Knowledge Compilation for Boolean Functional Synthesis

TL;DR

This paper introduces SynNNF, a new normal form for Boolean formulas that enables polynomial-time synthesis and quantification, improving the efficiency of Boolean functional synthesis and solving benchmarks beyond current tools.

Contribution

The paper presents SynNNF, a novel normal form that guarantees efficient synthesis and quantification, and proposes an algorithm to convert CNF formulas into SynNNF for practical synthesis.

Findings

SynNNF can be more succinct than existing normal forms.

The conversion algorithm enables solving complex benchmarks.

Prototype implementation outperforms state-of-the-art tools.

Abstract

Given a Boolean formula F(X,Y), where X is a vector of outputs and Y is a vector of inputs, the Boolean functional synthesis problem requires us to compute a Skolem function vector G(Y)for X such that F(G(Y),Y) holds whenever \exists X F(X,Y) holds. In this paper, we investigate the relation between the representation of the specification F(X,Y) and the complexity of synthesis. We introduce a new normal form for Boolean formulas, called SynNNF, that guarantees polynomial-time synthesis and also polynomial-time existential quantification for some order of quantification of variables. We show that several normal forms studied in the knowledge compilation literature are subsumed by SynNNF, although SynNNFcan be super-polynomially more succinct than them. Motivated by these results, we propose an algorithm to convert a specification in CNF to SynNNF, with the intent of solving the Boolean functional synthesis problem. Experiments with a prototype implementation show that this approach solves several benchmarks beyond the reach of state-of-the-art tools.

Tables

Table 1. TABLE I : Compilation into SynNNF

Benchmarks	Compiled By $\mathrm{𝖢𝟤𝖲𝗒𝗇}$			BDD	Total
(Total)	Stage I	Stage II	Total	compilation	in SynNNF
QBFEval (402)	103	82	185	153	283
FA.QD (6)	0	6	6	6	6

Table 2. TABLE II : Comparison Results of 𝖢𝟤𝖲𝗒𝗇 𝖢𝟤𝖲𝗒𝗇 \mathsf{C2Syn}

Bench	$\mathrm{𝖢𝟤𝖲𝗒𝗇}$ vs Cadet		$\mathrm{𝖢𝟤𝖲𝗒𝗇}$ vs bfss		$\mathrm{𝖢𝟤𝖲𝗒𝗇}$ $\setminus$
	$\mathrm{𝖢𝟤𝖲𝗒𝗇}$$\setminus$	$\text{Cadet}\setminus$	$\mathrm{𝖢𝟤𝖲𝗒𝗇}$$\setminus$	$\text{bfss}\setminus$	($\text{Cadet}\cup$
mark	Cadet	$\mathrm{𝖢𝟤𝖲𝗒𝗇}$	bfss	$\mathrm{𝖢𝟤𝖲𝗒𝗇}$	bfss)
QBFEval	77	105	83	78	74
FA.QD	2	0	3	0	2

Equations

$$(x_1{\mathsf{op}}_1^{\prime }{f}_1({\mathbf{X}}_2^n,\mathbf{Y})){\mathsf{op}}_1(x_2{\mathsf{op}}_2^{\prime }{f}_2({\mathbf{X}}_3^n,\mathbf{Y})){\mathsf{op}}_2\dots {\mathsf{op}}_{n-1}(x_n{\mathsf{op}}_n^{\prime }{f}_n(\mathbf{Y})){\mathsf{op}}_n{f}_{n+1}(\mathbf{Y})$$

$$\exists x_1,\dots ,x_i\hat{F}(x_1,\dots ,x_i,{\Psi }_{i+1}^n,\neg x_1,\dots \neg x_i,\neg {\Psi }_{i+1}^n,\mathbf{Y})=1$$

$$\exists {\mathbf{X}}^{\prime }F({\mathbf{X}}^{\prime },{\mathbf{Y}}^{\ast })\wedge \exists 1\le i\le n\neg \exists {\mathbf{X}}_1^{i-1}F({\mathbf{X}}_1^{i-1},{\Psi }_i^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })$$

$${\Psi }_k({\mathbf{Y}}^{\ast })=\hat{F}(1^{k-1},1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-1},0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=1$$

$$\hat{F}(1^{k-1},1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-1},0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })\vee \hat{F}(1^{k-1},0,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-1},1,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=1$$

$$\exists {\mathbf{X}}_1^{k-1}F({\mathbf{X}}_1^{k-1},1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\Rightarrow \exists {\mathbf{X}}_1^{k-1}\hat{F}({\mathbf{X}}_1^{k-1},1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),\neg {\mathbf{X}}_1^{k-1},0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\hat{F}(1^{k-2},1,1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-2},0,0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\text{and}\hat{F}(1^{k-2},0,1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-2},1,0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\hat{F}(1^{k-2},x_{k-1},1,{{\psi }^{\prime }}_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-2},{\bar{x}}_{k-1},0,\neg {{\psi }^{\prime }}_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })\leftrightarrow x_{k-1}\wedge {\bar{x}}_{k-1}$$

$$\hat{F}(1^{k-2},1,1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-2},0,0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=1$$

$$\exists {\mathbf{X}}_1^{k-2}\hat{F}({\mathbf{X}}_1^{k-2},1,1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),\neg {\mathbf{X}}_1^{k-2},0,0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\hat{F}(1^{k-2},1,1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-2},0,0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0={\Psi }_{k-1}({\mathbf{Y}}^{\ast })$$

$$\text{and }\hat{F}(1^{k-2},0,1,{\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),1^{k-2},1,0,\neg {\Psi }_{k+1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=1$$

$$\text{i.e., }\hat{F}(1^{k-2},{\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),1^{k-2},\neg {\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=1$$

$$\hat{F}(1^{k-3},1,{\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),1^{k-3},0,\neg {\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\hat{F}(1^{k-3},0,{\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),1^{k-3},1,\neg {\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\hat{F}(1^{k-3},1,{\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),1^{k-3},0,\neg {\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=1$$

$${\exists }_1^{k-3}\hat{F}({\mathbf{X}}_1^{k-3},1{\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),\neg {\mathbf{X}}_1^{k-3},0,\neg {\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\hat{F}(1^{k-3},1,{\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),1^{k-3},0,\neg {\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=0$$

$$\hat{F}(1^{k-3},0,{\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),1^{k-3},1,\neg {\Psi }_{k-1}^n({\mathbf{Y}}^{\ast }),{\mathbf{Y}}^{\ast })=1$$

Full text

I Introduction

Boolean functional synthesis is the problem of synthesizing outputs as Boolean functions of inputs, while satisfying a declarative relational specification between inputs and outputs. Also called Skolem function synthesis, this problem has numerous applications including certified QBF solving, reactive control synthesis, circuit and program repair and the like. While variants of the problem have been studied since long [17, 3], there has been significant recent interest in designing practically efficient algorithms for Boolean functional synthesis. The resulting breed of algorithms [14, 23, 22, 11, 25, 18, 13, 2, 1, 15, 7, 24] have been empirically shown to work well on large collections of benchmarks. Nevertheless, there are not-so-large examples that are currently not solvable within reasonable resources by any known algorithm. To make matters worse, it is not even fully understood what properties of a Boolean relational specification or of its representation make it amenable to efficient synthesis. In this paper, we take a step towards answering this question. Specifically, we propose a new sub-class of negation normal form called SynNNF, such that every Boolean relational specification in SynNNF admits polynomial-time synthesis. Furthermore, a Boolean relational specification admits polynomial-time synthesis (by any algorithm) if and only if there exists a polynomial-sized refinement of the specification in SynNNF.

To illustrate the hardness of Boolean functional synthesis, consider the specification $F(\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{Y})\equiv(\mathbf{Y}=(\mathbf{X}_{1}\times_{[n]}\mathbf{X}_{2}))\wedge(\mathbf{X}_{1}\neq 0\cdots 01)\wedge(\mathbf{X}_{2}\neq 0\cdots 01)$, where $|\mathbf{Y}|=2n$, $|\mathbf{X}_{1}|=|\mathbf{X}_{2}|=n$ and $\times_{[n]}$ denotes multiplication of $n$-bit unsigned integers. This specification asserts that $\mathbf{Y}$, viewed as a $2n$-bit unsigned integer, is the product of $\mathbf{X}_{1}$ and $\mathbf{X}_{2}$, each viewed as an $n$-bit unsigned integer different from $1$. The specification $F(\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{Y})$ can be easily represented as a circuit of AND, OR, NOT gates with $\mathcal{O}(n^{2})$ gates. However, synthesizing $\mathbf{X}_{1}$ and $\mathbf{X}_{2}$ as functions of $\mathbf{Y}$ requires us to obtain a circuit that factorizes a $2n$-bit unsigned integer into factors different from $1$, whenever possible. It is a long-standing open question whether such a circuit of size polynomial in $n$ exists. Thus, although the relational specification is succinctly representable, the outputs expressed as functions of the inputs may not have any known succinct representation.

It was recently shown [1] that unless some long-standing complexity theoretic conjectures are falsified, Boolean functional synthesis must necessarily require super-polynomial (or even exponential) space and time. In the same work [1], it was also shown that if a specification is represented in weak decomposable negation normal form wDNNF, synthesis can be accomplished in time polynomial in the size of the specification. While this was a first step towards identifying a normal form with the explicit objective of polynomial-time synthesis, experimental results in [1] indicate that wDNNF doesn’t really characterize specifications that admit efficient synthesis. Specifically, experiments in [1] showed that a polynomial-time algorithm intended for synthesis from wDNNF specifications ends up solving the synthesis problem for a large class of specifications not in wDNNF. This motivates us to ask if there exists a weaker (than wDNNF) sub-class of Boolean relational specifications that admit polynomial-time synthesis.

We answer the above question affirmatively in this paper, the polynomial dependence being quadratic in the number of outputs and the size of the specification. En route, we also show that the weaker normal form, viz. SynNNF, admits polynomial-time existential quantifier elimination of a set of variables for some (not all) order of quantification of variables. Applications of such quantifier elimination abound in practice, viz. image computation in symbolic model checking, synthesis of QBF certificates, computation of interpolants etc. Note that ensuring efficient quantifier elimination for some ordering of variables is simpler than ensuring efficient quantifier elimination for all orderings of variables – the latter having been addressed by normal forms like DNNF [9].

Our primary contributions can be summarized as follows:

• •

We present a new sub-class of negation normal form, called SynNNF, that admits polynomial-time synthesis and quantifier elimination for a set of variables.

• •

We show that SynNNF is super-polynomially (in some cases, exponentially) more succinct than several other sub-classes studied in the literature (viz. wDNNF, dDNNF, DNNF, FBDD, ROBDD), unless some long-standing complexity theoretic conjectures are falsified.

• •

We show that by suitably weakening SynNNF, we can precisely characterize the class of Boolean specifications that admit polynomial-time synthesis by a simple algorithm originally proposed in [1].

• •

We define a natural notion of refinement of specifications w.r.t synthesis and show that every specification that admits polynomial-time synthesis necessarily has a polynomial-sized refinement that is in SynNNF.

• •

We present a novel algorithm for compiling a Boolean relational specification in $\mathsf{CNF}$ to a refined specification in SynNNF. We call this knowledge compilation for synthesis and quantifier elimination.

• •

Finally, we present experimental results that show that synthesis by compiling to SynNNF solves a large set of benchmarks, including several benchmarks beyond the reach of existing tools.

Related Work: The literature on knowledge compilation of Boolean functions is rich and extensive [6, 9, 20, 10]. While existential quantification or forgetting of propositions has been studied in [16, 10], neither Boolean functional synthesis nor existential quantification for some (not all) ordering of variables has received attention in earlier work on knowledge compilation. Sub-classes of negation normal forms like DNNF and other variants [10] admit efficient existential quantification for all orders in which variables are quantified. However, if we are interested in only the result of existentially quantifying a given set of variables, these forms can be unnecessarily restrictive and exponentially larger. Recent work on Boolean functional synthesis [13, 14, 18, 24, 11, 2, 1, 8] has focused more on algorithms to directly synthesize outputs as functions of inputs. Some of these algorithms (viz. [11, 1, 8]) exploit properties of specific input representations for optimizing the synthesis process. This has led to the articulation of sufficient conditions on representation of specifications for efficient synthesis. For example, [15] suggested using input-first ROBDDs for efficient synthesis, and a quadratic-time algorithm for synthesis from input-first ROBDDs was presented in [11]. This result was subsequently generalized in [1], where it was shown that specifications in wDNNF (which strictly subsumes ROBDDs) suffice to give a quadratic-time algorithm for synthesis. As we show later, wDNNF can itself be generalized to SynNNF. In another line of investigation, it was shown [8] that if a $\mathsf{CNF}$ specification is decomposed into an input-part and an output-part, then synthesis can be achieved in time linear in the size of the $\mathsf{CNF}$ specification and $k$, where $k$ is the smaller of the count of maximal falsifiable subsets (MFS) of the input-part and the count of maximal satisfiable subsets (MSS) of the output-part. However, this does not yield an algorithm whose running time is polynomial in the size of the representation of $F(\mathbf{X},\mathbf{Y})$.

