r,1)}$. That means that x is contained in at least one antecedent clause (cube) after (and including) $p_{(r,1)}$. In particular, we need $\bar{x}\in \mathcal {T}$ (resp. $x\in \mathcal {T}$). Because x is still contained in $C_{(r,1)}$, it cannot have been resolved away during learning, hence $\bar{x}\in \mathcal {T}[r,1]$ (resp. $x\in \mathcal {T}[r,1]$). This is a contradiction to the definition of E.

We conclude that E can only contain universal (existential) literals, hence $\text {red}_{\Phi }^\forall (E)=(\bot )$ (resp. $\text {red}_{\Phi }^\exists (E)=[\top ]$). But then we would have got a conflict directly after $d_r$, which is impossible by the NCC. That means that $C_{(r,1)}$ must a new clause (cube).

We can backtrack to the point where we undo the rightmost existential (universal) literal in $\mathcal {T}$ that is contained in $C_{(r,1)}$. At this point, $C_{(r,1)}$ will not become unit since it still includes at least this one literal.

Case 2: $\mathcal {T}$ does not run into a conflict, but we assigned all variables in $\mathcal {T}$.

Assume that we cannot find such a C. Then there exists a $C\in \mathfrak {L}(\mathcal {T})$ such that $C\in \mathfrak {D}(\Phi _a)$, where $\Phi _a$ is the current formula for $\mathcal {T}$. That means there exists a cube $E\subseteq \mathcal {T}$ such that $\text {red}_{\Phi _a}^\exists (E)=C$ and E satisfies $\mathfrak {C}(\Phi _a)$. In particular, we have $\text {red}_{\Phi _a}^\exists (C|_{\mathcal {T}})=[\top ]$, which means that $\mathcal {T}$ should have run into a conflict. This is a contradiction.

We can backtrack to the point where we undo the rightmost universal literal in $\mathcal {T}$, that is contained in C. Then C will just propagate this universal literal and not an empty one. If this point is on the last decision level, we can alternatively restart. $\square$

Remark 3

To illustrate the importance of the NCC, we give an example of a $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ trail—violating the NCC—from which we cannot learn a new clause. Consider the trail $\mathcal {T}=(\textbf{x},\bot )$ for the false QCNF $\forall u\exists x\cdot (u\vee x)\wedge (u\vee \bar{x})\wedge (\bar{u}\vee x)\wedge (\bar{u}\vee \bar{x})$. The trail violates the NCC, as we got a conflict directly after the decision x. The only learnable clause is $\text {ante}_{\mathcal {T}}(\bot )=\bar{u}\vee \bar{x}$, which is obviously already known.

Another example illustrates the case where we can learn a new clause but no asserting clause. Let the trail be $\mathcal {U}:=( \textbf{x},y;\textbf{u},\bar{z},\bot )$ for the false QCNF $\forall u\exists x,y,z\cdot (\bar{x}\vee y)\wedge (x\vee y)\wedge (u\vee \bar{y}\vee \bar{z})\wedge (\bar{u}\vee \bar{y}\vee \bar{z})\wedge (u\vee \bar{y}\vee z)\wedge (\bar{u}\vee \bar{y}\vee z)$. There are two new clauses we could learn: $\bar{u}\vee \bar{y}$ or $\bar{u}\vee \bar{x}$. None of the two can become unit after backtracking since we used the decision heuristic for $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$, although we followed the NCC.

As a special case we obtain for our base model $\mathsf{{QCDCL}}$ the following situation.

Corollary 4

[Folklore, cf. [23]] If $\mathcal {T}$ is a trail in a $\mathsf{{QCDCL}}$ proof for a formula $\Phi$, that has run into a conflict, then $\mathfrak {L}({\mathcal {T}})$ contains an asserting or empty clause or cube. If $\mathcal {T}$ has not run into a conflict, but we have assigned all variables in $\mathcal {T}$, then $\mathfrak {L}(\mathcal {T})$ contains at least a new cube C.

Furthermore, if C is non-empty, there exists a point [i, j] in the trail to which we can backtrack after learning C such that the NCC continues to hold.

Figure 2 provides an overview of the four systems and their ability to learn asserting clauses and cubes. As a consequence of always learning new constraints, we infer that our models are all complete and terminating proof methods.

Theorem 5

$\mathsf{{QCDCL}}$, $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ are sound and complete proof systems.⁶ Additionally, as long as we follow the rules of decision making (especially the NCC), we will always learn the empty clause or cube at some point, no matter what decisions were made.

Proof

By Propositions 2 and 3 as well as Corollary 4 we conclude that from each trail (that has either run into a conflict or assigned all variables) we can always learn a new clause or cube. Note that these results have to be interpreted in the context of Proposition 1.

Since a given formula only consists of finitely many variables, we can only learn finitely many new clauses and cubes. We finish the proof as soon as we learn the empty clause or cube, which will happen at some point. Therefore all four systems are complete.

The soundness results from the fact that from each $\mathsf{{QCDCL}}$, $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ proof $\iota$ we can extract a long-distance Q-resolution or long-distance Q-consensus proof $\mathfrak {R}(\iota )$ for the same formula. $\square$

Fig. 2

Overview of guaranteed learnable constraints after a trail conflict in the corresponding models

5 Separations of QCDCL Systems

In this section, we will exponentially separate our three new models—where decisions do not necessarily follow the prefix order—from the plain model $\mathsf{{QCDCL}}$.

We will use the gauge lower bound technique, introduced in [9], which we will first review. This technique works on $\Sigma _3^b$ QCNFs. To ease notation, we will assume that prefixes of $\Sigma _3^b$ QCNFs have the form $\exists X\forall U\exists T$, for sets of literals X, U, T, and we will use the notions of X-, U- and T-variables and -literals. Further, we define certain types of clauses:

• X-clauses consist of X-literals only (analogously we define U-clauses and T-clauses),

• XT-clauses consist of at least one X- and at least one T-literal, but no U-literals,

• XUT-clauses consist of at least one X-, U- and T-literal, respectively.

The gauge lower bound method works for a specific class of $\Sigma _3^b$ QCNFs with the XT-property.

Definition 7

([5]) We say that $\Phi$ fulfills the XT-property, if $\mathfrak {C}(\Phi )$ contains no XT-clauses, no T-clauses that are unit (or empty) and no two T-clauses from $\mathfrak {C}(\Phi )$ are resolvable.

The XT-property extends to entire QCDCL proofs, as stated in the next lemma.

Lemma 6

([5]) If $\Phi$ is a $\Sigma _3^b$ QCNF that fulfills the XT-property, then it is not possible to derive XT-clauses or new T-clauses via long-distance Q-resolution from $\Phi$.

The gauge lower-bound method from [9] uses the next two notions of fully reduced and primitive proofs (they were implicit in [9] and stated explicitly in [4]).

Definition 8

(fully reduced proofs [4, 9]) A long-distance Q-resolution refutation $\pi$ of a QCNF $\Phi$ is fully reduced, if for each clause $C\in \pi$ that contains universal literals that are reducible, the reduction step is performed immediately and C is not used otherwise in the proof.

Fully reduced proofs are not much of a limitation. In fact, all long-distance Q-resolution proofs that we extract from a $\mathsf{{QCDCL}}$ run are already fully reduced by default. Also, we can always shorten a given long-distance Q-resolution proof by making it fully reduced.

Definition 9

(primitive proofs [4, 9]) A long-distance Q-resolution proof $\pi$ from a $\Sigma _3^b$ formula is primitive, if there are no two XUT-clauses in $\pi$ that are resolved over an X-variable.

Unlike the fully reduced property, not all proofs extracted from $\mathsf{{QCDCL}}$ are primitive, in general.

Our lower bound method will not work for all $\mathsf{{QCDCL}}$ proofs, but needs fully reduced primitive Q-resolution proofs, which are better suited for a proof-complexity analysis. Later, the challenge will be to show that certain extracted proofs from $\mathsf{{QCDCL}}$ are primitive. Note that fully reduced primitive long-distance Q-resolution proofs are always Q-resolution proofs.

The main measure for the lower bound technique is the gauge of a formula, defined in [9].

Definition 10

([9]) Let $\Phi$ be a $\Sigma _3^b$ QCNF with prefix $\exists X\forall U\exists T$. We define $W_\Phi$ as the set of all Q-resolution proofs $\pi$ from $\Phi$ of X-clauses $C_\pi$, such that $\pi$ consists of resolutions over T-literals and reductions only. We define

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

Intuitively, $\text {gauge}(\Phi )$ is the minimal number of X-literals that are piled up during the process of deriving an X-clause without using resolutions over X-literals. In other words: to get rid of all T-literals from $\Phi$, we have to pile up at least $\text {gauge}(\phi )$ many different X-literals.

All notions we introduced so far are combined into the following lower bound method:

Theorem 7

([9]) Each fully reduced primitive Q-resolution refutation of a $\Sigma _3^b$ QCNF $\Phi$ that fulfils the XT-property has size $2^{\Omega (\text {gauge}(\Phi ))}$.

Proof

We refer to the notion of quasi level-ordered proofs from [9] (there is no need to define this here). In that paper, it is explained how we can transform a $\mathsf{{QCDCL}}$ refutation of a formula that fulfils the XT-property into a quasi level-ordered $\mathsf{{Q-resolution}}$ refutation in polynomial time via an algorithm. However, the input proof does not need to be a $\mathsf{{QCDCL}}$ proof necessarily. It suffices that this proof is fully reduced and it does not contain an X-resolution over two XUT-clauses (these are the only two properties that were needed for proving Theorem 2 in [9]). In other words, this algorithm can be used to transform fully reduced primitive Q-resolution refutations into quasi level-ordered Q-resolution refutations in polynomial times.

The lower bound then follows from Theorem 5 of [9]. $\square$

Our goal is to find formulas that separate $\mathsf{{QCDCL}}$ from $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\mathsf{{QCDCL}}$ from $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$, respectively. We start with the latter.

5.1 Separation on True Formulas

The advantage of $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ (compared to $\mathsf{{QCDCL}}$) is to decide existential literals out of order while still learning asserting cubes. Since cubes are important for certificates of true formulas, it makes sense to use true QBFs for the separation.

First, we discuss two generic modifications for QBFs. The twin construction doubles all universal variables. For all clauses with universal variables a copy is created in the twin variables.

Definition 11

(twin formulas) Let $\Phi =\exists X\forall U\exists T\cdot \mathfrak {C}(\Phi )$ be a QCNF. Let $U=\{u_1,\ldots ,u_m\}$ and let $v_1,\ldots ,v_m$ be variables not occuring in $\Phi$. Then the twin formula of $\Phi$ is the QCNF $\texttt{Twin}\Phi$ defined as

$$\begin{aligned} \texttt{Twin}\Phi :=\exists X\forall (U\cup \{v_1,\ldots ,v_m\})\exists T\cdot \mathfrak {C}(\Phi )\wedge \bigwedge _{C\in \mathfrak {C}(\Phi )}C[u_1/ v_1,\ldots ,u_m/ v_m], \end{aligned}$$

where $u_i/v_i$ indicates that all occurrences of $u_i$ are substituted by $v_i$.

The second modification is the reversion of a formula.

Definition 12

If $\Phi =\mathcal {Q}_1V_1\ldots \mathcal {Q}_kV_k\cdot \bigwedge _{j=1}^mC_j$ is a QCNF with $\mathcal {Q}_i\in \{\exists ,\forall \}$ and disjoint sets of variables $V_i$ for $i=1,\ldots ,k$, then the reversion $\text {Rev}(\Phi )$ of $\Phi$ is the QCNF

$$\begin{aligned} \mathcal {Q}'_1V_1\ldots \mathcal {Q}'_kV_k\forall w\exists c_1,\ldots ,c_m\cdot (\bar{c}_1\vee \ldots \vee \bar{c}_m)\wedge \bigwedge _{j=1}^m\bigwedge _{\ell \in C_j}(\bar{\ell }\vee w\vee c_j)\wedge (\bar{\ell }\vee \bar{w}\vee c_j) \end{aligned}$$

where $\mathcal {Q}'_i=\forall$ if $\mathcal {Q}_i=\exists$, and $\mathcal {Q}'_i=\exists$ if $\mathcal {Q}_i=\forall$, and $w,c_1,\ldots ,c_m$ are new variables not contained in $\Phi$.

It is easy to prove that there exists a duality between the truth values of $\Phi$ and $\text {Rev}(\Phi )$.

Lemma 8

If $\Phi$ is a QCNF, then $\text {Rev}(\Phi )$ is true if and only if $\Phi$ is false.

Proof

Case 1: $\Phi$ is false.

Then there exists a winning strategy for the universal player of $\Phi$. We will show that $\text {Rev}(\Phi )$ has an existential winning strategy.

The existential player for $\text {Rev}(\Phi )$ can just follow the universal winning strategy for $\Phi$. That means there is at least one clause $C_j\in \mathfrak {C}(\Phi )$ falsified by the total assignment that consists of the assignment from the universal player and the corresponding response determined by the winning strategy. Then the clauses $\bar{\ell }\vee w\vee c_j$ and $\bar{\ell }\vee \bar{w}\vee c_j$ for each $\ell \in C_j$ are satisfied (for this particular j) by this assignment. Note that it does not matter how w was assigned. Therefore, the existential player for this modified formula can just set $c_j$ to true and all the other $c_i$ to false.

Case 2: $\Phi$ is true.

This case is analogous to Case 1. The universal player for the modified $\text {Rev}(\Phi )$ version follows the existential winning strategy for $\Phi$. Then the universal player can set w to true (it does not matter, actually). For each $j\in \{ 1,\ldots ,m \}$ the clause $C_j$ is satisfied, hence at least one literal $\ell \in C_j$ is set to true. Therefore for each $c_j$, the clause $\bar{\ell }\vee \bar{w}\vee c_j$ becomes the unit clause $(c_j)$ at some point under the assignment determined by the strategy. That means the existential player for $\text {Rev}(\Phi )$ has to set each $c_j$ to true, falsifying the clause $\bar{c}_1\vee \ldots \vee \bar{c}_m$.

We now have constructed a universal winning strategy for $\text {Rev}(\Phi )$. $\square$

We will use the reversion to lift hardness from false to true QCNFs. To verify a true formula, we need to create a proof using cubes. We will show that $\text {Rev}(\Phi )$ is designed such that its initial cubes are basically the negated axiom clauses of $\Phi$. Thus, a certificate of $\text {Rev}(\Phi )$ can be transformed into a refutation of $\Phi$.

Our reversion was inspired by the notion of the negation from [19]. The only change we made is adding the variable w. We did this to prevent a direct connection between an X- or U-block and an auxiliary variable $c_j$ from the last block. Our lower bound technique is based on the fact that on certain formulas we cannot have direct connections (hence: cannot directly propagate) between outer and inner quantifier blocks. The added variable w helps to maintain this property.

The next two results shows how we can transform certificates of $\text {Rev}(\Phi )$ into refutations of $\Phi$ by interpreting the cubes from the certificate as negated clauses of a refutation.

Lemma 9

Let $\Phi =\mathcal {Q}\cdot \bigwedge _{j=1}^mC_j$ be a QCNF and let $\sigma$ be an assignment that satisfies $\mathfrak {C}(\text {Rev}(\Phi ))$. Then there exists some $C\in \mathfrak {C}(\Phi )$ with $\overline{C}\subseteq \sigma$.

Proof

Since we have to satisfy the clause $(\bar{c}_1\vee \ldots \vee \bar{c}_m)$, there is some $j\in \{1,\ldots ,m\}$ with $\bar{c}_j\in \sigma$. Then the clauses $\bar{\ell }\vee w\vee c_j$ and $\bar{\ell }\vee \bar{w}\vee c_j$ have to be satisfied for each $\ell \in C_j$. We do not need to assign w, but we need to set each $\ell$ to false, hence $\overline{C}_j\subseteq \sigma$. $\square$

Proposition 10

If $\Phi$ is a false QCNF and $\rho$ is a long-distance Q-consensus certificate of $\text {Rev}(\Phi )$, then $\rho$ can be transformed into a fully reduced long-distance Q-resolution refutation $\pi$ of $\Phi$ with $|\pi |\le |\rho |$.

More precisely, for each clause $C\in \pi$ there is a cube $C'\in \rho$ with $\overline{C}\subseteq C'$. Furthermore, for each two clauses C, D that are resolved in $\pi$, the corresponding cubes $C',D'$ are resolved in $\rho$, as well.

Proof

By Lemma 9, for each initial cube $D\in \rho$ there is a clause $C\in \mathfrak {C}(\Phi )$ with $\text {red}_{\text {Rev}(\Phi )}^\exists (\overline{C})\subseteq D$ (note that the assignment from Lemma 9 can still be reduced). We substitute each initial cube D with its corresponding $\text {red}_{\text {Rev}(\Phi )}^\exists (\overline{C})$ and shorten the proof, if necessary (i.e., delete redundant resolutions and reductions). We receive a subproof $\pi '\subseteq \rho$, that is still a certificate.

After that, we negate all cubes in $\pi '$ and receive a proof $\pi$ that consists of clauses. If we interpret $\pi$ as a proof for $\Phi$ (or $\text {red}(\Phi )$ to be precise), all resolutions and reductions are still sound because the quantifiers were flipped, as well.