The paper is organized as follows. After preliminaries, we present the new normal form SynNNF and its properties in Section III. In Section IV, we introduce the idea of refinement, which allows us to simplify the specification. In Section V, we describe an algorithm to compile any function into our normal form, followed in Section VI by experimental results, before ending with a conclusion. Proofs of lemmas and theorems are mostly deferred to the appendix.

II Preliminaries and notations

A Boolean formula $F(z_{1},\ldots z_{p})$ on $p$ variables is a mapping $F:\{0,1\}^{p}\rightarrow\{0,1\}$. The set of variables $\{z_{1},\ldots z_{p}\}$ is called the support of the formula, and denoted $\mathsf{sup}({F})$. We normally use $\mathbf{Z}$ to denote the sequence $(z_{1},\ldots z_{p})$. For notational convenience, we will also use $\mathbf{Z}$ to denote a set of variables, when there is no confusion. A satisfying assignment or model of $F$ is a mapping of variables in $\mathsf{sup}({F})$ to $\{0,1\}$ such that $F$ evaluates to $1$ under this assignment. If $\pi$ is a model of $F$, we write $\pi\models F$ and use $\pi(z_{i})$ to denote the value assigned to $z_{i}\in\mathsf{sup}({F})$ by $\pi$. If $\mathbf{Z}^{\prime}$ is a subsequence of $\mathbf{Z}$, we use ${\pi}\!\!\downarrow\!\!{\small{\mathbf{Z}^{\prime}}}$ to denote the projection of $\pi$ on $\mathbf{Z}^{\prime}$, i.e. $(\pi({z^{\prime}}_{1}),\ldots\pi({z^{\prime}}_{k}))$, where $k=|\mathbf{Z}^{\prime}|$. We use $\mathsf{form}({{\pi}\!\!\downarrow\!\!{\small{\mathbf{Z}^{\prime}}}})$ to denote the conjunction of literals (i.e. variables or their negation) corresponding to ${\pi}\!\!\downarrow\!\!{\small{\mathbf{Z}^{\prime}}}$. For example, if $\pi$ assigns $1$ to $z_{1},z_{3}$ and [math] to $z_{2},z_{4}$ and $\mathbf{Z}^{\prime}=(z_{1},z_{4})$, then $\mathsf{form}({{\pi}\!\!\downarrow\!\!{\small{\mathbf{Z}^{\prime}}}})=z_{1}\wedge\neg z_{4}$.

II-1 Negation normal form ($\mathsf{NNF}$)

This is the class of Boolean formulas in which (i) the only operators used are conjunction ($\wedge$), disjunction ($\vee$) and negation ($\neg$), and (ii) negation is applied only to variables. Every Boolean formula can be converted to a semantically equivalent $\mathsf{NNF}$ formula. Moreover, this conversion can be done in linear time for representations like $\mathsf{AIG}$s, ROBDDs, Boolean circuits etc.

II-2 Unate formulas

Let $F|_{z_{i}=0}$ (resp. $F|_{z_{i}=1}$) denote the positive (resp. negative) cofactor of $F$ with respect to $z_{i}$. Then, $F$ is positive unate in $z_{i}\in\mathsf{sup}({F})$ iff $F|_{z_{i}=0}\Rightarrow F|_{z_{i}=1}$. Similarly, $F$ is negative unate in $z_{i}$ iff $F|_{z_{i}=1}\Rightarrow F|_{z_{i}=0}$. A literal $\ell$ is said to be pure in an $\mathsf{NNF}$ formula $F$ iff $F$ has at least one instance of $\ell$ but no instance of $\neg\ell$. If $z_{i}$ (resp. $\neg z_{i}$) is pure in $F$, then $F$ is positive (resp. negative) unate in $z_{i}$.

II-3 Independent support and functionally defined variables

A subsequence $\mathbf{Z}^{\prime}$ of $\mathbf{Z}$ is said to be an independent support of $F$ iff every pair of satisfying assignments $\pi,\pi^{\prime}$ of $F$ that agree on the assignment of variables in $\mathbf{Z}^{\prime}$ also agree on the assignment of all variables in $\mathbf{Z}$. Variables not in $\mathbf{Z}^{\prime}$ are said to be functionally defined by the independent support. Effectively, the assignment of variables in $\mathbf{Z}^{\prime}$ uniquely determine that of functionally defined variables, when satisfying $F$. $\mathsf{CNF}$ encodings of Boolean functions originally specified as circuits, ROBDDs, $\mathsf{AIG}$s etc. often use Tseitin encoding [26], which introduces a large number of functionally defined variables.

II-4 Boolean functional synthesis

Unless mentioned otherwise, we use $\mathbf{X}=(x_{1},\ldots x_{n})$ to denote a sequence of Boolean outputs, and $\mathbf{Y}=(y_{1},\ldots y_{m})$ to denote a sequence of Boolean inputs. The Boolean functional synthesis problem, henceforth denoted $\mathsf{BFnS}$, asks: given a Boolean formula $F(\mathbf{X},\mathbf{Y})$ specifying a relation between inputs $\mathbf{Y}$ and outputs $\mathbf{X}$, determine functions $\mathbf{\Psi}=(\psi_{1}(\mathbf{Y}),\ldots\psi_{n}(\mathbf{Y}))$ such that $F(\mathbf{\Psi},\mathbf{Y})$ holds whenever $\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})$ holds. Thus, $\forall\mathbf{Y}(\exists\mathbf{X}\,F(\mathbf{X},\mathbf{Y})\Leftrightarrow\left.F(\mathbf{\Psi},\mathbf{Y})\right)$ must be a tautology. The function $\psi_{i}$ is called a Skolem function for $x_{i}$ in $F$, and $\mathbf{\Psi}$ is called a Skolem function vector for $\mathbf{X}$ in $F$.

For $1{\leq}i{\leq}j{\leq}n$, we use $\mathbf{X}_{i}^{j}$ to denote the subsequence $(x_{i},x_{i+1},\ldots x_{j})$. If $i\leq k<j$, we sometimes use $(\mathbf{X}_{i}^{k},\mathbf{X}_{k+1}^{j})$ interchangeably with $\mathbf{X}_{i}^{j}$ for notational convenience. Let $F^{(i-1)}(\mathbf{X}_{i}^{n},\mathbf{Y})$ denote $\exists\mathbf{X}_{1}^{i-1}F(\mathbf{X}_{1}^{i-1},\mathbf{X}_{i}^{n},\mathbf{Y})$. It has been argued in [14, 11, 2, 12] that the $\mathsf{BFnS}$ problem for $F(\mathbf{X},\mathbf{Y})$ can be solved by first ordering the outputs, say as $x_{1}\prec x_{2}\cdots\prec x_{n}$, and then synthesizing a function $\psi_{i}(\mathbf{X}_{i+1}^{n},\mathbf{Y})\equiv F^{(i-1)}(\mathbf{X}_{i}^{n},\mathbf{Y})[x_{i}\mapsto 1]$ for each $x_{i}$. This ensures that $F^{(i-1)}(\psi_{i},\mathbf{X}_{i+1}^{n},\mathbf{Y})\Leftrightarrow\exists x_{i}F^{(i-1)}(x_{i},\mathbf{X}_{i+1}^{n},\mathbf{Y})$. Once all such $\psi_{i}$’s are obtained, one can substitute $\psi_{i+1}$ through $\psi_{n}$ for $x_{i+1}$ through $x_{n}$ respectively, in $\psi_{i}$ to obtain a Skolem function for $x_{i}$ as a function of $\mathbf{Y}$. The primary problem of using this approach as-is is the exponential blow-up incurred in the size of the Skolem functions.

II-5 DAG representations

For an $\mathsf{NNF}$ formula $F$, its DAG representation is naturally induced by the structure of $F$. Specifically, if $F$ is simply a literal $\ell$, its DAG representation is a leaf labeled $\ell$. If $F$ is $F_{1}~{}\mathsf{op}~{}F_{2}$ where $\mathsf{op}\in\{\vee,\wedge\}$, its DAG representation is a node labeled $\mathsf{op}$ with two children, viz. the DAG representations of $F_{1}$ and $F_{2}$. W.l.o.g. we assume that a DAG representation of $F$ is always in a simplified form, where $t\wedge 1$, $t\vee 0$, $t\wedge t$ and $t\vee t$ are replaced by $t$, $t\wedge 0$ is replaced by 0 and $t\vee 1$ is replaced by $1$ for every node $t$. We use $|F|$ for the node count in the DAG representation of $F$.

$\mathsf{FBDD}$ and $\mathsf{ROBDD}$ are well-known representations of Boolean formulas and we skip their definitions. We briefly recall the definitions of DNNF, dDNNF and wDNNF below. Let $\alpha$ be the subformula represented by an internal node $N$ (labeled by $\wedge$ or $\vee$) in a DAG representation of an NNF formula $F$. We use $lits({\alpha})$ to denote the set of literals labeling leaves that have a path to the node $N$ representing $\alpha$ in the DAG representation of $F$. We also use $atoms({\alpha})$ to denote the underlying set of variables in $\mathsf{sup}({F})$ that appear in $lits({\alpha})$. For each $\wedge$-labeled internal node $N$ in the DAG of $F$ with $\alpha=\alpha_{1}\wedge\ldots\wedge\alpha_{k}$ being the subformula represented by $N$, if for all distinct indices $r,s\in\{1,\ldots k\}$, $atoms({\alpha_{r}})\cap atoms({\alpha_{s}})=\emptyset$, then $F$ is said to be in DNNF [9]. If, instead, for all distinct indices $r,s\in\{1,\ldots k\}$, $lits({\alpha_{r}})\cap\{\neg\ell\mid\ell\in lits({\alpha_{s}})\}=\emptyset$, then $F$ is said to be in wDNNF [1]. Finally $F(\mathbf{X},\mathbf{Y})$ is said to be in deterministic DNNF(or dDNNF) [10] if $F$ is in DNNF and for each $\vee$-labeled internal node $D$ in the DAG of $F$ with $\beta=\beta_{1}\vee\ldots\vee\beta_{k}$ being the subformula represented by $D$, $\beta_{r}\wedge\beta_{s}$ is a contradiction for all distinct indices $r,s$.

II-6 Positive form of input specification

Given a specification $F(\mathbf{X},\mathbf{Y})$ in $\mathsf{NNF}$, we denote by $\widehat{{F}}(\mathbf{X},\overline{{\mathbf{X}}},\mathbf{Y})$ the formula obtained by replacing every occurrence of $\neg x_{i}~{}(x_{i}\in\mathbf{X})$ in $F$ with a fresh variable $\overline{{x_{i}}}$. This is also called the positive form of the specification and has been used earlier in [2]. Observe that for any $F$ in $\mathsf{NNF}$, $\widehat{{F}}$ is positive unate (or monotone) in all variables in $\mathbf{X}$ and $\overline{{\mathbf{X}}}$. For $i\in\{1,\ldots n\}$, we sometimes split $\mathbf{X}$ into two parts, $\mathbf{X}_{1}^{i}$ and $\mathbf{X}_{i+1}^{n}$, and represent $\widehat{{F}}(\mathbf{X},\overline{{\mathbf{X}}},\mathbf{Y})$ as $\widehat{{F}}(\mathbf{X}_{1}^{i},\mathbf{X}_{i+1}^{n},\overline{{\mathbf{X}}}_{1}^{i},\overline{{\mathbf{X}}}_{i+1}^{n},\mathbf{Y})$. For $b,c\in\{0,1\}$, let $\mathbf{b}^{i}$ (resp. $\mathbf{c}^{i}$) denote a vector of $i$ $b$’s (resp. $c$’s). For notational convenience, we use $\widehat{{F}}(\mathbf{b}^{i},\mathbf{X}_{i+1}^{n},\mathbf{c}^{i},\overline{{\mathbf{X}}}_{i+1}^{n},\mathbf{Y})$ to denote $\widehat{{F}}(\mathbf{X}_{1}^{i},\mathbf{X}_{i+1}^{n},\overline{{\mathbf{X}}}_{1}^{i},\overline{{\mathbf{X}}}_{i+1}^{n},\mathbf{Y})|_{\mathbf{X}_{1}^{i}=\mathbf{b}^{i},\overline{{\mathbf{X}}}_{1}^{i}=\mathbf{c}^{i}}$.

III A New Normal Form for Efficient Synthesis

In [1], it was shown that if $F(\mathbf{X},\mathbf{Y})$ is represented as a ROBDD/FBDD or in DNNF or in wDNNF form, Skolem functions can be synthesized in time polynomial in $|F|$. In this section, we define a new normal form called SynNNF that subsumes and is more succinct than these other normal forms, and yet guarantees efficient synthesis of Skolem functions.

Definition 1**.**

Given a specification $F(\mathbf{X},\mathbf{Y})$, for every $i\in\{1,\ldots n\}$ we define the $i^{th}$-reduct of $\widehat{{F}}$, denoted $[\widehat{{F}}]_{i}$, to be $\widehat{{F}}(1^{i-1},{\mathbf{X}}_{i}^{n},1^{i-1},\overline{{\mathbf{X}}}_{i}^{n},\mathbf{Y})$. We also define $[\widehat{{F}}]_{n+1}$ to be $\widehat{{F}}(1^{n},1^{n},\mathbf{Y})$.

Note that $[\widehat{{F}}]_{1}$ is the same as $\widehat{{F}}$, and $\mathsf{sup}({[\widehat{{F}}]_{i}})=\mathbf{X}_{i}^{n}\cup\overline{{\mathbf{X}}}_{i}^{n}\cup\mathbf{Y}$ for $i\in\{1,\ldots n\}$.

Example 1**.**

Consider the $\mathsf{NNF}$ formula $K(x_{1},x_{2},y_{1},y_{2})=(x_{1}\vee x_{2})\wedge(\neg x_{2}\vee y_{1})\wedge(\neg y_{1}\vee y_{2})$. Then $\widehat{{K}}=((x_{1}\vee x_{2})\wedge(\overline{{x_{2}}}\vee y_{1})\wedge(\neg y_{1}\vee y_{2}))$. Thus, we have $[\widehat{{K}}]_{1}=\widehat{{K}}$ and $[\widehat{{K}}]_{2}=\widehat{{K}}[x_{1}\mapsto 1,\overline{{x_{1}}}\mapsto 1]=(\overline{{x_{2}}}\vee y_{1})\wedge(\neg y_{1}\vee y_{2})$.

Next, we define a useful property for the $i^{th}$-reduct, which will be crucial for efficient synthesis of Skolem functions.

Definition 2**.**

Given $F(\mathbf{X},\mathbf{Y})$, let $\alpha_{i}^{jk}$ denote $[\widehat{{F}}]_{i}[x_{i}\mapsto j,\overline{{x}}_{i}\mapsto k,\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]$, where $j,k\in\{0,1\}$. We say that $[\widehat{{F}}]_{i}$ is $\wedge_{i}$-unrealizable if $\zeta=\alpha_{i}^{11}\wedge\neg\alpha_{i}^{10}\wedge\neg\alpha_{i}^{01}$ is unsatisfiable.

Intuitively, we wish to say that there is no assignment to $\mathbf{X}_{i+1}^{n}$ and $\mathbf{Y}$ such that $[\widehat{{F}}]_{i}$ is equivalent to $x_{i}\wedge\overline{x}_{i}$. The formula $\zeta$ captures this semantic condition. Indeed, if an assignment makes $\zeta$ true, then it also makes $[\widehat{{F}}]_{i}$ equivalent to $x_{i}\wedge\overline{{x}}_{i}$ (i.e., $[\widehat{{F}}]_{i}=1$ for $x_{i},\overline{{x_{i}}}$ having values $(1,1)$, but not for $(0,1)$, $(1,0)$, $(0,0)$). Note that since $[\widehat{{F}}]_{i}$ is positive unate in $x_{i}$ and $\overline{{x_{i}}}$, $\zeta$ is satisfiable iff $\zeta\wedge\neg\alpha_{i}^{00}$ is satisfiable; we need not conjoin $\neg\alpha_{i}^{00}$ in the definition of $\zeta$.

A sufficient condition for $[\widehat{{F}}]_{i}$ to be $\wedge_{i}$-unrealizable is that in the DAG representation of $[\widehat{{F}}]_{i}$, there is no pair of paths – one from $x_{i}$ and the other from $\overline{{x_{i}}}$ – which meet for the first time at an $\wedge$-labeled node. In Example 1, $[\widehat{{K}}]_{1}$ is $\wedge_{1}$-unrealizable since there is no leaf labeled $\overline{{x_{1}}}$ in its DAG representation. Similarly, $[\widehat{{K}}]_{2}=(\overline{{x_{2}}}\vee y_{1})\wedge(\neg y_{1}\vee y_{2})$ is $\wedge_{2}$-unrealizable as there is no leaf labeled $x_{2}$ in the DAG representation of $[\widehat{{K}}]_{2}$ (although such a leaf exists in the DAG representation of $[\widehat{{K}}]_{1}$).

Example 2**.**

Let $H(x_{1},x_{2},y_{1},y_{2})=(x_{1}\vee x_{2}\vee y_{1})\wedge(\neg x_{1}\vee(\neg x_{2}\wedge y_{2}))$. Then $\widehat{{H}}(\mathbf{X},\overline{\mathbf{X}},\mathbf{Y})=(x_{1}\vee x_{2}\vee y_{1})\wedge(\overline{{x_{1}}}\vee(\overline{{x_{2}}}\wedge y_{2}))$. Using the notation in Definition 2, $\alpha_{1}^{11}=1$, $\alpha_{1}^{10}=\neg{x}_{2}\wedge y_{2}$ and $\alpha_{1}^{01}=(x_{2}\vee y_{1})$. There is an assignment ($x_{2}=0,y_{2}=0,y_{1}=0)$ such that $(\alpha_{1}^{11}\wedge\neg\alpha_{1}^{10}\wedge\neg\alpha_{1}^{01})$ is satisfiable. Hence $[\widehat{{H}}]_{1}$ is not $\wedge_{1}$-unrealizable (equivalently, it is $\wedge_{1}$-realizable). However, $[\widehat{{H}}]_{2}=\widehat{{H}}[x_{1}\mapsto 1,\overline{{x_{1}}}\mapsto 1]=1$; hence it is vacuously $\wedge_{2}$-unrealizable.

Definition 3**.**

A formula $F(\mathbf{X},\mathbf{Y})$ is said to be in synthesizable $\mathsf{NNF}$ (or SynNNF ) wrt the sequence $\mathbf{X}$ if $F$ is in $\mathsf{NNF}$, and for all $1\leq i\leq n$, $[\widehat{{F}}]_{i}$ is $\wedge_{i}$-unrealizable.

In Examples 1, 2, $K$ is in SynNNF, while $H$ is not. Also neither of them are in DNNF or wDNNF. Additionally, the functions as presented do not correspond to ROBDD/FBDD representations either. We now show three important properties of SynNNF which motivate our proposal of SynNNF as a normal form for synthesis and existential quantification.

III-1 SynNNF leads to efficient quantification and synthesis

Our first result is that existentially quantifying $\mathbf{X}$ and synthesizing $\mathbf{X}$ are easy for SynNNF.

Theorem 1**.**

Suppose $F(\mathbf{X},\mathbf{Y})$ is in SynNNF. Then,

1. (i)

$\exists\mathbf{X}_{1}^{i}F(\mathbf{X},\mathbf{Y})\Leftrightarrow[\widehat{{F}}]_{i+1}[\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]$* for $i\in\{1,\ldots,n\}$,* 2. (ii)

Skolem function vector $\Psi_{1}^{n}$ for $\mathbf{X}_{1}^{n}$ can be computed in $\mathcal{O}(n^{2}\cdot|F|)$ time and $\mathcal{O}(n\cdot|F|)$ space, where $|\mathbf{X}|=n$.

Proof.

The proof of Part (i) is similar to that of Theorem 2(a) in [1], and follows by induction on $i$. For $i=1$, $\exists\mathbf{X}_{1}^{1}F(\mathbf{X},\mathbf{Y})\Leftrightarrow\widehat{{F}}(1,\mathbf{X}_{2}^{n},0,\neg\mathbf{X}_{2}^{n},\mathbf{Y})\vee\widehat{{F}}(0,\mathbf{X}_{2}^{n},1,\neg\mathbf{X}_{2}^{n},\mathbf{Y})\Rightarrow\widehat{{F}}(1,\mathbf{X}_{2}^{n},1,\neg\mathbf{X}_{2}^{n},\mathbf{Y})=[\widehat{{F}}]_{2}[\overline{{\mathbf{X}}}_{2}^{n}\mapsto\neg\mathbf{X}_{2}^{n}]$ (by positive unateness of $\widehat{{F}}$ in $x_{1},\overline{{x_{1}}}$). Conversely, as $F$ is in SynNNF, $[\widehat{{F}}]_{2}$ is $\wedge_{2}$-unrealizable, which implies that with notation as in Definition 2, $\alpha_{1}^{11}\Rightarrow\alpha_{1}^{10}\vee\alpha_{1}^{01}$, i.e., $\widehat{{F}}(1,\mathbf{X}_{2}^{n},1,\neg\mathbf{X}_{2}^{n},\mathbf{Y})\Rightarrow\widehat{{F}}(1,\mathbf{X}_{2}^{n},0,\neg\mathbf{X}_{2}^{n},\mathbf{Y})\vee\widehat{{F}}(0,\mathbf{X}_{2}^{n},1,\neg\mathbf{X}_{2}^{n},\mathbf{Y})$. This give us the proof in the reverse direction, i.e., $[\widehat{{F}}]_{2}[\overline{{\mathbf{X}}}_{2}^{n}\mapsto\neg\mathbf{X}_{2}^{n}]\Rightarrow\exists\mathbf{X}_{1}^{1}F(\mathbf{X},\mathbf{Y})$.

Suppose the statement holds for $1\leq i<n$. We will show that it holds for $i+1$ as well. By inductive hypothesis and definition of existential quantification, $\exists\mathbf{X}_{1}^{i+1}F(\mathbf{X},\mathbf{Y})\Leftrightarrow\exists x_{i+1}[\widehat{{F}}]_{i+1}[\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]\Leftrightarrow[\widehat{{F}}]_{i+1}[x_{i}\mapsto 1,\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]\vee[\widehat{{F}}]_{i+1}[x_{i}\mapsto 0,\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]$. Again, using unateness of $[\widehat{{F}}]_{i+1}$ in $x_{i+1}$ and $\overline{{x_{i+1}}}$ in one direction, and using the defining property of SynNNF ($\alpha_{i+1}^{11}\Rightarrow\alpha_{i+1}^{10}\vee\alpha_{i+1}^{01}$) in the other direction, we obtain $\exists\mathbf{X}_{1}^{i+1}F(\mathbf{X},\mathbf{Y})\Leftrightarrow[\widehat{{F}}]_{i+2}[\overline{{\mathbf{X}}}_{i+2}^{n}\mapsto\neg\mathbf{X}_{i+2}^{n}]$.

Part(ii): For $i\in\{1,\ldots n\}$, let $\psi^{\prime}_{i}(\mathbf{X}_{i+1}^{n},\mathbf{Y})$ denote $[\widehat{{F}}]_{i}[x_{i}\mapsto 1,\overline{{x}}_{i}\mapsto 0,\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]=\alpha_{i}^{10}$. Further, from $n$ to $1$, we recursively define $\psi_{n}(\mathbf{Y})=\psi^{\prime}_{n}(\mathbf{Y})$ and $\psi_{i}(\mathbf{Y})=\psi^{\prime}_{i}({\Psi}_{i+1}^{n}(\mathbf{Y}),\mathbf{Y})$. We can now show that $\psi_{i}(\mathbf{Y})$ is indeed a correct Skolem function for $x_{i}$ in $F$. Starting from $n$ to $1$, we know from the preliminaries that $F^{(n-1)}[x_{n}\mapsto 1]$ gives a correct Skolem function for $x_{n}$ in $F$. From part (i) above, $F^{(n-1)}\Leftrightarrow[\widehat{{F}}]_{n}[\overline{{\mathbf{X}_{n}^{n}}}\mapsto\neg\mathbf{X}_{n}^{n}]$. Hence $\alpha_{n}^{10}=\psi_{n}=\psi^{\prime}_{n}$ gives a correct Skolem function for $x_{n}$ in $F$. For any $i\in\{1,\ldots n-1\}$, assuming that $\Psi_{i+1}^{n}$ gives a correct Skolem function vector for $\mathbf{X}_{i+1}^{n}$ in $F$, the same argument shows that $\psi^{\prime}_{i}({\psi}_{i+1}^{n}(\mathbf{Y}),\mathbf{Y})$ is a correct Skolem function for $x_{i}$ in $F$.

Finally, note that $|\psi_{n}|$ is at most $|\widehat{{F}}|$, which is in $\mathcal{O}(|F|)$. A DAG representation of $\psi_{n-k}$ requires a fresh copy of $[\widehat{{F}}]_{n-k}$, but can re-use the DAG representations of $\psi_{j}$ for $j\in\{n-k+1,\ldots n\}$ as sub-DAGs. Thus, $|\psi_{n-k}|$ is in $\mathcal{O}(k\cdot|F|)$. Hence, if we use a multi-rooted DAG to represent all Skolem functions together, we need only $\mathcal{O}(n\cdot|F|)$ nodes. The time required is in $\mathcal{O}(n^{2}\cdot|F|)$ since the resulting DAG has $\sum_{k=1}^{n}k$ edges (root of $\psi_{j}$ connects to a leaf of every $\psi_{i}$ for $i<j$). ∎

The above polynomial-time strategy based on $[\widehat{{F}}]_{i}$ was used in [1] for computing over-approximations of Skolem functions $\psi_{i}(\mathbf{X}_{i+1},\mathbf{Y})$ for each $x_{i}\in\mathbf{X}$. Specifically, it was shown that $[\widehat{{F}}]_{i}[x_{i}\mapsto 1,\overline{{x_{i}}}\mapsto 1]$ over-approximates $\exists\mathbf{X}_{1}^{i}F(\mathbf{X},\mathbf{Y})$ and $[\widehat{{F}}]_{i}[x_{i}\mapsto 1,\overline{{x_{i}}}\mapsto 0]$ over-approximates a Skolem function for $x_{i}$ in $F$. In the remainder of this paper, we refer to the functions $\psi_{i}$ used in the proof of Part (ii) above as $\mathsf{GACKS}{}$ functions (after the author names of [1]). We use $\Psi_{1}^{n}$ to denote the $\mathsf{GACKS}{}$ (Skolem) function vector $(\psi_{1},\ldots,\psi_{n})$.