We can assume that in $\pi$ we will reduce as soon as possible, otherwise we could shorten the proof even more. Obviously, the last clause in $\pi$ has not received any additional literals, therefore $\pi$ is a long-distance Q-resolution refutation of $\Phi$. $\square$

For our next results, we need an even stronger property than the XT-property: We require, that clauses from the formula contain at least one U- and T-literal.

Lemma 11

If $\Phi$ is a $\Sigma _3^b$ QCNF, in which all clauses contain at least one U- and T-literal, then $\Phi$ fulfils the XT-property.

Proof

Obviously, $\Phi$ does not contain any XT- or T-clauses and therefore the XT-property is fulfilled. $\square$

We combine the results above and obtain a new lower bound technique for true formulas, which builds on the gauge technique for false formulas.

Theorem 12

Let $\Phi$ be a false $\Sigma _3^b$. Additionally, let all clauses $C\in \mathfrak {C}(\Phi )$ contain at least one U- and one T-literal. If the QCNF $\texttt{Twin}\Phi$ needs fully reduced primitive Q-resolution refutations of size s, then $\mathsf{{QCDCL}}$ certificates for $\text {Rev}(\texttt{Twin}\Phi )$ also need size s.

Proof

Let $\iota$ be a $\mathsf{{QCDCL}}$ certificate for $\text {Rev}(\texttt{Twin}\Phi )$. We will show that there exists a fully reduced primitive Q-resolution refutation $\pi$ for $\texttt{Twin}\Phi$ with $|\pi |\le |\mathfrak {R}(\iota )|$.

Let $\pi$ be the long-distance Q-resolution refutation of $\texttt{Twin}\Phi _n$ as described in Proposition 10. Then $\pi$ is fully reduced. We will show that $\pi$ is primitive.

Assume not. Then there are two XUT-clauses $B_1,B_2\in \pi$ that are resolved over some $x\in X$. By the construction of $\pi$ described in Proposition 10, we can find two cubes $D_1,D_2\in \mathfrak {R}(\iota )$ such that $\text {var}(D_i)\cap U\ne \emptyset$ and $\text {var}(D_i)\cap T\ne \emptyset$ for $i=1,2$ which are resolved over x. One of these cubes was an antecedent cube for x in some trail $\mathcal {T}\in \mathfrak {T}(\iota )$, say $D_1=\text {ante}_\mathcal {T}({x})$ (that means $\bar{x}\in D_1$).

In particular, there is some T-literal $t\in D_1$ such that ${t}<_\mathcal {T}x$ because $D_1$ must become unit. Remember that t is universal in $\text {Rev}(\texttt{Twin}\Phi )$ and we can only reduce cubes existentially. Then either t was a regular decision, or a propagation.

Case 1: t was decided.

This is only possible if all U-variables were assigned before. Hence, for each $u\in U$ there is a literal $\ell _u$ with $\text {var}(\ell _u)=u$ and $\ell _u<_\mathcal {T}t<_\mathcal {T}x$. Because decisions have to be level-ordered in $\mathsf{{QCDCL}}$, all $\ell _u$ had to have been propagated.

Let $\ell _u$ be the leftmost U-literal in $\mathcal {T}$. Consider its antecedent clause $A:=\text {ante}_\mathcal {T}(\ell _u)$.

Claim: If $\ell _u$ is the leftmost U-literal in $\mathcal {T}$, then there exists an $i\in \{1,\ldots ,m\}$ such that $c_i\in \text {var}(\text {ante}_{\mathcal {T}}(\ell _u))$ (where $c_1,\ldots ,c_m$ are the variables from $\text {Rev}(\texttt{Twin}\Phi )$ as in Definition 12).

Proof of the claim

Assume not. We will show that $A:=\text {ante}_{\mathcal {T}}(\ell _u)$ has to contain at least two different U-literals.

Assume that A only contains one U-literal, namely $\ell _u$ itself. Let $\Phi$ consist of the clauses $C_1, \ldots ,C_{m'}$ and let $\texttt{Twin}\Phi$ consist of the clauses $C_1,\ldots ,C_m$ with $m> m'$. We can assume that $\ell _u$ is a copy of a literal from $\Phi$ by the construction of a twin formula. In particular, $\ell _u$ (and $\bar{\ell }_u$) cannot be contained in the clauses $C_1,\ldots ,C_{m'}$.

Let $\rho$ be the long-distance Q-resolution derivation of A that was constructed in $\iota$, but not used for $\mathfrak {R}(\iota )$ since certificates can only make use of cubes. By assumption, A does not contain any $c_i$ or $\bar{c}_i$. However, each axiom clause from $\text {Rev}(\texttt{Twin}\Phi )$ includes at least one $c_i$ or $\bar{c}_i$. Hence, we have to resolve over these variables somehow. In particular, we need $\bar{c}_1\vee \ldots \vee \bar{c}_m\in \rho$ since this is the only axiom clause where these variables occur in a negative polarity.

We will now construct another long-distance Q-resolution derivation $\rho '$ by substituting $\bar{c}_1\vee \ldots \vee \bar{c}_m$ with $\bar{c}_1\vee \ldots \vee \bar{c}_{m'}$ in $\rho$ and gradually deleting all redundant clauses. In particular, all clauses from $\text {Rev}(\texttt{Twin}\Phi )$ that contain $\ell _u$ or $\bar{\ell }_u$ will be deleted because the corresponding $c_i$ is missing. Let $A'$ be the last clause in $\rho '$, hence $\rho '$ is a long-distance Q-resolution proof of $A'$ from $\text {Rev}(\Phi )$. Obviously, we get $A'\subseteq A$ and $\ell _u\not \in A'$ as well as $c_i,\bar{c}_i\not \in A'$ for all $i=1,\ldots ,m$. Since $\ell _u$ was the only U-literal in A, the clause $A'$ cannot have any U-literals. Therefore $A'$ is a clause consisting of universal literals only. Reducing $A'$ universally gives us the empty clause $(\bot )$, which means that we can extend $\rho '$ to a refutation of $\text {Rev}(\Phi )$. But this is a contradiction to the fact that $\text {Rev}(\Phi )$ is a true formula (by Lemma 8).

That shows that A must contain more than one U-literal. Let $\ell _u\ne z\in A$ be another U-literal. Then we need $\bar{z}<_\mathcal {T}\ell _u$ since z is existential. However, this contradicts the choice of $\ell _u$, which finishes the proof of the claim. $\square$

We want to create a contradiction by applying the claim, for which we need to show that A does not contain any literal from $\{ c_r,\bar{c}_r|\,r=1,\ldots ,m \}$.

Assume that there is such a literal. That means we can find the leftmost literal $c\in \{ c_r,\bar{c}_r|\,r=1,\ldots ,m \}$ in $\mathcal {T}$, hence $c<_\mathcal {T}\ell _u<_\mathcal {T}t<_\mathcal {T}x$. Now, c cannot have been a decision since decisions must be level-ordered. That means that c has been propagated by an antecedent clause $F:=\text {ante}_\mathcal {T}(c)$. Because c was leftmost, F cannot be the clause $\bar{c}_1\vee \ldots \vee \bar{c}_m$. It is easy to see that F then has to contain either w or $\bar{w}$ by the structure of a reversion (see Definition 12). W.l.o.g. let $w\in F$. Then we need $\bar{w}<_\mathcal {T}c<_\mathcal {T}\ell _u$. Because of the quantification order, $\bar{w}$ cannot be a decided literal. Hence $\bar{w}$ must have been propagated by some antecedent cube $E:=\text {ante}_{\mathcal {T}}(\bar{w})$. Let $\rho$ be the subproof of E from $\mathfrak {R}(\iota )$. Then there exists an initial cube $G\in \rho$ with ${w}\in G$, which is not getting resolved away in $\rho$. Furthermore, G is also an initial cube in $\mathfrak {R}(\iota )$. By Lemma 9, there exists some $H\in \mathfrak {C}(\texttt{Twin}\Phi )$ such that $\overline{H}\subseteq G$. Since each clause of $\Phi$ contains a U-literal, there is such a U-literal $v\in \overline{H}\subseteq G$ and also $v\in E$ because it cannot be resolved or reduced away. This means we need $v<_\mathcal {T}\bar{w}<_\mathcal {T}\ell _u$, which is a contradiction to the choice of $\ell _u$.

We have now shown that A does not contain any $c_r,\bar{c}_r$, $r\in \{ 1,\ldots ,m \}$. However, this is impossible by our claim. We conclude that Case 1 cannot occur.

Case 2: t was propagated.