III-2 Succinctness of SynNNF

SynNNF strictly subsumes many known representations used for efficient analysis of Boolean functions. In the following theorem, sizes and times are in terms of the number of input and output variables.

Theorem 2**.**

1. (i)

Every specification in ROBDD/FBDD, dDNNF, DNNF or wDNNF form is either already in SynNNF or can be compiled in linear time to SynNNF. 2. (ii)

There exist poly-sized SynNNF specifications that only admit

1. (a)

exponential sized FBDD representations. 2. (b)

super-polynomial sized dDNNF representations, unless $\mathsf{P}=\mathsf{VNP}$ 3. (c)

super-polynomial sized wDNNF and DNNF representations, unless $\mathsf{P}=\mathsf{NP}$. 3. (iii)

There exist poly-sized NNF-representations that only admit super-polynomial sized SynNNF representations, unless the polynomial hierarchy collapses.

In the above, $\mathsf{VNP}$ is the algebraic analogue of $\mathsf{NP}$ [27]. Also, (iii) shows that we cannot always hope to obtain a succinct SynNNF representation.

III-3 SynNNF “almost” characterizes efficient synthesis using $\mathsf{GACKS}{}$  functions

We now show that SynNNF precisely characterizes specifications that admit linear-time existential quantification of output variables strengthening Theorem 1(i). Further, a slight weakening of SynNNF condition by restricting assignments on $\mathbf{X}_{i+1}^{n}$ gives us a necessary and sufficient condition for poly-time synthesis using $\mathsf{GACKS}{}$ functions.

Theorem 3**.**

Given a relational specification $F(\mathbf{X},\mathbf{Y})$,

1. (i)

$F$* is in SynNNF iff $\exists\mathbf{X}_{1}^{i}F(\mathbf{X},\mathbf{Y})\Leftrightarrow[\widehat{{F}}]_{i+1}[\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]$* 2. (ii)

The $\mathsf{GACKS}{}$-function vector $\Psi_{1}^{n}$ is a Skolem function vector for $\mathbf{X}_{1}^{n}$ in $F(\mathbf{X},\mathbf{Y})$ iff $[\widehat{{F}}]_{i}[\mathbf{X}_{i+1}^{n}\mapsto\Psi_{i+1}^{n},\overline{{\mathbf{X}_{i+1}^{n}}}\mapsto\neg\Psi_{i+1}^{n}]$ is $\wedge_{i}$-unrealizable for all $i\in\{1\ldots n\}$.

In [14], it was shown that an error formula $\varepsilon$ for $\Psi_{1}^{n}$, defined as $F(\mathbf{X},\mathbf{Y})\wedge\neg F(\mathbf{X}^{\prime},\mathbf{Y})\wedge\bigwedge_{i=1}^{n}(x_{i}^{\prime}\leftrightarrow\Psi_{i})$ is unsatisfiable iff $\Psi_{1}^{n}$ is a Skolem function vector for $F$. Therefore, an (un)satisfiability check for $\varepsilon$ serves to check if $[\widehat{{F}}]_{i}[\mathbf{X}_{i+1}^{n}\mapsto\Psi_{i+1}^{n}]$ is $\wedge_{i}$-unrealizable for all $i\in\{1\ldots n\}$. Further, in [1], it was observed experimentally, that $\mathsf{GACKS}{}$ functions give correct Skolem functions, even when the specifications are not in wDNNF. This surprising behavior, which was left unexplained in [1], can now be explained using SynNNF, thanks to Theorem 3(ii).

Note that Theorem 3(ii) weakens the requirement of SynNNF since $\mathbf{X}_{i+1}^{n}$ are constrained to take only the values defined by $\Psi_{i+1}^{n}$. For an example of a specification not in SynNNF  for which $\mathsf{GACKS}{}$  functions are correct Skolem functions, consider again $H$ from Example 2, which we saw was not in SynNNF. In this case, $\psi^{\prime}_{1}(x_{2},\mathbf{Y})=[\widehat{{H}}]_{1}[x_{1}\mapsto 1,\overline{x}_{1}\mapsto 0,\overline{{x}}_{2}\mapsto\neg x_{2}]=\neg x_{2}\wedge y_{2}$ and $\psi_{2}(\mathbf{Y})=\psi^{\prime}_{2}(\mathbf{Y})=[\widehat{{H}}]_{2}[x_{2}\mapsto 1,\overline{x}_{2}\mapsto 0]=1$. Therefore, $\psi_{1}(\mathbf{Y})=\psi^{\prime}_{1}[x_{2}\mapsto\psi_{2}(\mathbf{Y})]=0$. It can be verified that $x_{1}=\psi_{1}(\mathbf{Y})=0,x_{2}=\psi_{2}(\mathbf{Y})=1$ is indeed a correct Skolem function vector for $\mathbf{X}$ in $H$. Also, $H$ satisfies the condition of Theorem 3(ii) since [\widehat{{H}}]_{1}[x_{2}\mapsto\psi_{2},\overline{x}_{2}\mapsto\neg\psi_{2}]=\overline{{x}}_{1}\mathrel{{\ooalign{\not\phantom{"}$\cr$\Leftrightarrow}}}(x_{1}\wedge\overline{{x}}_{1}), and $[\widehat{{H}}]_{2}=1$.

IV Refinement for Synthesis

Given a specification $F(\mathbf{X},\mathbf{Y})$, sometimes it is easier to solve the $\mathsf{BFnS}$ problem for a “simpler” specification $\mathsf{\widetilde{F}}(\mathbf{X},\mathbf{Y})$ such that a solution for $\mathsf{\widetilde{F}}$ also serves as a solution for $F$. While “simplifications” of this nature have been used in earlier work [14, 1, 22, 7], we formalize this notion below as one of refinement.

Definition 4**.**

Let $F(\mathbf{X},\mathbf{Y})$ and $\mathsf{\widetilde{F}}(\mathbf{X},\mathbf{Y})$ be Boolean relational specifications on the same input and output vectors. We say that $\mathsf{\widetilde{F}}$ refines $F$ w.r.t. synthesis, denoted $\mathsf{\widetilde{F}}\preceq_{syn}F$, iff the following conditions hold: (a) $\forall\mathbf{Y}\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\Rightarrow\exists\mathbf{X}^{\prime}{\mathsf{\widetilde{F}}}(\mathbf{X}^{\prime},\mathbf{Y}))\right)$, and (b) $\forall\mathbf{Y}\forall\mathbf{X}^{\prime}\left(\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\wedge{\mathsf{\widetilde{F}}}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow F(\mathbf{X}^{\prime},\mathbf{Y})\right)$.

Informally, condition (a) specifies that $\mathsf{\widetilde{F}}$ doesn’t restrict the set of input valuations (i.e. $\mathbf{Y}$) over which the specification $F$ can be satisfied, and condition (b) specifies that for all such input valuations $\mathbf{Y}$, any $\mathbf{X}^{\prime}$ that satisfies $\mathsf{\widetilde{F}}$ also satisfies $F$.

Lemma 4**.**

If $\mathsf{\widetilde{F}}\preceq_{syn}F$, every Skolem function vector for $\mathbf{X}$ in $\mathsf{\widetilde{F}}$ is also a Skolem function vector for $\mathbf{X}$ in $F$.

We say $\mathsf{\widetilde{F}}$ refines $F$ w.r.t. synthesis because the set of all Skolem function vectors for $\mathbf{X}$ in $\mathsf{\widetilde{F}}$ is a subset of that for $\mathbf{X}$ in $F$. Note that Definition 4 provides a direct 2QBF-SAT based check of whether $\mathsf{\widetilde{F}}$ refines $F$ without referring to the details of how $\mathsf{\widetilde{F}}$ is obtained from $F$.

Example 3**.**

Let $G(x_{1},x_{2},y_{1},y_{2})\equiv(\neg x_{1}\vee x_{2}\vee y_{1})\wedge(x_{1}\vee\neg x_{2})\wedge(x_{1}\vee\neg y_{1})\wedge(x_{2}\vee y_{2})$ and $\mathsf{\widetilde{G}}(x_{1},x_{2},y_{1},y_{2})\equiv x_{2}\wedge x_{1}$. Although $G\not\Leftrightarrow\mathsf{\widetilde{G}}$, both conditions (a) and (b) of Definition 4 are satisfied; hence $\mathsf{\widetilde{G}}\preceq_{syn}G$.

The following are easy consequences of Definition 4.

Proposition 5**.**

$\preceq_{syn}$* is a reflexive and transitive relation on all Boolean relational specifications on $\mathbf{X}\cup\mathbf{Y}$.* 2. 2.

If $\bigwedge_{y_{j}\in\mathbf{Y}}\left(F|_{y_{j}=0}\Leftrightarrow F|_{y_{j}=1}\right)$ and $\pi\models F(\mathbf{X},\mathbf{Y})$, then $\mathsf{form}({{\pi}\!\!\downarrow\!\!{\small{\mathbf{X}}}})\preceq_{syn}F$. 3. 3.

If $\bigwedge_{x_{i}\in\mathbf{X}}\left(F|_{x_{i}=0}\Leftrightarrow F|_{x_{i}=1}\right)$, then $1\preceq_{syn}F$. 4. 4.

If $F$ is positive (resp. negative) unate in $x_{i}\in\mathbf{X}$, then $x_{i}\wedge F|_{x_{i}=1}$ (resp. $\neg x_{i}\wedge F|_{x_{i}=0}$) $\preceq_{syn}F$. 5. 5.

If $\mathsf{\widetilde{F}}_{1}\preceq_{syn}F_{1}$ and $\mathsf{\widetilde{F}}_{2}\preceq_{syn}F_{2}$, then

1. (a)

$(\mathsf{\widetilde{F}}_{1}\vee\mathsf{\widetilde{F}}_{2})\preceq_{syn}(F_{1}\vee F_{2})$. 2. (b)

$(\mathsf{\widetilde{F}}_{1}\wedge\mathsf{\widetilde{F}}_{2})\preceq_{syn}(F_{1}\wedge F_{2})$* if the output supports of $F_{1}$ and $F_{2}$, and similarly of $\mathsf{\widetilde{F}}_{1}$ and $\mathsf{\widetilde{F}}_{2}$, are disjoint.*

Propositions 5(2) and 5(3) effectively require $F(\mathbf{X},\mathbf{Y})$ to be semantically (but not necessarily syntactically) independent of $\mathbf{Y}$ and $\mathbf{X}$ respectively. While these may appear to be degenerate cases, we will soon see that both these propositions turn out to be useful when recursively compiling a $\mathsf{CNF}$ specification into refined SynNNF specification. Interestingly, a version of Proposition 5(4) was used in a pre-processing step of BFSS [1], although the precise notion of refinement w.r.t. synthesis was not defined there. Thanks to Definition 4, we can now generalize Proposition 5(4) to refine a specification even when $F$ is not unate in any output variable. We discuss below how this can be done.

Suppose the specification $F(\mathbf{X},\mathbf{Y})$ uniquely defines an output variable as a function of other input and output variables. For example, if $F(\mathbf{X},\mathbf{Y})\equiv(\neg x_{i}\vee x_{j})\wedge(\neg x_{i}\vee y_{k})\wedge(x_{i}\vee\neg x_{j}\vee\neg y_{k})\wedge\cdots$, then $F(\mathbf{X},\mathbf{Y})\Rightarrow\left(x_{i}\Leftrightarrow(x_{j}\wedge y_{k})\right)$. Such specifications arise naturally when a non-$\mathsf{CNF}$ Boolean formula is converted to $\mathsf{CNF}$ via Tseitin encoding [26]. Variables like $x_{i}$ above are said to be functionally determined (henceforth called $\mathsf{FD}$) in $F$, and implied functional dependencies like $\left(x_{i}\leftrightarrow(x_{j}\wedge y_{k})\right)$ are called functional definitions (henceforth called f-defs) of $\mathsf{FD}$ variables in $F$.

Let $\mathbf{T}\subseteq\mathbf{X}$ be a set of $\mathsf{FD}$ output variables in $F$, and let ${\mathsf{Fun}}_{\mathbf{T}}$ be the conjunction of f-defs of all variables in $\mathbf{T}$. We say that $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ is an acyclic system of f-defs if no variable in $\mathbf{T}$ transitively depends on itself via the functional definitions in ${\mathsf{Fun}}_{\mathbf{T}}$. In other words, ${\mathsf{Fun}}_{\mathbf{T}}$ induces an acyclic system of functional dependencies between variables in $\mathbf{T}$. For $x_{i}\in\mathbf{X}\setminus{\mathbf{T}}$, define $\theta_{F,\mathbf{T},x_{i},a}$ to be the formula $\left(F(\mathbf{X},\mathbf{Y})|_{x_{i}=a}\wedge\bigwedge_{x_{j}\in\mathbf{X}\setminus(\mathbf{T}\cup\{x_{i}\})}(x_{j}\Leftrightarrow x_{j}^{\prime})\right.$ $\wedge$ $\left.{\mathsf{Fun}}_{\mathbf{T}}(\mathbf{X}^{\prime},\mathbf{Y})|_{x_{i}^{\prime}=1-a}\right)$ $\Rightarrow F(\mathbf{X}^{\prime},\mathbf{Y})|_{x_{i}^{\prime}=1-a}$ , where $a\in\{0,1\}$ and $\mathbf{X}^{\prime}$ is a sequence of fresh variables $(x_{1}^{\prime},\ldots x_{n}^{\prime})$. Informally, $\theta_{F,\mathbf{T},x_{i},a}$ asserts that if the specification $F$ can be satisfied by setting a non-$\mathsf{FD}$ output $x_{i}$ to $a$, then it can also be satisfied by setting $x_{i}$ to the complement value ($1-a$), while preserving the values of all other non-$\mathsf{FD}$ outputs. The $\mathsf{FD}$ outputs in ${\mathbf{T}}$ must of course be set as per the functional definitions in ${\mathsf{Fun}}_{\mathbf{T}}$.

Lemma 6**.**

Let $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ be an acyclic system of f-defs in $F$.

If $\mathbf{X}=\mathbf{T}$, then ${\mathsf{Fun}}_{\mathbf{T}}\preceq_{syn}F$. 2. 2.

If $\mathbf{X}\setminus\mathbf{T}\neq\emptyset$, then for every $x_{i}\in\mathbf{X}\setminus\mathbf{T}$, we have:

If $\theta_{F,\mathbf{T},x_{i},0}$ is a tautology, then $(x_{i}\wedge F|_{x_{i}=1})\preceq_{syn}F$. Similarly, if $\theta_{F,\mathbf{T},x_{i},1}$ is a tautology, then $(\neg x_{i}\wedge F|_{x_{i}=0})\preceq_{syn}F$.

If $\mathbf{T}=\emptyset$, Lemma 6(2) simply reduces to Proposition 5(4). However, if $\mathbf{T}\neq\emptyset$ (as is often the case), Lemma 6(2) shows that $x_{i}\wedge F|_{x_{i}=1}$ (resp. $\neg x_{i}\wedge F|_{x_{i}=0}$) can refine $F$ even if $F$ is not positive (resp. negative) unate in $x_{i}$. As an illustration, the specification $G(x_{1},x_{2},y_{1},y_{2})$ in Example 3 is not unate in either $x_{1}$ or $x_{2}$. However, with $\mathbf{T}=\{x_{1}\}$ and ${\mathsf{Fun}}_{\mathbf{T}}\equiv(x_{1}\Leftrightarrow(x_{2}\vee y_{1}))$, we have $\theta_{F,\mathbf{T},x_{2},0}\equiv 1$. Hence, $x_{2}\wedge G|_{x_{2}=1}\equiv(x_{1}\wedge x_{2})\preceq_{syn}G$. When $F$ is refined by an application of Lemma 6(2), we say that $F$ is refined by pivoting on $x_{i}$.

Lemma 7**.**

Let $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ and $(\mathbf{T}^{\prime},{\mathsf{Fun}}_{\mathbf{T}^{\prime}})$ be acyclic systems of f-defs in $F$, where $\mathbf{T}^{\prime}\subseteq\mathbf{T}\subseteq\mathbf{X}$ and ${\mathsf{Fun}}_{\mathbf{T}}\equiv{\mathsf{Fun}}_{\mathbf{T}^{\prime}}\wedge{\mathsf{Fun}}_{\mathbf{T}\setminus\mathbf{T}^{\prime}}$. For $a\in\{0,1\}$, if $\theta_{F,\mathbf{T}^{\prime},x_{i},a}$ is a tautology, then so is $\theta_{F,\mathbf{T},x_{i},a}$.

Lemma 7, along with Lemma 6(2), shows that if $\mathbf{T}^{\prime}\subsetneq\mathbf{T}\subseteq\mathbf{X}$, the system of acyclic f-defs $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ potentially provides more opportunities for refinement compared to $(\mathbf{T}^{\prime},{\mathsf{Fun}}_{\mathbf{T}^{\prime}})$. Hence, it is advantageous to augment the set $\mathbf{T}$ of $\mathsf{FD}$ outputs (and correspondingly ${\mathsf{Fun}}_{\mathbf{T}}$) whenever possible.

The following theorem suggests that compiling a given specification to a refined SynNNF specification (as opposed to an equivalent SynNNF specification) holds promise for Boolean functional synthesis.

Theorem 8**.**

*For every relational specification $F(\mathbf{X},\mathbf{Y})$, there exists a polynomial-sized Skolem function vector for $\mathbf{X}$ in $F$ iff there exists a SynNNF specification $\mathsf{\widetilde{F}}(\mathbf{X},\mathbf{Y})$ such that $\mathsf{\widetilde{F}}\preceq_{syn}F$ and $\mathsf{\widetilde{F}}$ is polynomial-sized in $F$. *

Theorem 8 guarantees that whenever a polynomial-sized Skolem function vector exists for a specification $F(\mathbf{X},\mathbf{Y})$, there is also a polynomial-sized refined specification in SynNNF. It is therefore interesting to ask if we can compile $F(\mathbf{X},\mathbf{Y})$ to a “small enough” SynNNF specification $\mathsf{\widetilde{F}}(\mathbf{X},\mathbf{Y})$ that refines $F$. In the next two sections, we present such a compilation algorithm and results of our preliminary experiments using this algorithm. Note that as shown in [1], there exist problem instances for which there are no polynomial-sized Skolem function vectors, unless the Polynomial Hierarchy ($\mathsf{PH}$) collapses. Thus, any algorithm for compilation to SynNNF must incur super-polynomial blow-up (unless $\mathsf{PH}$ collapses). Nevertheless, as our experiments show, the compilation-based approach works reasonably well in practice, even solving benchmarks beyond the reach of existing state-of-the-art $\mathsf{BFnS}$ tools.

V A Refining $\mathsf{CNF}$ to SynNNF Compiler

We now describe $\mathsf{C2Syn}$ – an algorithm that takes as input a $\mathsf{CNF}$ specification $F(\mathbf{X},\mathbf{Y})$ given as a set of clauses, and outputs a DAG representation of a SynNNF specification $\widetilde{F}(\mathbf{X},\mathbf{Y})$ that refines $F(\mathbf{X},\mathbf{Y})$ w.r.t. synthesis. Given a set $\mathcal{S}$ of clauses, we use $\varphi_{{\mathcal{S}}}$ to denote the formula $\bigwedge_{C_{i}\in\mathcal{S}}C_{i}$.

Let $\mathcal{S}=\{C_{1},\ldots C_{r}\}$ be a set of clauses. Abusing notation introduced in Section II, let $atoms({C_{i}})=\{z\mid z\in\mathbf{X}\cup\mathbf{Y},lits({C_{i}})\cap\{z,\neg z\}\neq\emptyset\}$. We define an undirected graph $G_{\mathcal{S}}=(V_{\mathcal{S}},E_{\mathcal{S}})$, where $V_{\mathcal{S}}=\{C_{1},\ldots C_{r}\}$ and $(C_{i},C_{j})\in E_{\mathcal{S}}$ iff $i\neq j$ and $atoms({C_{i}})\cap atoms({C_{j}})\cap\mathbf{X}\neq\emptyset$. Thus, there exists an edge $(C_{i},C_{j})$ iff $C_{i}$ and $C_{j}$ share an output atom. Let $\{\mathcal{S}_{1},\ldots\mathcal{S}_{q}\}$ be the set of maximally connected components (henceforth called $\mathsf{MCC}$s) of $G_{\mathcal{S}}$. It is easy to see that $\varphi_{{\mathcal{S}}}~{}\equiv~{}\bigwedge_{k=1}^{q}\varphi_{{\mathcal{S}_{k}}}$; moreover, the output supports of $\varphi_{{\mathcal{S}_{k}}}$ for $k\in\{1,\ldots q\}$ are mutually disjoint. We use $C_{i}\sim_{\mathcal{S}}C_{j}$ to denote that clauses $C_{i}$ and $C_{j}$ are in the same $\mathsf{MCC}$ of $G_{\mathcal{S}}$. We will soon see how factoring $\varphi_{{\mathcal{S}}}$ based on $\mathsf{MCC}$s of $G_{\mathcal{S}}$ allows us to decompose the $\mathsf{CNF}$-to-SynNNF compilation problem into independent sub-problems, thanks to Proposition 5(5)b. Note that factoring based on $\mathsf{MCC}$s has also been used in DSharp [20] for converting a $\mathsf{CNF}$ formula to dDNNF. However, unlike $G_{\mathcal{S}}$ above, the underlying graph in DSharp has an edge between every pair of clauses that shares any atom, including input variables. Thus, $G_{\mathcal{S}}$ has potentially fewer edges, and hence smaller $\mathsf{MCC}$s, than the corresponding graph constructed by DSharp.

Before delving into Algorithm $\mathsf{C2Syn}$, we first discuss some important sub-routines used in the algorithm. Sub-routine FDRefine takes as inputs a set $\mathcal{S}$ of clauses and a (possibly empty) acyclic system of f-defs $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ in $\varphi_{{\mathcal{S}}}$. It returns a (possibly augmented) acyclic system of f-defs $(\mathbf{T}^{\prime},{\mathsf{Fun}}_{\mathbf{T}^{\prime}})$ and a set of clauses $\mathcal{S}^{\prime}$ such that $\varphi_{{\mathcal{S}^{\prime}}}\preceq_{syn}\varphi_{{\mathcal{S}}}$ and $\varphi_{{\mathcal{S}^{\prime}}}\Rightarrow{\mathsf{Fun}}_{\mathbf{T}^{\prime}}$. Sub-routine FDRefine works by iteratively finding new $\mathsf{FD}$ ouptut variables and refining the specification using Lemma 6(2) whenever possible. In the pseudo-code of FDRefine (see Algorithm 1), sub-routine FindFD matches a pre-defined set of clause-patterns in $\mathcal{S}^{\prime}$ to identify new $\mathsf{FD}$ output variables not already in $\mathbf{T}^{\prime}$. The patterns currently matched correspond to $\mathsf{CNF}$ encodings of the input-output relation of common Boolean functions, viz. $\mathsf{and}$, $\mathsf{or}$, $\mathsf{nand}$, $\mathsf{nor}$, $\mathsf{xor}$, $\mathsf{xnor}$, $\mathsf{not}$ and $\mathsf{identity}$. For example, we match the pattern $(\neg\alpha\vee\beta_{1})\wedge(\neg\alpha\vee\beta_{2})\wedge(\neg\beta_{1}\vee\neg\beta_{2}\vee\alpha)$, where $\alpha,\beta_{1},\beta_{2}$ are place-holders, to identify the functional definition $(\alpha\leftrightarrow(\beta_{1}\wedge\beta_{2}))$. Each new $\mathsf{FD}$ output variable thus identified is added to $\mathbf{T}^{\prime}$ and the corresponding functional definition is added to ${\mathsf{Fun}}_{\mathbf{T}^{\prime}}$ unless this introduces a cyclic dependency among the f-defs already in ${\mathsf{Fun}}_{\mathbf{T}^{\prime}}$. Assuming all patterns used by FindFD to determine functional dependencies are sound, the (possibly augmented) $(\mathbf{T}^{\prime},{\mathsf{Fun}}_{\mathbf{T}^{\prime}})$ computed by FindFD is a system of acyclic f-defs in $\varphi_{{\mathcal{S}^{\prime}}}$. In lines $6$-$12$ of Algorithm 1, we next check if Lemma 6(2) can be applied to refine $\varphi_{{\mathcal{S}^{\prime}}}$ by pivoting on some variable $x_{i}\in\mathbf{Out}\setminus\mathbf{T}^{\prime}$. The refinement, if applicable, is easily done by replacing each clause $C_{i}\in\mathcal{S}^{\prime}$ by $C_{i}|_{x_{i}=1}$ (resp. $C_{i}|_{x_{i}=0}$) and by adding the unit clause $x_{i}$ (resp. $\neg x_{i}$) to $\mathcal{S}^{\prime}$. The pivot $x_{i}$ is also added to $\mathbf{T}^{\prime}$ and the corresponding functional definition ($x_{i}\Leftrightarrow 1$ or $x_{i}\Leftrightarrow 0$ as the case may be) is added to ${\mathsf{Fun}}_{\mathbf{T}^{\prime}}$.