Consider the antecedent cube $J:=\text {ante}_{\mathcal {T}}(t)$. Let $\tau$ be the subproof of J in $\mathfrak {R}(\iota )$. Then the first cubes in $\tau$ were (reduced) satisfying assignments for $\text {Rev}(\texttt{Twin}\Phi _n)$. At least one of these initial cubes in $\tau$ contains $\bar{t}$ which will not get resolved away since it appears in J. Let $I\in \tau$ be an initial cube with $\bar{t}\in I$ that does not get resolved away in $\tau$. By Lemma 9, there exists a clause $K\in \mathfrak {C}(\texttt{Twin}\Phi _n)$ such that $\overline{K}\subseteq I$. By our assumption, K contains at least one U- and one T-literal. But then also I contains at least one U-literal $\ell$. Because $\ell$ is blocked by $\bar{t}$ all the time, it does not get reduced away in $\tau$, hence $\ell \in J$.

Due to $\ell <_{\text {Rev}(\texttt{Twin}\Phi _n)}t$, we need $\ell <_\mathcal {T}t$ in order for J to become unit. W.l.o.g. let $\ell$ be the leftmost U-literal in $\mathcal {T}$ (the fact that $\ell \in J$ is not important anymore from this point on). Because of $x<_{\text {Rev}(\texttt{Twin}\Phi _n)}\ell$, the literal $\ell$ cannot be a regular decision. That means it must have been propagated.

We can repeat the argument from Case 1. We conclude that such an $\ell$ does not exist. Thus Case 2 does not occur and we get a contradiction regarding our assumption that $\pi$ was not primitive. $\square$

We now construct specific QBFs that meet the conditions of Theorem 12. We already know from [9] that the equality formulas $\texttt{Eq}_n$ of [6] have linear gauge and therefore need exponential-size fully reduced primitive Q-resolution refutations. However, not all clauses from $\texttt{Eq}_n$ contain a U-literal. We modify the formulas by adding an artificial U-literal p to the relevant clauses:

Definition 13

The QCNF $\texttt{ModEq}_n$ consists of the prefix $\exists x_1,\ldots ,x_n\forall u_1,\ldots ,u_n,p\exists t_1,\ldots ,t_n$

and the matrix $x_i\vee u_i\vee t_i$, $\bar{x}_i\vee \bar{u}_i\vee t_i$, $p\vee \bar{t}_1\vee \ldots \vee \bar{t}_n$, $\bar{p}\vee \bar{t}_1\vee \ldots \vee \bar{t}_n$ for $i=1,\ldots ,n$.

Neither this nor the $\texttt{Twin}$ modification changes the gauge of the formulas. Hence we get:

Proposition 13

It holds $\text {gauge}(\mathtt {\texttt{Twin}\texttt{ModEq}_n})= n$. Hence, $\mathtt {\texttt{Twin}\texttt{ModEq}_n}$ needs exponential-size fully reduced primitive Q-resolution refutations.

Proof

Since all axiom clauses contain T-literals, we have to get rid of them somehow. The only four clauses that contain T-literals in a negative polarity are the clauses $p\vee \bar{t}_1\vee \ldots \vee \bar{t}_n$, $\bar{p}\vee \bar{t}_1\vee \ldots \vee \bar{t}_n$, $q\vee \bar{t}_1\vee \ldots \vee \bar{t}_n$ and $\bar{q}\vee \bar{t}_1\vee \ldots \vee \bar{t}_n$, where q is the copy of p. Hence, we have to use at least one of them in order to derive an X-clause. In particular, we have to resolve over each $t_i$. The only four clauses in which $t_i$ occurs in a positive polarity are $x_i\vee u_i\vee t_i$, $\bar{x}_i\vee \bar{u}_i\vee t_i$, $x_i\vee v_i\vee t_i$ and $\bar{x}_i\vee \bar{v}_i\vee t_i$, where $v_i$ is the copy of $u_i$. In each case we will pile up $x_i$ or $\bar{x}_i$ for each resolution over $t_i$. Therefore, our X-clause at the end will contain at least n different X-literals.

Hence $\text {gauge}(\mathtt {\texttt{Twin}\texttt{ModEq}_n})= n$. The second claim then follows from Theorem 7. $\square$

The lower bound for the true QBFs then follows with Theorem 12.

Corollary 14

$\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$ needs exponential-size $\mathsf{{QCDCL}}$ certificates.

We now use a direct construction to show that $\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$ is easy for $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$.

Proposition 15

$\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$ has polynomial-size $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ certificates.

Proof

Let us first list all the clauses of $\texttt{Twin}\texttt{ModEq}_n$. It consists of the prefix

$$\begin{aligned} \exists x_1,\ldots ,x_n\forall u_1,\ldots ,u_n,p,v_1,\ldots ,v_n,q\exists t_1,\ldots ,t_n \end{aligned}$$

and the matrix

$$\begin{aligned} C_{(i,1)}:= x_i\vee u_i\vee t_i \,\,\,\,\, C_1:= p\vee \bar{t}_1\vee \ldots \vee \bar{t}_n \\ C_{(i,2)}:= \bar{x}_i\vee \bar{u}_i\vee t_i \,\,\,\,\, C_2:= \bar{p}\vee \bar{t}_1\vee \ldots \vee \bar{t}_n \\ C_{(i,3)}:= x_i\vee v_i\vee t_i \,\,\,\,\, C_3:=q\vee \bar{t}_1\vee \ldots \vee \bar{t}_n \\ C_{(i,4)}:= \bar{x}_i\vee \bar{v}_i\vee t_i \,\,\,\,\, C_4:= \bar{q}\vee \bar{t}_1\vee \ldots \vee \bar{t}_n \\ \end{aligned}$$

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

Then the true QCNF $\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$ consists of the prefix

$$\begin{aligned} \forall x_1,\ldots ,x_n\exists u_1,\ldots ,u_n,p,v_1,\ldots ,v_n,q\forall t_1,\ldots ,t_n,w\exists M, \end{aligned}$$

with $M:=\{ c_{(i,j)},c_j|\, i=1,\ldots ,n,\, j=1,\ldots ,4 \}$, and the matrix

$$\begin{aligned} E:= \bigvee _{i=1}^n\bigvee _{j=1}^4\bar{c}_{(i,j)}\vee \bigvee _{k=1}^4\bar{c}_k \\&\bar{x}_i\vee w\vee c_{(i,1/3)}\,\,\,\,\, \bar{u}_i\vee w\vee c_{(i,1)} \,\,\,\,\, \bar{v}_i\vee w\vee c_{(i,3)} \\&\bar{x}_i\vee \bar{w}\vee c_{(i,1/3)}\,\,\,\,\, \bar{u}_i\vee \bar{w}\vee c_{(i,1)} \,\,\,\,\, \bar{v}_i\vee \bar{w}\vee c_{(i,3)} \\&{x}_i\vee w\vee c_{(i,2/4)}\,\,\,\,\, {u}_i\vee w\vee c_{(i,2)} \,\,\,\,\, {v}_i\vee {w}\vee c_{(i,4)} \\&{x}_i\vee \bar{w}\vee c_{(i,2/4)}\,\,\,\,\, {u}_i\vee \bar{w}\vee c_{(i,2)}\,\,\,\,\, {v}_i\vee \bar{w}\vee c_{(i,4)} \\&\bar{t}_i\vee w\vee c_{(i,1/2/3/4)}\\&\bar{t}_i\vee \bar{w}\vee c_{(i,1/2/3/4)}\\&\bar{p}\vee w\vee c_1\,\,\,\,\, {p}\vee w \vee c_2 \,\,\,\,\, \bar{q}\vee w\vee c_3 \,\,\,\,\, {q}\vee {w}\vee c_4 \\&\bar{p}\vee \bar{w}\vee c_1\,\,\,\,\, {p}\vee \bar{w}\vee c_2\,\,\,\,\, \bar{q}\vee \bar{w}\vee c_3 \,\,\,\,\, {q}\vee \bar{w}\vee c_4 \\&{t}_i\vee w\vee c_{1/2/3/4}\\&{t}_i\vee \bar{w}\vee c_{1/2/3/4} \end{aligned}$$

for $i=1,\ldots ,n$, where variables like $c_{(i,1/3)}$ decode two versions of this clause: One clause with $c_{(i,1)}$ and the other with $c_{(i,3)}$ (analogously with $c_{(i,2/4)}$, $c_{(i,1/2/3/4)}$ and $c_{1/2/3/4}$).

Let us now construct a polynomial size $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ certificate. At first, we would like to learn the cubes

$$\begin{aligned} D_{(i,1)}&:=\bar{x}_i\wedge \bar{u}_i\wedge \bar{t}_i\\ D_{(i,2)}&:={x}_i\wedge {u}_i \wedge \bar{t}_i\\ D_1&:=\bar{p}\wedge {t}_1\wedge \ldots \wedge {t}_n \end{aligned}$$

for $i=1,\ldots ,n$. In order to learn $D_{(i,1)}$, we will make (level-ordered) decisions that satisfy all literals from $D_{(i,1)}$, but falsify all the other $D_{(i',1)}$ for $i'\ne i$. For example, we set $x_i$, $u_i$ and $t_i$ to false, and we can assign all the other variables left of w arbitrarily. Note that until we reach w, we will never make any propagations since w or $\bar{w}$ is blocking them. After having decided all variables left of w, we will decide w and potentially trigger some propagations. However, the variable $c_{(i,1)}$ will never be propagated because all clauses containing it are already satisfied. After this we will set $c_{(i,1)}$ to false and all the remaining variables to true.