In general, identifying an acyclic system of f-defs in $F$ potentially enables refinement of $F$ via Lemma 6(2), which in turn, can lead to augmenting the acyclic system of f-defs further. Therefore, the loop in lines $3$-$13$ of Algorithm 1 is iterated until no new $\mathsf{FD}$ outputs or additional refinements are obtained. Once this happens, subroutine FDRefine returns the resulting acyclic system of f-defs $(\mathbf{T}^{\prime},{\mathsf{Fun}}_{\mathbf{T}^{\prime}})$ and the resulting set of refined clauses $\mathcal{S}^{\prime}$.

Two other important sub-routines used in $\mathsf{C2Syn}$ are GetCkt and GetDefCkt. Sub-routine GetCkt takes as input an $\mathsf{NNF}$ formula $G(\mathbf{X},\mathbf{Y})$ and returns the DAG representation of $G(\mathbf{X},\mathbf{Y})$. Sub-routine GetDefCkt takes as input a system of acyclic f-defs $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$, where $\mathbf{X}\cap\mathsf{sup}({{\mathsf{Fun}}_{\mathbf{T}}})=\mathbf{T}$ (i.e. $\mathbf{T}$ is the entire output support of ${\mathsf{Fun}}_{\mathbf{T}}$). It returns a DAG representation of a SynNNF specification equivalent to ${\mathsf{Fun}}_{\mathbf{T}}$. Without loss of generality, let $x_{1}\sqsubset\ldots\sqsubset x_{n}$ be a linear ordering of the output variables in $\mathbf{T}$ such that the functional definition of $x_{i}$ in ${\mathsf{Fun}}_{\mathbf{T}}$ does not depend on any $x_{j}$ for $j\geq i$. Such an ordering always exists since $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ is an acyclic system of f-defs. Let $x_{i}\Leftrightarrow\mathsf{op}_{i}(u_{1},\ldots u_{n_{i}})$ be the functional definition of $x_{i}$ in ${\mathsf{Fun}}_{\mathbf{T}}$, where $\mathsf{op}_{i}$ is a Boolean function identified via clause-pattern matching in sub-routine FindFD. For each $i$ in $\sqsubset$-order in $\{1,\ldots n\}$, we now construct a DAG ${\mathcal{D}}_{i}$ representing $\mathsf{op}_{i}(u_{1},\ldots u_{n_{i}})$ in $\mathsf{NNF}$. While constructing ${\mathcal{D}}_{i}$, we ensure that every $x_{j}\in\mathbf{T}$ that is also an argument of $\mathsf{op}_{i}$ is replaced by the root, say $t_{j}$, of the DAG ${\mathcal{D}}_{j}$. Since $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ is an acyclic system of f-defs, this is always possible. Finally, we construct the overall DAG, say $\mathcal{D}$, representing $\bigwedge_{x_{i}\in\mathbf{T}}\left((x_{i}\wedge t_{i})\vee(\neg x_{i}\wedge\neg t_{i})\right)$. It is easy to see that for every $x_{i}\in\mathbf{T}$, there are no paths from $x_{i}$ and $\neg x_{i}$ that meet for the first time at an $\wedge$-labeled node in $\mathcal{D}$. Abusing notation and using $\mathcal{D}$ to denote the specification represented by the above DAG, we therefore have $[\widehat{{\mathcal{D}}}]_{i}$ is $\wedge_{i}$-unrealizable for all $i\in\{1,\ldots n\}$; hence ${\mathcal{D}}$ is in SynNNF.

We are now in a position to describe Algorithm $\mathsf{C2Syn}$. The algorithm is recursive and takes as inputs a set $\mathcal{S}$ of clauses, a (possibly empty) system of acyclic f-defs $(\mathbf{T},\mathsf{Fun}_{\mathbf{T}})$ in $\varphi_{{\mathcal{S}}}$, and the recursion level $\ell$. Initially, $\mathsf{C2Syn}$ is invoked with $\mathcal{S}=$ given set of $\mathsf{CNF}$ clauses, $\mathbf{T}=\emptyset$, ${\mathsf{Fun}}_{\mathbf{T}}=1$ and $\ell=0$. The pseudocode of $\mathsf{C2Syn}$, shown in Algorithm 2, first computes the output support $\mathbf{Out}$ of $\varphi_{{\mathsf{S}}}$, and then checks a few degenerate cases (lines $2$-$8$) to determine if a refined SynNNF specification can be easily obtained. In case these checks fail, sub-routine FDRefine is invoked to augment the set $\mathbf{T}^{\prime}$ of functionally dependent outputs and their corresponding acyclic f-defs ${\mathsf{Fun}}_{\mathbf{T}^{\prime}}$, and also to obtain a (possibly) refined set $\mathcal{S}^{\prime}$ of clauses. If all outputs in $\mathbf{Out}$ get functionally determined by this, Lemma 6(1) guarantees that ${\mathsf{Fun}}_{\mathbf{Out}}\preceq_{syn}\varphi_{{\mathcal{S^{\prime}}}}$; hence an invocation of GetDefCkt($\mathbf{Out},{\mathsf{Fun}}_{\mathbf{Out}}$) gives the desired result in line $12$. Otherwise, we check in lines $14$-$17$ if Theorem 3(ii) can be applied. Recall that Theorem 3(ii) relaxes the requirements of the SynNNF definition by requiring $\wedge_{i}$-unrealizability only when $\mathsf{GACKS}{}$ functions are substituted for the $\mathbf{X}$ variables. As discussed in Section III-3, the relaxed requirement can be checked by testing the unsatisfiability of the error formula $\varepsilon$ for the $\mathsf{GACKS}{}$ function vector $\Psi$. If $\varepsilon$ is indeed unsatisfiable, $\Psi$ is a Skolem function vector for $\mathbf{Out}$ in $\varphi_{{\mathcal{S}^{\prime}}}$, and hence $\bigwedge_{x_{i}\in\mathbf{Out}}(x_{i}\Leftrightarrow\Psi_{i})$ refines $\varphi_{{\mathcal{S}^{\prime}}}$.

If $\varepsilon$ is satisfiable, we use a sub-routine ChooseOutputVar that heuristically chooses an output variable $x\in\mathbf{Out}\setminus\mathbf{T}^{\prime}$ on which to branch. Currently, we use a $\mathsf{VSIDS}$ [19] score based heuristic, similar to that used in DSharp [20], to rank variables in $\mathbf{Out}\setminus\mathbf{T}^{\prime}$, and then choose the variable with the highest score. This allows us to represent $\varphi_{{\mathcal{S}^{\prime}}}$ as $x_{i}\wedge\varphi_{{\mathcal{S}^{\prime}|_{x=1}}}\vee\neg x_{i}\wedge\varphi_{{\mathcal{S}^{\prime}|_{x=0}}}$, so that we can refine the two disjuncts independently, thanks to Proposition 5(5)a. However, this may lead to some duplicate processing of clauses. We can avoid this by factoring out the subset of clauses whose satisfiability is independent of whether $x_{i}$ is set to $1$ or [math]. Let $\mathcal{S}_{1}$ (resp. $\mathcal{S}_{2}$) be the subset of clauses in $\mathcal{S}^{\prime}$ that are in the same $\mathsf{MCC}$ of $G_{\mathcal{S}^{\prime}}$ as some $C_{j}$ that has $x$ (resp. $\neg x$) as a literal. Let $\mathcal{S}_{3}$ be the set of all clauses in $\mathcal{S}^{\prime}$ that are neither in $\mathcal{S}_{1}$ nor $\mathcal{S}_{2}$. By definition of $G_{\mathcal{S}^{\prime}}$, the sub-specifications $\varphi_{{\mathcal{S}_{1}}}$ and $\varphi_{{\mathcal{S}_{3}}}$ (and similarly, $\varphi_{{\mathcal{S}_{2}}}$ and $\varphi_{{\mathcal{S}_{3}}}$) do not share any output variable in their supports, and can be refined independently. This is exactly what algorthm $\mathsf{C2Syn}$ does in lines $19$-$30$. The roots of the DAGs resulting from the recursive calls in lines $27$, $28$ and $29$ are finally combined as in line $30$ to yield the desired DAG representation.

Theorem 9**.**

For every set $\mathcal{S}$ of clauses, \mathsf{C2Syn}$$\left(\mathcal{S},\emptyset,1,0\right) always terminates and returns a DAG representation of a SynNNF specification $\mathsf{\widetilde{F}}$ such that $\mathsf{\widetilde{F}}\preceq_{syn}\varphi_{{\mathcal{S}}}$.

VI Experimental results

We ran Algorithm $\mathsf{C2Syn}$ on a suite of $\mathsf{CNF}$ specifications comprised of benchmarks from the Prenex 2QBF track of QBFEval 2018 [21], and the .qdimacs version of Factorization benchmarks [1], which we will refer to as FA.QD. By Theorem 2(i), a ROBDD/FBDD specification can be compiled to an equivalent SynNNF specification in linear time. Therefore, any algorithm that compiles a $\mathsf{CNF}$ specification to an ROBDD can be viewed as an alternative to $\mathsf{C2Syn}$ for compiling a $\mathsf{CNF}$ specification to SynNNF (albeit without refinement). We compare the performance of $\mathsf{C2Syn}$ with that of a BDD compiler and two state-of-the-art boolean function synthesis tools, namely, $(i)$ the AIG-NNF pipeline of bfss [1] with ABC’s MiniSat as the SAT solver and $(ii)$ Cadet [22, 24]. For the BDD Compiler, the .qdimacs input was converted to an AIG using simple Tseitin variable detection; this AIG was then simplified and ROBDDs built using dynamic variable ordering (of all input and output variables) – this is part of the BDD pipeline of bfss [1], henceforth called $\textsc{BDD}^{\textsc{bfss}}$. We also ran DSharp [20] which compiles a $\mathsf{CNF}$ formula into dDNNF (and hence SynNNF by Theorem 2(i)), but it was successful on very few of our benchmarks; hence we do not present its performance. Each tool took as input the same .qdimacs file. Experiments were performed on a cluster with $20$ cores and $64$ GB memory per node, each core being a $2.2$ GHz Intel Xeon processor running CentOS6.5. Each run was performed on a single core, with timeout of $1$ hour and main memory limited to $16$GB.

For $\mathsf{C2Syn}$, several benchmarks were solved in the initial part of the Algorithm 2 before line 17, i.e., before any recursive calls are made. Table I presents the results for $\mathsf{C2Syn}$, divided into those that succeeded at recursion level zero (Stage-I) and those that required recursions (Stage-II), as well as the comparison with $\textsc{BDD}^{\textsc{bfss}}$. Since BDDs are also in SynNNF, the total number of benchmarks in QBFEval which could be compiled into SynNNF (by either compiler) is a whopping $283/402$.

Figure 1 compares the run-times of $\mathsf{C2Syn}$ and $\textsc{BDD}^{\textsc{bfss}}$: for most QBFEval benchmarks that were solved by both, $\mathsf{C2Syn}$ took less time, while for FA.QD, $\mathsf{C2Syn}$ took more time. There were $130$ QBFEval benchmarks that $\mathsf{C2Syn}$ solved by $\textsc{BDD}^{\textsc{bfss}}$ couldn’t, whereas $98$ were solved by $\textsc{BDD}^{\textsc{bfss}}$ but not $\mathsf{C2Syn}$. This indicates that the two approaches to SynNNF compilation have orthogonal strengths.

We next compare $\mathsf{C2Syn}$ with Cadet and bfss. Cadet (resp. bfss) solved $213$ (resp. $181$) benchmarks in QBFEval and $4$ (resp. $3$) in FA.QD. Table II gives a comparison in terms of number of benchmarks solved by each tool but not by others. Figure 2 (left, right) compares the run-times of $\mathsf{C2Syn}$ and those of Cadet and bfss, respectively. As expected, since $\mathsf{C2Syn}$ does complete compilation, it takes more time than Cadet and marginally more than bfss on many benchmarks, though for most of these, the time taken is less than a minute. In fact for FA.QD, $\mathsf{C2Syn}$ takes less time than bfss on all benchmarks. Overall, $\mathsf{C2Syn}$ appears to have strengths orthogonal to $\textsc{BDD}^{\textsc{bfss}}$, bfss and Cadet, and adds to the repertoire of state-of-the-art tools for Boolean functional synthesis.

To validate our experimental results, we also developed an independent approach to verify if the output of $\mathsf{C2Syn}$ is (i) in SynNNF and (ii) a refinement of the original specification (which by Theorem 1 and Lemma 4 suffices to efficiently generate Skolem functions). For (i), we check a stronger than required, syntactic condition for being in SynNNF, namely, for every output variable $x_{i}$, there is no pair of paths from $x_{i}$ and $\overline{x_{i}}$ in the DAG output by $\mathsf{C2Syn}$ that meet at an $\wedge$-node. Note that this is the sufficient that was described just after Definition 2 in Section III. While this requirement is stronger than the semantic requirement for SynNNF, we choose to use this because of the efficient manner in which this can be checked.

For (ii), we just check the two semantic conditions in Definition 4. Checking condition (a) requires the use of a 2QBF solver, while condition (b) can be checked using a propositional (un)satisfiability solver. Of the 185 benchmarks on which C2Syn was successful, our verifier successfully verified 183 benchmarks, ran out of memory on 1 and out of time on another benchmark (time limit: 1 hour, main memory limit : 16GB).

Finally, we note that pre-processing techniques are known to effectively simplify several QBF problem instances. Stage-I of $\mathsf{C2Syn}$ can be seen as subsuming several simple QBF preprocessing techniques, e.g., unit clause and pure literal detection, semantic unateness and identifying Tseitin variables. Using more aggressive QBF preprocessing could further improve the performance of our tool, and we leave this for future work.

VII Conclusion

We presented a new sub-class of $\mathsf{NNF}$ called SynNNF that admits quadratic-time synthesis and linear-time existential quantification of a set of variables. Our prototype compiler is able to handle several benchmarks that cannot be handled by other state-of-the-art tools. Since representations like ROBDDs, DNNF and the like are either already in or easily transformable to SynNNF, our work is widely applicable and can be used in tandem with other techniques. As future work, we intend to work on optimizing our SynNNF compiler.

Appendix A Material from Section III

A-A Proof of Theorem 2 of Section III

This section is dedicated to the proof of Theorem 2. We show that SynNNF is a space-efficient DAG-based representation of boolean functions, when compared with other representations using $\mathsf{FBDD}$, DNNF and dDNNF.

First, observe that Part(i) is easy. That is, it has been shown, e.g., in  [9] that $\mathsf{FBDD}$can be converted to DNNF with a linear complexity blowup. Now, focussing on dDNNF, DNNF, wDNNF, an examination of their definitions immediately gives us that each of these forms is already in SynNNF. Further, from the definition again it is clear that dDNNFis subsumed by DNNF, which is further subsumed by wDNNF, as depicted in Figure 3. To show strictness, it suffices to consider Example 1, which is in SynNNF but not in wDNNF since $x_{2}$ and $\neg x_{2}$ indeed meet up at an $\wedge$-node in $G$. This completes Part (i).

For part (ii), we start by noting that it has been shown in [9] that the DNNF representation is exponentially more succinct than $\mathsf{FBDD}$. We now show that SynNNF is super-polynomially (resp. exponentially) more succinct than dDNNF, DNNF and wDNNF (resp. $\mathsf{FBDD}$) representations, unless some long-standing complexity conjectures are falsified. To do this, we describe a family of specifications having a polynomial sized SynNNF representation, but for which the other representations are necessarily super-polynomially larger, unless these complexity conjectures are falsfied.

Consider the family $F(\mathbf{X},\mathbf{Y})$ of specifications defined as follows. Let $\mathbf{X}=\{x_{1},\dots,x_{n}\}$, and let $f_{i}(\mathbf{X}_{i+1}^{n},\mathbf{Y})$, $1\leq i\leq n-1$ be arbitrary boolean functions in $\mathsf{NNF}$ over $x_{i+1},\dots,x_{n},\mathbf{Y}$. We define the family $F(\mathbf{X},\mathbf{Y})_{(\mathsf{op}^{\prime}_{1},\mathsf{op}_{1},\dots,\mathsf{op}^{\prime}_{n},\mathsf{op}_{n})}$, parametrized by $\mathsf{op}_{i}\in\{\wedge,\vee\}$, and $\mathsf{op}^{\prime}_{i}\in\{\wedge,\vee,\oplus\}$ as

[TABLE]

Lemma 10**.**

Let $g$ be a function in the family of specifications $F(\mathbf{X},\mathbf{Y})_{(\mathsf{op}^{\prime}_{1},\mathsf{op}_{1},\dots,\mathsf{op}^{\prime}_{n},\mathsf{op}_{n})}$. Then

If $\mathsf{op}^{\prime}_{1}=\dots=\mathsf{op}^{\prime}_{n}=\vee$, then $g$ is in SynNNF. 2. 2.

If $\mathsf{op}^{\prime}_{1}=\dots=\mathsf{op}^{\prime}_{n}=\oplus$, then $g$ is in SynNNF.

Proof.

Let $g$ be any function in the family with all the $\mathsf{op}^{\prime}_{i}=\vee$. It is easy to see that $g$ is in SynNNF, using the sufficient condition in Section III. That is, in $[\widehat{{g}}]_{1}$, there is no $\overline{{x_{1}}}$, so we never have a $x_{1}$ and $\overline{{x}}_{1}$ meeting at the root. Further, $[\widehat{{g}}]_{2}$ after replacing $x_{1}$ with 1, the leftmost subtree rooted at $\vee$ having children $x_{1},f_{1}$ is no longer there after constant propagation. In the rest of the tree, we have no occurrences of $\overline{{x}}_{2}$, hence no way for $x_{2}$ and $\overline{{x}}_{2}$ to meet at the root. Thus, for each $[\widehat{{g}}]_{i}$, the argument is similar, since on replacing $x_{1},\dots,x_{i-1}$ with 1 and doing constant propagation, the remaining DAG will not have $x_{i+1}$ and $\overline{{x}}_{i+1}$ together, which shows that $g$ is in SynNNF. 2. 2.

Let $g$ be any function in the family with all the $\mathsf{op}^{\prime}_{i}=\oplus$. Note that in this case, we cannot use the sufficient condition as above, clearly, $x_{1},\overline{{x}}_{1}$ meet at a $\wedge$ in $[\widehat{{g}}]_{1}$. Nevertheless $g$ is in SynNNF, if we consider $[\widehat{{g}}]_{i}$ for $1\leq i\leq n$, and consider the root node with children $\alpha_{1}=x_{i}\vee f_{i}$ and $\alpha_{2}=\overline{{x}}_{i}\vee\neg f_{i}$, after substituting $x_{1},\dots,x_{i-1},\overline{{x}}_{1},\dots,\overline{{x}}_{i-1}$ to 1, $\overline{{x}}_{i+1},\dots,\overline{{x}}_{n}$ to $\neg x_{i+1},\dots,\neg x_{n}$, and constant propagation, it is easy to see that $\alpha_{1}^{11}\wedge\alpha_{2}^{11}=1~{}\mathsf{op}_{1}G$, $\alpha_{1}^{10}\wedge\alpha_{2}^{10}=\neg f_{i}\mathsf{op}_{1}G$ and $\alpha_{1}^{01}\wedge\alpha_{2}^{01}=f_{i}\mathsf{op}_{1}G$ for $\mathsf{op}_{1}\in\{\vee,\wedge\}$ and some $G$. Thus $((\alpha_{1}^{11}\wedge\alpha_{2}^{11})\wedge\neg(\alpha_{1}^{10}\wedge\alpha_{2}^{10})\wedge\neg(\alpha_{1}^{01}\wedge\alpha_{2}^{01}))$ is unsatisfiable. Thus, $g$ is $\wedge_{i}$ unrealizable for any $i$.

∎

Theorem 11** (Restatement of Theorem 2(ii)).**

1. (a)

There are functions which admit polynomial sized SynNNF representations, yet admit only exponential sized $\mathsf{FBDD}$ representations. 2. (b)

Unless $\mathsf{P}=\mathsf{VNP}$, there are functions which admit polynomial sized SynNNF representations, yet admit only super-polynomial sized dDNNF representations. 3. (c)

Unless $\mathsf{P}=\mathsf{NP}$, there are functions which admit polynomial sized SynNNF representations, yet admit only super-polynomial sized wDNNF and DNNF representations.

Proof.

We use the family of specifications $F(\mathbf{X},\mathbf{Y})$ defined above, with different instantiations to obtain all three results. Set $\mathsf{op}_{1}=\dots=\mathsf{op}_{n}=\wedge$, $\mathsf{op}^{\prime}_{1}=\dots=\mathsf{op}^{\prime}_{n}=\vee$, $f_{i}(\mathbf{X}_{i+1}^{n},\mathbf{Y})=\top$ for $2\leq i\leq n$, obtaining $g=x_{1}\vee f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$. Let $\mathbf{Y}=\{y_{1},\dots,y_{n-1}\}$. As seen in Lemma 10, $g$ is in SynNNF. In each of the subparts below, we define $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$ appropriately.

Item (a): Succinctness w.r.t $\mathsf{FBDD}$. Let $k=n-1$. We use the $k$-bit multiplier function over $\{x_{2},\dots,x_{n}\}\cup\{y_{1},\dots,y_{n-1}\}$ in the construction of $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$. The two $k$ bit arguments to the multiplier are respectively, $\{x_{2},\dots,x_{n}\}$ and $\{y_{1},\dots,y_{n-1}\}$ with $x_{n},y_{n-1}$ being the most significant bits, and $x_{2},y_{1}$ being the least significant bits. Let $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$ be the boolean function representing the $k$th bit of the $k$-bit multiplier function. The size of $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$ is quadratic in $k$, since the size of any multiplier circuit consisting of $\vee,\wedge$ gates is quadratic in $k$ (sum of $k^{2}$ partial products). For this $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$, the size of $g$ is $\mathcal{O}(k^{2}+1)$.

Let $\mathsf{rep}_{1}$ be a representation of $g$ using $\mathsf{FBDD}$,  by fixing a certain variable order. Set $x_{1}{=}0$. This assignment makes $g=f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$. Indeed, the $\mathsf{FBDD}$  representation obtained as a restriction of $\mathsf{rep}_{1}$ with respect to this truth assignment is simpler [4]. It is known [5] that any $\mathsf{FBDD}$, $\mathsf{OBDD}$  representations for $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$ is exponential in $k$. This establishes the exponential succinctness of SynNNF over $\mathsf{FBDD}$.

Item (b): **Succinctness w.r.t **dDNNF. We use a CNF encoding of the perfect matchings of a bipartite graph $G$ (denoted $\mathsf{pm}(G)$) in the construction of $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$. Given a bipartite graph $G$ with two parts $U=\{u_{1},\dots,u_{m}\}$ and $V=\{v_{1},\dots,v_{m}\}$, we can define a 0-1 matrix $A=(a_{ij}),1\leq i,j\leq m$ such that $a_{ij}=1$ iff there is an edge between $u_{i}\in U$ and $v_{j}\in V$. It is easy to see from the definition of the permanent of $A$ (denoted $\mathsf{perm}(A)$) that $\mathsf{perm}(A)=\mathsf{pm}(G)$. Likewise, the number of perfect matchings of a bipartite graph is the permanent of its incidence matrix. Set $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$ as the function which encodes $\mathsf{pm}(G)$.

Let $\mathsf{rep}_{2}$ be the dDNNF representation of $g$. As in the first case, choose an assignment $x_{1}{=}0$ obtaining $g=0\vee f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})=f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$. Then it can be seen that the number of solutions of $f_{1}$ is exactly the number of perfect matchings of the bipartite graph $G$. Fixing the assignment $x_{1}{=}0$ results in a simpler dDNNF  representation (say $\mathsf{rep}_{3}$) for $g$ (now $f_{1}$). Counting the models of $\mathsf{rep}_{3}$ can be done in time polynomial in the size of $\mathsf{rep}_{3}$ [10]. This implies that we can find the number of perfect matchings of the underlying bipartite graph $G$ in time polynomial in the size of $\mathsf{rep}_{3}$. Unless $\mathsf{P}=\mathsf{VNP}$, $\mathsf{rep}_{3}$ cannot have a polynomial representation, since otherwise, we would obtain a polynomial time solution for computing $\mathsf{perm}(A)$. This shows the super-polynomial succinctness of SynNNF over dDNNF, unless $\mathsf{P}=\mathsf{VNP}$.

Item(c): **Succinctness w.r.t wDNNF and DNNF ** Let $\mathsf{op}^{\prime}_{1}=\dots=\mathsf{op}^{\prime}_{n}=\vee$, $f_{i}(\mathbf{X}_{i+1}^{n},\mathbf{Y})=\top$ for $2\leq i\leq n$, obtaining $g=x_{1}\vee f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$. As shown in Lemma 10, $g$ is in SynNNF, where $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$ is an arbitrary SAT formula. If we can obtain a poly-sized DNNF representation for the function $g$, then using the assignment $x_{1}=0$ in $g$, we obtain a DNNF representation for $f_{1}(\mathbf{X}_{2}^{n},\mathbf{Y})$. But it is known [10] that consistency checking is poly-time for DNNF representations. A polynomial sized DNNF representation for $g$ would imply a polynomial time solution for the satisfiability checking of an arbitrary SAT formula. Thus, unless $\mathsf{P}=\mathsf{NP}$, any DNNF representation for $g$ will necessarily be super polynomial. ∎

This completes the proof of Part (ii) of Theorem 2.

Part(iii). By Theorem 1 of [1], we know that there exist instances of poly-sized NNF formulas whose Skolem functions are necessarily super-polynomial size (resp. exponential) unless the polynomial hierarchy collapses (resp. the non-uniform exponential hypothesis is falsified). For any such instance, suppose we were able to obtain a poly-sized SynNNF representation, then by Theorem 1, we will be able to synthesize polynomial-sized Skolem functions, which contradicts the above.