We now have satisfied the clause E. Furthermore, we have set all $c_{(i',j)}$ and $c_{k}$ to true except $c_{(i,1)}$. Hence we have satisfied all clauses except the four clauses containing $c_{(i,1)}$. But these two clauses were already satisfied because we have satisfied the cube $D_{(i,1)}$ with the decisions left of w.

Let $\mathcal {T}_{(i,1)}$ be the trail we have constructed now. We can extract the cube

$$\begin{aligned} \bar{x}_i\wedge \bar{u}_i\wedge \bar{t}_i\wedge \bar{c}_{(i,1)}\wedge \bigwedge _{(i',j)\in (\{1,\ldots ,n\}\times \{1,2,3,4\})\backslash \{(i,1)\}}c_{(i',j)}\wedge \bigwedge _{k=1}^4c_k, \end{aligned}$$

which, as an assignment, already satisfies all clauses from $\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$. This cube can be existentially reduced to $D_{(i,1)}$, which is the cube we learn from $\mathcal {T}_{(i,1)}$. Analogously, we can learn the cubes $D_{(i,2)}$ for $i=1,\ldots , n$ via some analogue trails $\mathcal {T}_{(i,2)}$.

It remains to learn the cube $D_1$, which represents the clause $C_1\in \mathfrak {C}(\texttt{Twin}\texttt{ModEq}_n)$. We will construct a trail $\mathcal {T}_1$ which includes (level-ordered) decisions that satisfy $D_1$. But now we have to make sure not to trigger propagations via $D_{(i,1)}$ or $D_{(i,2)}$ since we must not set $t_i$ to false. This can be done by setting all $x_i$ to false and all $u_i$ to true. Then we can set p to false and all $t_i$ to true. The remaining variables left of w can again be decided arbitrarily. Then we set w to true and potentially trigger some propagations of $c_{(i,j)}$ or $c_k$, which is not a problem since $c_1$ will never be propagated (the clauses containing $c_1$ are already satisfied). Then we set $c_1$ to false and all remaining variables can be set to true.

As with $\mathcal {T}{(i,1)}$, we have satisfied all clauses from $\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$. We can extract the cube

$$\begin{aligned} \bar{p}\wedge {t}_1\wedge \ldots \wedge {t}_n\wedge \bar{c}_{1}\wedge \bigwedge _{i=1}^n\bigwedge _{j=1}^4c_{(i,j)}\wedge \bigwedge _{k=2}^4c_k, \end{aligned}$$

from $\mathcal {T}_1$, which already satisfies the matrix and can be existentially reduced to $D_1$.

We will now define the cubes

$$\begin{aligned} R_i:=\bar{x}_i\wedge \bar{u}_i\wedge \bar{p}\wedge \bigwedge _{k=i+1}^n(u_k\wedge \bar{u}_k)\wedge \bigwedge _{\ell =1}^{i-1}{t}_\ell \\ L_i:={x}_i\wedge {u}_i\wedge \bar{p}\wedge \bigwedge _{k=i+1}^n(u_k\wedge \bar{u}_k)\wedge \bigwedge _{\ell =1}^{i-1}{t}_\ell \end{aligned}$$

for $i=2,\ldots ,n-1$. We will construct trails $\mathcal {U}_{n-1},\mathcal {V}_{n-1},\ldots ,\mathcal {U}_2,\mathcal {V}_2$ with which we will gradually learn the clauses $R_{n-1},L_{n-1},\ldots ,R_2,L_2$.

We start with

$$\begin{aligned} \mathcal {U}_{n-1}:=( \varvec{\bar{p}};\varvec{\bar{x}_1};\varvec{\bar{u}_1},{t}_1;\ldots ;\varvec{\bar{x}_{n-1}};\varvec{\bar{u}_{n-1}},{t}_{n-1},\bar{t}_n,\bar{x}_n,\top ) \end{aligned}$$

with antecedent cubes

$$\begin{aligned} \text {ante}_{\mathcal {U}_{n-1}}({t}_j)&=D_{(j,1)}\\ \text {ante}_{\mathcal {U}_{n-1}}(\bar{t}_n)&=D_1\\ \text {ante}_{\mathcal {U}_{n-1}}(\bar{x}_n)&=D_{(n,2)}\\ \text {ante}_{\mathcal {U}_{n-1}}(\top )&=D_{(n,1)} \end{aligned}$$

for $j=1,\ldots ,n-1$. We learn the cube $R_{n-1}=\left( \left( D_{(n,1)}{\mathop {\bowtie }\limits ^{x_n}}D_{(n,2)}\right) {\mathop {\bowtie }\limits ^{t_n}}D_1\right) {\mathop {\bowtie }\limits ^{t_{n-1}}}D_{(n-1,1)}$.

Analogously, by flipping some polarities, we construct the trail $\mathcal {V}_{n-1}$ and learn the cube $L_{n-1}=\left( \left( D_{(n,1)}{\mathop {\bowtie }\limits ^{x_n}}D_{(n,2)}\right) {\mathop {\bowtie }\limits ^{t_n}}D_1\right) {\mathop {\bowtie }\limits ^{t_{n-1}}}D_{(n-1,2)}$. Note that $R_{n-1}$ will not interfere with the assignments in $\mathcal {V}_{n-1}$.

Assume we have already learned the clauses $R_{n-1},L_{n-1},\ldots , R_i,L_i$ for some $i\in \{3,\ldots ,n-1\}$. Then we can construct the following trail:

$$\begin{aligned} \mathcal {U}_{i-1}:=(\varvec{\bar{p}};\varvec{\bar{x}_1};\varvec{\bar{u}_1},{t}_1;\ldots ;\varvec{\bar{x}_{i-1}};\varvec{\bar{u}_{i-1}},{t}_{i-1},{x}_i,\top ) \end{aligned}$$

with antecedent cubes

$$\begin{aligned} \text {ante}_{\mathcal {U}_{i-1}}({t}_j)&=D_{(j,1)}\\ \text {ante}_{\mathcal {U}_{i-1}}({x}_i)&=R_i\\ \text {ante}_{\mathcal {U}_{i-1}}(\top )&=L_i \end{aligned}$$

for $j=1,\ldots ,i-1$. We learn the cube $R_{i-1}=\left( L_i{\mathop {\bowtie }\limits ^{x_i}}R_i\right) {\mathop {\bowtie }\limits ^{t_{i-1}}}D_{(i-1,1)}$. Analogously, we can construct the trail $\mathcal {V}_{i-1}$ and learn $L_{i-1}=\left( L_i{\mathop {\bowtie }\limits ^{x_i}}R_i\right) {\mathop {\bowtie }\limits ^{t_{i-1}}}D_{(i-1,2)}$.

After having learned the cubes $R_{n-1},L_{n-1},\ldots ,R_2,L_2$, we construct two more trails, namely

$$\begin{aligned} \mathcal {U}_1:=(\varvec{\bar{p}};\varvec{\bar{x}_1};\varvec{\bar{u}_1},{t}_1,{x}_2,\top ) \end{aligned}$$

with antecedent cubes

$$\begin{aligned} \text {ante}_{\mathcal {U}_{1}}({t}_1)&=D_{(1,1)}\\ \text {ante}_{\mathcal {U}_{1}}({x}_2)&=R_2\\ \text {ante}_{\mathcal {U}_{1}}(\top )&=L_2, \end{aligned}$$

from which we learn $[\bar{x}_1]=\left( L_2{\mathop {\bowtie }\limits ^{x_2}}R_2\right) {\mathop {\bowtie }\limits ^{t_1}}D_{1,1}$, and the trail

$$\begin{aligned} \mathcal {V}_1:=({x}_1;\varvec{\bar{p}};\mathbf {{u}_1},{t}_1,{x}_2,\top ) \end{aligned}$$

with antecedent cubes

$$\begin{aligned} \text {ante}_{\mathcal {V}_{1}}({x}_1)&=[x_1]\\ \text {ante}_{\mathcal {V}_{1}}({t}_1)&=D_{(1,2)}\\ \text {ante}_{\mathcal {V}_{1}}({x}_2)&=R_2\\ \text {ante}_{\mathcal {V}_{1}}(\top )&=L_2, \end{aligned}$$

from which we learn the empty cube $[\top ]=\text {red}_{\text {Rev}(\texttt{Twin}\mathtt {ModEq_n})}^\exists \left( \left( L_2{\mathop {\bowtie }\limits ^{x_2}}R_2\right) {\mathop {\bowtie }\limits ^{t_1}}D_{(1,2)}\right) {\mathop {\bowtie }\limits ^{x_1}}[x_1]$.

All in all, we have constructed a $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ certificate using the $4n-1$ trails

$$\begin{aligned} \mathcal {T}_{(1,1)},\ldots ,\mathcal {T}_{(n,1)},\mathcal {T}_{(1,2)}\ldots ,\mathcal {T}_{(n,2)},\mathcal {T}_1,\mathcal {U}_{n-1},\mathcal {V}_{n-1},\ldots ,\mathcal {U}_1,\mathcal {V}_1. \end{aligned}$$

$\square$

Corollary 16

$\mathsf{{QCDCL}}$ and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ are exponentially separated on true formulas.

5.2 Separation on False Formulas

For separating $\mathsf{{QCDCL}}$ and $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, we recall the completion principle $\texttt{CR}_n$ of [18].

Definition 14

([18]) The false QCNF $\texttt{CR}_n$ consists of the prefix $\exists X\forall U\exists T$ with

$$\begin{aligned} X:=\{ x_{(i,j)}|\, i,j\in \{1,\ldots ,n\} \}, \,\,\,\,\, U:=\{ u \}, \,\,\,\,\, T:=\{ a_i,b_i|\, i\in \{1,\ldots ,n\} \} \end{aligned}$$

and the matrix

$$\begin{aligned}&x_{(i,j)} \vee u\vee a_i \,\,\,\,\, \bar{x}_{(i,j)}\vee \bar{u}\vee b_j \,\,\,\,\, \bar{a}_1\vee \ldots \vee \bar{a}_n \,\,\,\,\, \bar{b}_1\vee \ldots \vee \bar{b}_n \end{aligned}$$

for $i,j=1,\ldots ,n$.

For the lower bound, we will use the modification $\texttt{Twin}\texttt{CR}_n$. As we show, cube learning becomes rather useless with the $\texttt{Twin}$ modification. This fact helps us to ensure that $\mathsf{{QCDCL}}$ refutations of $\texttt{Twin}\texttt{CR}_n$ are primitive, and thus we can apply the gauge lower-bound method.

Similarly as in Proposition 13 we can compute the gauge.

Lemma 17

It holds $\text {gauge}(\texttt{Twin}\texttt{CR}_n)=n$.

Proof

For the derivation of an X-clause we need at least one of the clauses $\bar{a}_1\vee \ldots \vee \bar{a}_n$ or $\bar{b}_1\vee \ldots \vee \bar{b}_n$ since we have to get rid of all T-literals. In particular, w.l.o.g. we have to resolve over each $a_i$. For this, we need one of the clauses $x_{(i,j)} \vee u\vee a_i$ or $x_{(i,j)} \vee v\vee a_i$ for each i. That means for each i we will pile up at least one $x_{(i,j)}$ for some j. Therefore $\text {gauge}(\texttt{Twin}\texttt{CR}_n)=n$. $\square$

The main work is to check that $\mathsf{{QCDCL}}$ refutations of $\texttt{Twin}\texttt{CR}_n$ are primitive.

Proposition 18

If $\iota$ is a $\mathsf{{QCDCL}}$ refutation of $\texttt{Twin}\texttt{CR}_n$, then $\mathfrak {R}(\iota )$ is fully reduced and primitive.

Proof

It suffices to show that $\mathfrak {R}(\iota )$ is primitive. Assume not.

Then there exists two XUT-clauses $C,D\in \mathfrak {R}(\iota )$ that are resolved over an X-literal, say x. One of these two clauses has to be the antecedent clause of x by the definition of clause learning, say $C=\text {ante}_{\mathcal {T}}(x)$ for some trail $\mathcal {T}\in \mathfrak {T}(\iota )$. Let $t_1\in C$ be one of the T-literals. We want to show, that there exists a U-literal w with $w<_\mathcal {T}x$.

Assume that no such w exists. Since C had to become unit at the propagation of x, we need $\bar{t}_1<_\mathcal {T}x$. The literal $\bar{t}_1$ cannot be a decision in $\mathcal {T}$, since this would mean that we assigned all U-variables earlier in the trail, which contradicts our assumption. Hence $\bar{t}_1$ must have been a propagation.

Starting with $i=1$, we define $F_i:=\text {ante}_{\mathcal {T}}(\bar{t}_i)$. Now, $F_i$ cannot contain U-literals since we cannot falsify these literals before assigning $\bar{t}_i$. Because of the XT-property (and Lemma 6), $F_i$ cannot contain X-literals, as well (otherwise it would be an XT-clause). But if the XT-property is fulfilled, we cannot derive unit T-clauses, therefore $F_i$ has to contain at least one additional T-literal, say $t_{i+1}\in F_i$.

This argument can be repeated for each $i\in \mathbb {N}$, which means we could find an infinite amount of T-literals $\bar{t}_i$ that must be all contained in $\mathcal {T}$, which is obviously not possible. This shows that our assumption was false and we can indeed find such a U-literal $w<_\mathcal {T}\bar{t}_1<_\mathcal {T}x$.

W.l.o.g. let w be the first (leftmost) U-literal in $\mathcal {T}$. Define $A:=\text {ante}_\mathcal {T}(w)$. Clearly, A is a cube. We will show that A contains at least two different U-literals. Then, since w was the first U-literal in $\mathcal {T}$, A cannot become unit until at least one U-literal was assigned, which would be a contradiction.

Now, A is a cube that was derived during cube learning from cubes that represent satisfying (partial) assignments of the matrix of $\mathtt {\texttt{Twin}\texttt{CR}_n}$. Let D be a cube that satisfies the matrix of $\mathtt {\texttt{Twin}\texttt{CR}_n}$. Because we have to satisfy the clauses $\bar{a}_1\vee \ldots \vee \bar{a}_n$ and $\bar{b}_1\vee \ldots \vee \bar{b}_n$, there exists an $r\in \{1,\ldots ,n\}$ with $\bar{a}_r\in D$ and an $s\in \{1,\ldots ,n\}$ with $\bar{b}_s\in D$. Furthermore, we have to satisfy the clauses $x_{(r,s)} \vee u\vee a_r$, $x_{(r,s)} \vee v\vee a_r$, $\bar{x}_{(r,s)}\vee \bar{u}\vee b_s$ and $\bar{x}_{(r,s)}\vee \bar{v}\vee b_s$. That means we have to assign u in some polarity. W.l.o.g. let $u\in D$. Then we have to set $x_{(r,s)}$ to false, hence $\bar{x}_{(r,s)}\in D$. In order to satisfy $x_{(r,s)} \vee v\vee a_r$, we have to set v to true, as well. Therefore we get $v\in D$.

We conclude, that $u\in D$ if and only if $v\in D$, and analogously $\bar{u}\in D$ if and only if $\bar{v}\in D$. This means that we will never be able to resolve such two learned cubes in $\iota$ since we cannot create universal tautologies in cubes. In particular, we have proven that A contains at least two U-literals, which leads to a contradiction as described above. $\square$

Applying Theorem 7 then yields the lower bound.

Corollary 19

$\texttt{Twin}\texttt{CR}_n$ needs exponential-sized $\mathsf{{QCDCL}}$ refutations.

On the other hand, $\texttt{Twin}\texttt{CR}_n$ is easy for $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$. Basically, we can simulate the Q-resolution refutation of $\texttt{CR}_n$ from [17], because we can decide universal literals out of order.

Proposition 20

$\texttt{Twin}\texttt{CR}_n$ has polynomial-sized $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ refutations.

Proof

For each $k=1,\ldots ,n$ we construct the trail

$$\begin{aligned} \mathcal {T}_k:=( \varvec{\bar{x}_{(1,k)}};\ldots ;\varvec{\bar{x}_{(n,k)}}; \varvec{\bar{u}}, a_1,\ldots ,a_n,\bot ) \end{aligned}$$

with antecedent clauses

$$\begin{aligned} \text {ante}_{\mathcal {T}_k}(a_i)= x_{(i,k)}\vee u\vee a_i, \,\,\,\,\, \text {ante}_{\mathcal {T}_k}(\bot )=\bar{a}_1\vee \ldots \vee \bar{a}_n, \end{aligned}$$

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

Resolving $\bar{a}_1\vee \ldots \vee \bar{a}_n$ over each $\text {ante}_{\mathcal {T}_k}(a_i)$ gives us the clause $E_k:=x_{(1,k)}\vee \ldots \vee x_{(n,k)}$, which we will learn. Note that the trails and the learned clauses will not affect each other, hence the order in which we construct these n trails does not matter. Next, we construct the trails $\mathcal {U}_1,\ldots , \mathcal {U}_{n-1}$ (in that order). From each $\mathcal {U}_k$ we learn the clause $C_k:=\bar{u}\vee b_k$. While constructing $\mathcal {U}_k$, we assume that $C_1,\ldots ,C_{k-1}$ were already learned. Then, $\mathcal {U}_k$ looks as follows:

$$\begin{aligned} \mathcal {U}_k:=( \textbf{u},b_1,\ldots ,b_{k-1};\textbf{v};\varvec{\bar{b}_k },\bar{x}_{(1,k)},\ldots ,\bar{x}_{(n,k)},\bot ) \end{aligned}$$

with antecedent clauses

$$\begin{aligned} \text {ante}_{\mathcal {U}_k}(b_j)=C_j, \,\,\,\,\, \text {ante}_{\mathcal {U}_k}(\bar{x}_{(i,k)})=\bar{x}_{(i,k)}\vee \bar{u}\vee b_k, \,\,\,\,\, \text {ante}_{\mathcal {U}_k}(\bot )=E_k, \end{aligned}$$

for $i=1,\ldots ,n$ and $j=1,\ldots ,k-1$. Resolving $E_k$ over each $\text {ante}_{\mathcal {U}_k}(\bar{x}_{(i,k)})$ leads to the learnable clause $C_k$. Having learned the clauses $C_1,\ldots ,C_{n-1}$, we continue with the trail $\mathcal {V}$, which will be the last one. It looks as follows:

$$\begin{aligned} \mathcal {V}:=( \textbf{u},b_1,\ldots ,b_{n-1},\bar{b}_n,\bar{x}_{(1,n)},\ldots ,\bar{x}_{(n,n)},\bot ) \end{aligned}$$

with antecedent clauses

$$\begin{aligned}&\text {ante}_{\mathcal {V}}(b_j)=C_j, \,\,\,\,\, \text {ante}_{\mathcal {V}}(\bar{b}_n)=\bar{b}_1\vee \ldots \vee \bar{b}_n, \,\,\,\,\, \text {ante}_{\mathcal {V}}(\bar{x}_{(i,n)})=\bar{x}_{(i,n)}\vee \bar{u}\vee b_n, \\ {}&\text {ante}_{\mathcal {V}}(\bot )=E_n, \end{aligned}$$

for $i=1,\ldots ,n$ and $j=1,\ldots ,n-1$. Since we only made a universal decision, we can learn the empty clause $(\bot )$ from $\mathcal {V}$ by resolving over everything.

Thus we constructed a $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ refutation using $2n+1$ trails. $\square$

Besides $\texttt{TwinCR}_n$, we can find further separations between $\mathsf{{QCDCL}}$ and $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$. The QCNFs $\texttt{MirrorCR}_n$ were introduced in [4] as a modification of $\texttt{CR}_n$, where it was shown that the formula is hard for several variants of QCDCL, including our base model $\mathsf{{QCDCL}}$. It is notable that the matrix of $\texttt{MirrorCR}_n$ is unsatisfiable, and therefore we will never perform cube learning.

Definition 15

The false QCNF $\texttt{MirrorCR}_n$ consists of the prefix

$$\begin{aligned} \exists x_{(1,1)},\ldots ,x_{(n,n)}\forall u\exists a_1,\ldots ,a_n,b_1,\ldots ,b_n \end{aligned}$$

and the matrix

$$\begin{aligned} x_{(i,j)}\vee u\vee a_i&\,\,\,\,\, \bar{a}_1\vee \ldots \vee \bar{a}_n\\ \bar{x}_{(i,j)}\vee \bar{u}\vee b_j&\,\,\,\,\, \bar{b}_1\vee \ldots \vee \bar{b}_n\\ x_{(i,j)}\vee \bar{u}\vee \bar{a}_i&\,\,\,\,\,{a}_1\vee \ldots \vee {a}_n\\ \bar{x}_{(i,j)}\vee u\vee \bar{b}_j&\,\,\,\,\, {b}_1\vee \ldots \vee {b}_n \quad \text {for}\,\, i,j\in \{1,\ldots ,n\}. \end{aligned}$$

Proposition 21

([4]) $\texttt{MirrorCR}_n$ needs exponential-sized $\mathsf{{QCDCL}}$ refutations.

Proposition 22

$\texttt{MirrorCR}_n$ has polynomial-sized $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ refutations.

Proof

At first, we will derive the clauses $A_k:=x_{(1,k)}\vee \ldots \vee x_{(n,k)}$ for each $k=1,\ldots ,n$. Suppose, we have already learned $A_1,\ldots ,A_{k-1}$. We construct the trail $\mathcal {T}_k$ as follows:

$$\begin{aligned} \mathcal {T}_k:=( \varvec{\bar{x}_{(1,k)}};\ldots ;\varvec{\bar{x}_{(n,k)}};\varvec{\bar{u}},a_1,\ldots ,a_n,\bot ) \end{aligned}$$

with

$$\begin{aligned} \text {ante}_{\mathcal {T}_k}(a_i)&=x_{(i,k)}\vee u\vee a_i\\ \text {ante}_{\mathcal {T}_k}(\bot )&=\bar{a}_1\vee \ldots \vee \bar{a}_n \end{aligned}$$

for $i=1,\ldots ,n$. From this trail we can learn $E_k$ by resolving over all $a_i$ and then we restart.

Our next goal is to learn the clauses $B_k:=\bar{u}\vee b_k$ for each $k=1,\ldots ,n-1$. We now suppose that we have already learned $A_1,\ldots ,A_n$ and $B_1,\ldots ,B_{k-1}$. We construct the trail $\mathcal {U}_k$ as follows:

$$\begin{aligned} \mathcal {U}_k:= ( \textbf{u},b_1,\ldots ,b_{k-1};\varvec{\bar{b}_k},\bar{x}_{(1,k)},\ldots ,\bar{x}_{(n,k)},\bot ) \end{aligned}$$

with

$$\begin{aligned} \text {ante}_{\mathcal {U}_{k}}(b_j)&=B_j\\ \text {ante}_{\mathcal {U}_{k}}(\bar{x}_{(i,k)})&=\bar{x}_{(i,k)}\vee \bar{u}\vee b_k\\ \text {ante}_{\mathcal {U}_{k}}(\bot )&=A_k \end{aligned}$$

for $j=1,\ldots ,k-1$ and $i=1,\ldots ,n$. We learn $B_k$ by resolving $A_k$ over all $x_{(i,k)}$. After this we backtrack back to the point where we decided $\bar{b}_k$.

Our last trail, from which we plan to learn the empty clause, looks as follows:

$$\begin{aligned} \mathcal {U}_n:=(\textbf{u},b_1,\ldots ,b_n,\bot ) \end{aligned}$$

with

$$\begin{aligned} \text {ante}_{\mathcal {U}_{n}}(b_j)&=B_j\\ \text {ante}_{\mathcal {U}_n}(\bot )&=\bar{b}_1\vee \ldots \vee \bar{b}_n. \end{aligned}$$

We resolve over all $b_j$ and obtain $(\bot )$. $\square$

Corollary 23

$\texttt{MirrorCR}_n$ is hard for $\mathsf{{QCDCL}}$, but easy for $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$.

Corollary 24

$\mathsf{{QCDCL}}$ and $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ are exponentially separated on false formulas.

We combine both separations into our main result:

Theorem 25

• a) $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ is exponentially stronger than $\mathsf{{QCDCL}}$ on false formulas.

• b) $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ is exponentially stronger than $\mathsf{{QCDCL}}$ on true formulas.

• c) $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ is exponentially stronger than $\mathsf{{QCDCL}}$ both on false and true formulas.

6 Experiments

One of the aspirations of proof complexity is to explain and predict solver behaviour, in particular running time. In this section, we evaluate how well our proof-complexity results transfer to the ‘real world’ of QCDCL implemented in a solver.

For our experiments we picked the QCDCL solver Qute⁷ [29, 33], and implemented each of the aforementioned QCDCL variants: $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$, and $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ (Qute could already run in a mode that corresponds to $\mathsf{{QCDCL}}$). In order to ensure compliance with the NCC (Definition 4), we needed to adapt some of Qute’s internal data structures, and so for the sake of a fair comparison we also report on a version called $\textsf{QCDCL3}$: algorithmically plain $\mathsf{{QCDCL}}$ but with the new data structures that are required for the other variants (up to 3 watched literals rather than the usual 2, hence the name).⁸

We performed two experiments. In the first, we evaluated each QCDCL variant on the first 100 formulas from each separation family—$\texttt{Twin}\texttt{CR}$, $\texttt{MirrorCR}$, and $\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$—running the solver with a time limit of 600 s on each individual formula on a machine with two 16-core Intel${\circledR }$ Xeon${\circledR }$ E5-2683 v4@2.10GHz CPUs and 512GB RAM running Ubuntu 20.04.3 LTS on Linux 5.4.0-48, organizing the computation with the help of GNU Parallel [35].

In the second, we additionally evaluated each QCDCL variant on the formulas from the latest two QBF Evaluations, 2020⁹ and 2022¹⁰ (there was no evaluation of QBF solvers in 2021), in both PCNF and QCIR categories, with the same time limit of 10 min. This was executed on a different cluster with heterogeneous machines powered by different Intel${\circledR }$ Xeon${\circledR }$ CPUs and AMD${\circledR }$ EPYC${\circledR }$ 7402@2.80GHz.

For all of our experiments we executed Qute with the same, default parameters for all heuristics. However, in order to obtain any meaningful results on the separation formulas, we had to tweak the initialization process of Qute’s decision heuristic, which determines the next branching variable. Previously, the heuristic was initialized in prefix order, giving higher preference to variables earlier in the prefix. As a result, on our separation formulas, the solver kept branching on and learning clauses involving only outermost variables, and never made any decisions out of the prefix order even when allowed to, because the heurstic never suggested to. We changed this default initialization to go in reverse prefix order, and adopted this change for both experiments and all runs. We emphasize that this manual change affects only the initialization values (afterwards the heuristic updates according to the same rules as before), and also that the previous setting of in-order default initialization was an arbitrary choice.

6.1 Separation Formulas

In Figs. 3 and 4 we plot running times of the different QCDCL versions as a function of n. Any gaps in the plotted lines indicate the solver timed out at 600 seconds for that particular formula. In general, the proof complexity results are closely mirrored in solver performance, though there is occasionally a bit of surprise.

In Fig. 3, we see that for $\texttt{Twin}\texttt{CR}$ the configuration $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ is best and scales reasonably well up to $n=100$. But there are also gaps—for some reason the solver’s heuristics appear to be fooled for some particular formulas and fail to navigate towards the short proof. Overall, $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ manages to solve 87 out the first 100 $\texttt{Twin}\texttt{CR}$ formulas. $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$, which should theoretically be at least as good as $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, comes a distant second and fails to solve anything beyond $n=16$. $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ appears to be off to a good start, but also quickly loses breath solving nothing after $n=10$. The two vanilla variants $\mathsf{{QCDCL}}$ and $\textsf{QCDCL3}$ scale exponentially all the way as they should.

Fig. 3

Performance on $\texttt{Twin}\texttt{CR}_n$ (above) and $\texttt{MirrorCR}_n$ (below) Legends are sorted best-to-worst

The picture on the related $\texttt{MirrorCR}$ formulas (Fig. 3 below) is boring in comparison and perfectly corresponds to our theoretical results. The two variants that have short proofs—$\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$—are also fast in practice, and everything else is dead exponential.

Finally, $\text {Rev}(\texttt{Twin}\texttt{ModEq})$ in Fig. 4 paint a picture somewhat similar to $\texttt{Twin}\texttt{CR}$, though with a different set of peculiarities. The best variant is $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$, and unlike $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ on $\texttt{Twin}\texttt{CR}$, it solves all formulas up to $n=100$ very fast. The second best is $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$, but once again it drops out relatively early (last solved is $n=26$) in spite of its theoretical superiority. An interesting thing seems to happen to $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, which appears to be helplessly off to an exponential path, but somehow recovers and solves $n=15, 16$ fast, only to completely drop out afterwards. The two vanilla variants $\mathsf{{QCDCL}}$ and $\textsf{QCDCL3}$ are again dead exponential, as they should be.

The recurring theme in Figs. 3 and 4 is that the theoretically strongest system $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ is outperformed by the specialized version for each formula type. One appealing explanation would be that the specialized systems $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ and $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ profit from their ability to guarantee learning asserting clauses and cubes respectively. But this does not appear to be the real reason: $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ also (like $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$) learns almost exclusively asserting clauses on $\texttt{Twin}\texttt{CR}$ ($96\%$ on average, more than $99\%$ in over $70\%$ of cases), and similarly $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$ (like $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$) learns almost exclusively asserting cubes on $\text {Rev}(\texttt{Twin}\texttt{ModEq})$ ($98\%$ on average, more than $99\%$ in over $70\%$ of cases). Thus, the advantage of the specialized systems is unlikely to be explicable solely by the quantity of asserting constraints, but rather by their quality. This is also supported by the erratic performance of several of the variants on both $\texttt{Twin}\texttt{CR}$ and $\text {Rev}(\texttt{Twin}\texttt{ModEq})$—it appears that the existing short runs are hard for the solver to discover. Investigating this properly might require opening up the solver even more, and recording decisions and other details of the search path. We want to keep this paper focused on the theory part, and leave further investigation of this behaviour to future work.

Fig. 4

Running time in seconds on $\text {Rev}(\texttt{Twin}\texttt{ModEq}_n)$. The legend is sorted from best downwards

6.2 QBF Evaluations

Table 1 and Figs. 5, 6, 7, and 8 show the performance on PCNF and QCIR (circuit) formulas from the QBF Evaluations (QBFEval) 2020 and 2022. Even though the theoretical part is concerned with PCNF formulas only, here we evaluate the algorithms on circuit formulas as well, as the circuit format is a standard part of QBF Evaluations (in fact, it is preferred due to its greater flexibility for both encoding and solving). Circuit formulas are internally translated into a pair of PCNF formulas by Qute.

Table 1 Results of the QCDCL variants on QBF Eval 2020 and 2022. VBS stands for the virtual best solver, the best performing solver on each instance. $\top$ gives the number of true solved formulas, $\bot$ the number of false solved formulas, $\Sigma = \top + \bot$. Column maxima are in bold (excluding VBS)

In Fig. 5, we see a decisive victory of $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, which beat the second $\textsf{QCDCL3}$ by a margin of 32 solved instances. $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$ solved 27 instances from this benchmark set uniquely, of which 26 were false formulas. These 27 uniquely solved formulas include application formulas encoding bounded model checking problems, as well as several crafted formulas.

In all other cases, the winner is vanilla QCDCL; twice $\mathsf{{QCDCL}}$, once $\textsf{QCDCL3}$. This, as well as the relative ranking of $\mathsf{{QCDCL}}$ and $\textsf{QCDCL3}$ in Fig. 5, proves that the 3-watched-literal scheme, a by-product of the implementation, considered in its own right, is in fact competitive with the traditional 2-watched literal scheme, at least on these formulas.

With the already mentioned exception of $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, no other QCDCL variant beats vanilla QCDCL in any of the other cases. Each of $\textsf{QCDCL}^\textsf {{A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{U\tiny {\MakeUppercase {ni}}-A\tiny {\MakeUppercase {ny}}}}$, $\textsf{QCDCL}^\textsf {{E\tiny {\MakeUppercase {xi}}-A\tiny {\MakeUppercase {ny}}}}$ beats the other two on at least one benchmark set. In all cases except PCNF 22, the virtual best solver (VBS) is strictly better than any individual algorithm, meaning that there were always formulas not solver by the best variant, which were solved by another variant.

Such mixed results should perhaps not surprise. The act of performing out-of-order decisions amounts to revealing a future move in the game earlier than forced to. This should be advantageous, philosophically speaking, when strong moves exist that can already be played early. It is not clear how often such situations should arise in application formulas, into which they are not baked the way they are into separation formulas.

Fig. 5

Performance on QBFEval 2020 PCNF instances. Cactus plot: (x, y) means the configuration solved x instances in y seconds. Lower and right is better

In any case, the experiments show that both the new algorithms as well as the technical implementation are competitive. Further analysis would be needed to determine whether there are application formula families on which one QCDCL variant is significantly better then others. We provide all of our experimental data as supplementary material.

Fig. 6

Performance on QBFEval 2020 QCIR instances. Cactus plot: (x, y) means the configuration solved x instances in y seconds. Lower and right is better

Fig. 7

Performance on QBFEval 2022 PCNF instances. Cactus plot: (x, y) means the configuration solved x instances in y seconds. Lower and right is better

Fig. 8

Performance on QBFEval 2022 QCIR instances. Cactus plot: (x, y) means the configuration solved x instances in y seconds. Lower and right is better

7 Conclusion

We have laid the theoretical foundations for new flavours of QCDCL with the ability to ignore all quantification order for decisions. In this paper we focused on proof complexity, showing exponential advantage for the new systems over vanilla QCDCL. We complemented this with a proof-of-concept implementation in Qute, which validates the feasibility of our approach. Our preliminary experiments on crafted formulas already raise some interesting questions about poor solver performance on theoretically easy formulas.

As part of future work, we plan to advance on the practical front, polishing and possibly improving the implementation technically, performing a more thorough experimental evaluation, and combining the approaches presented here with other techniques like Qute’s native dependency learning (and possibly dependency schemes). We would also like to dive deeper into the analysis of how learning asserting constraints affects solver performance.