To see a concrete example where SynNNF is not likely to be succinct, we refer to Theorem 1 of [1], where a constructive reduction of the parameterized clique problem to $\mathsf{BFnS}$ was given. The specification, in this case, has a polynomial-sized representation, but unless some long-standing complexity-theory conjectures are violated, it was shown that any Skolem function must have exponential/super-polynomial size. Thus, unless these conjectures are violated, the same specification in SynNNF must also be exponential/super-polynomial sized.

This proves Part (iii) and completes the proof of this theorem.

Essentially this means that though we obtain succinctness with respect to several known forms (using classical complexity-theoretic results), it is not the case that SynNNF will always be able to produce a poly-sized representation.

A-B Proof of Theorem 3

Let us recall the characterization theorem from Section III. See 3

Proof.

Part 1): The forward direction follows from Theorem 1. For the reverse direction, we will prove the contrapositive: if $F$ is not in SynNNF, i.e., if $[\widehat{{F}}]_{i}[\mathbf{X}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]$ is not $\wedge_{i}$-unrealizable for some $i\in\{1\ldots n\}$, we will show that for some $i$, $\exists\mathbf{X}_{1}^{i}F(\mathbf{X},\mathbf{Y})\not\Leftrightarrow[\widehat{{F}}]_{i+1}[\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]$. Fix any $\mathbf{Y}\in\{\mathbf{Y}^{\prime}\mid\exists\mathbf{X}^{\prime},F(\mathbf{X}^{\prime},\mathbf{Y}^{\prime})\}$, i.e., it is a realizable valuation of inputs. Consider $i$ to be the largest index such that $[\widehat{{F}}]_{i}$ is not $\wedge_{i}$-unrealizable, i.e., the corresponding $\zeta$ is satisfiable. As a result, we have $\alpha^{11}=1$, i.e., $[\widehat{{F}}]_{i+1}[\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]=1$. On the other hand $\alpha^{01}=\widehat{{F}}(1^{i-1},0,{\mathbf{X}}_{i+1}^{n},1^{i-1},1,\neg{\mathbf{X}}_{i+1}^{n},\mathbf{Y})=0$ and $\alpha^{10}=\widehat{{F}}(1^{i-1},1,{\mathbf{X}}_{i+1}^{n},1^{i-1},0,\neg{\mathbf{X}}_{i+1}^{n},\mathbf{Y})=0$. By monotonicity, every assignment of $x_{1},\ldots x_{i-1},x_{i}$ will also result in 0 in $\widehat{{F}}$, which implies that $\exists\mathbf{X}_{1}^{i}F(\mathbf{X},\mathbf{Y})=0$. Thus for this $i$, $\exists\mathbf{X}_{1}^{i}F(\mathbf{X},\mathbf{Y})\not\Leftrightarrow[\widehat{{F}}]_{i+1}[\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg\mathbf{X}_{i+1}^{n}]$, which completes the proof.

Part 2): Forward direction: We will prove the contrapositive, i.e., if $[\widehat{{F}}]_{i}[\mathbf{X}_{i+1}^{n}\mapsto\Psi_{i+1}^{n}]$ is not $\wedge_{i}$-unrealizable for some $i\in\{1\ldots n\}$, we will show that $\Psi_{1}^{n}$ is not a correct Skolem function vector for $\mathbf{X}_{1}^{n}$ in $F(\mathbf{X},\mathbf{Y})$. Fix any $\mathbf{Y}\in\{\mathbf{Y}^{\prime}\mid\exists\mathbf{X}^{\prime},F(\mathbf{X}^{\prime},\mathbf{Y}^{\prime})\}$, i.e., it is a realizable valuation of inputs. Consider $i$ to be the largest index such that $[\widehat{{F}}]_{i}[\mathbf{X}_{i+1}^{n}\mapsto{\Psi}_{i+1}^{n},\overline{{\mathbf{X}}}_{i+1}^{n}\mapsto\neg{\Psi}_{i+1}^{n}]$ is not $\wedge_{i}$-unrealizable, i.e., the corresponding $\zeta$ is satisfiable.

We claim that one of the ${\Psi}_{i+1}^{n}$ must be an incorrect skolem function for this $\mathbf{Y}$. Suppose not, i.e., suppose all of them are correct. Then we have

[TABLE]

However, because at $i$, $\zeta$ is satisfiable, we have $\widehat{{F}}(1^{i-1},0,{\Psi}_{i+1}^{n},1^{i-1},1,\neg{\Psi}_{i+1}^{n},Y)=0$ and $\widehat{{F}}(1^{i-1},1,{\Psi}_{i+1}^{n},1^{i-1},0,\neg{\Psi}_{i+1}^{n},Y)=0$. By monotonicity, every assignment of $x_{1},\ldots x_{i-1}$ will also result in 0 in $\widehat{{F}}$. But this contradicts (1). Hence all the skolem functions cannot be correct for this $\mathbf{Y}$, proving the forward direction.

Reverse direction: Again, we prove by taking the contrapositive. Suppose, ${\Psi}_{i+1}^{n}$ is not a correct Skolem function vector. In [14], it was shown that for any function vector $\varphi_{1}^{n}$, it is a Skolem function vector for $F$ iff the error formula $\varepsilon_{\varphi}\equiv F(\mathbf{X},\mathbf{Y})\wedge\neg F(\mathbf{X}^{\prime},\mathbf{Y})\wedge\bigwedge_{i=1}^{n}(x_{i}^{\prime}\leftrightarrow\varphi_{i})$ is unsatisfiable. We will use this characterization now, i.e., since ${\Psi}_{i+1}^{n}$ is not a correct Skolem function vector, the error formula $\varepsilon_{\Psi}$ must be satisfiable.

Hence, we start by considering $\mathbf{Y}^{*}$ which gives a satisfying assignment for the error formula $\varepsilon_{\Psi}$. That is,

[TABLE]

Let $k$ be the highest such $i$ such that the above statement holds. That is, after $k$, the Skolem functions given by $\Psi$ are correct, and at $k$ they are incorrect. Then, we observe that the value at $k$ must be 1, i.e.,

[TABLE]

To see this, observe that $\exists\mathbf{X}^{\prime}F(\mathbf{X}^{\prime},\mathbf{Y}^{*})$ along with maximality of $k$ implies that $\exists\mathbf{X}_{1}^{k}F(\mathbf{X}_{1}^{k},{\Psi}_{k+1}^{n}(\mathbf{Y}^{*}),\mathbf{Y}^{*})=1$, which in turn implies that

[TABLE]

Now, if $\Psi_{k}(\mathbf{Y}^{*})=0$, this implies $\widehat{{F}}({\mathbf{1}}^{k-1},0,{\psi^{\prime}}_{k+1}^{n}(\mathbf{Y}^{*}),{\mathbf{1}}^{k-1},1,\neg{\psi^{\prime}}_{k+1}^{n}(\mathbf{Y}^{*}),\mathbf{Y}^{*})=1$. But then, setting $x_{k}=1$ is indeed correct, which would imply that there is no error at $k$, which violates the assumption on $k$. Thus we must have $\Psi_{k}(\mathbf{Y}^{*})=1$.

Now, we know that this is an incorrect assignment to $x_{k}$, which implies that the correct assignment is a [math] and we know that $\exists\mathbf{X}_{1}^{k-1}F(\mathbf{X}_{1}^{k-1},1,{\Psi}_{k+1}^{n}(\mathbf{Y}^{*}),\mathbf{Y}^{*})$ is a correct assignment to $x_{k}$. Hence, we must have

[TABLE]

The fact that equations (3), (4) hold together imply that the Skolem function $\Psi$ is wrong at level $k$, since it gives value 1, but fixing $x_{k}=1$, there is no way to set lower variables to get 1. The rest of the proof is a careful case-analysis, where we either show that $\zeta$ (with Skolem functions assigned according to $\Psi$) at level $k$ is satisfiable, i.e., $[\widehat{{F}}]_{k+1}[\mathbf{X}_{k+1}^{n}\mapsto\Psi_{k+1}^{n}]$ is not $\wedge_{k}$-unrealizable and hence the proof terminates, or we show that these equations are satisfied at a lower level (i.e., there is an error at a lower level). Since number of levels is finite this procedure will terminate. We describe the different cases now:

• $\bullet$ Case 1:

The first case is if

[TABLE]

then, $x_{k-1}$ behaves as an AND gate, i.e.,

[TABLE]

which implies that $\zeta$ (with the Skolem functions assigned according to $\Psi$) will be satisfiable at $k-1$ and hence this terminates the proof.

• $\bullet$ Case 2:

This case is if

[TABLE]

In this case, note that $\Psi_{k-1}(\mathbf{Y}*)=1$ and from Equation  (4), we have

[TABLE]

Thus the Skolem function $\Psi$ is wrong at level $k-1$, since it gives 1 but fixing $x_{k-1}=1$, there is no way to set lower variables to 1. In other words, we have reduced the problem by one level and can recursively apply this argument at level $k-1$.

• $\bullet$ Case 3:

[TABLE]

Note that this case is possible only if $k-2\geq 1$. But if this is not the case, i.e., if $k-2=0$, and $\widehat{{F}}({\Psi}_{1}^{n}(\mathbf{Y}^{*}),{\mathbf{1}}^{k-2},\neg{\Psi}_{1}^{n}(\mathbf{Y}^{*}),\mathbf{Y}^{*})=1$, this implies that there exists no counter-example which contradicts Equation (2). Now we have three subcases:

• $\bullet$ Case 3(a):

[TABLE]

But this case reduces to Case 1 above, i.e., we can see that $x_{k-2}$ behaves as an AND gate (i.e., it is not $\wedge_{k-1}$-unrealizable), and so it terminates.

• $\bullet$ Case 3(b):

[TABLE]

which as in Case 2, reduces the problem by two levels.

• $\bullet$ Case 3(c):

[TABLE]

But reduces to Case 3 at level $k-3$, thus ensuring strict progress in this case as well.

Together this completes the proof.

∎

A-C Proofs from Section IV

See 4

Proof.

Let $\mathbf{G}(\mathbf{Y})$ be a Skolem function vector for $\mathbf{X}$ in $\mathsf{\widetilde{F}}(\mathbf{X},\mathbf{Y})$. From condition (a) of Definition 4, we know that $\forall\mathbf{Y}\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\Rightarrow{\mathsf{\widetilde{F}}}(\mathbf{G}(\mathbf{Y}),\mathbf{Y})\right)$. Further, from condition (b) of Definition 4 and using $\mathbf{G}(\mathbf{Y})$ for $\mathbf{X}^{\prime}$, we have $\forall\mathbf{Y}\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\Rightarrow F(\mathbf{G}(\mathbf{Y}),\mathbf{Y})\right)$. This shows that $\mathbf{G}(\mathbf{Y})$ is a Skolem function vector for $\mathbf{X}$ in $F$. ∎

See 5

Proof.

The reflexivity of $\preceq_{syn}$ follows trivially from Definition 4. To see why $\preceq_{syn}$ is transitive, suppose $F_{1}\preceq_{syn}F_{2}$ and $F_{2}\preceq_{syn}F_{3}$. It follows from transitivity of $\Rightarrow$ that $\forall\mathbf{Y}\left(\exists\mathbf{X}F_{3}(\mathbf{X},\mathbf{Y})\Rightarrow\exists\mathbf{X}^{\prime}F_{1}(\mathbf{X},\mathbf{Y})\right)$. This proves condition (a) of $F_{1}\preceq_{syn}F_{3}$. To prove condition (b) of $F_{1}\preceq_{syn}F_{3}$, notice that $\forall\mathbf{Y}\forall\mathbf{X}^{\prime}\left(\left(\exists\mathbf{X}F_{3}(\mathbf{X},\mathbf{Y})\wedge F_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow\left(\exists\mathbf{X}F_{3}(\mathbf{X},\mathbf{Y})\wedge\exists\mathbf{X}^{\prime\prime}F_{2}(\mathbf{X}^{\prime\prime},\mathbf{Y})\wedge F_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\right)$ by condition (a) of $F_{2}\preceq_{syn}F_{3}$. Additionally, $\forall\mathbf{Y}\forall\mathbf{X}^{\prime}\left(\left(\exists\mathbf{X}F_{3}(\mathbf{X},\mathbf{Y})\wedge\exists\mathbf{X}^{\prime\prime}F_{2}(\mathbf{X}^{\prime\prime},\mathbf{Y})\wedge F_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow\left(\exists\mathbf{X}F_{3}(\mathbf{X},\mathbf{Y})\wedge F_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\right)$ by condition (b) of $F_{1}\preceq_{syn}F_{2}$. Finally, by condition (b) of $F_{2}\preceq_{syn}F_{3}$, it follows that $\forall\mathbf{Y}\forall\mathbf{X}^{\prime}\left(\left(\exists\mathbf{X}F_{3}(\mathbf{X},\mathbf{Y})\wedge F_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow F_{3}(\mathbf{X}^{\prime},\mathbf{Y})\right)$. Putting all the parts together and by transitivity of $\Rightarrow$, we have $\forall\mathbf{Y}\forall\mathbf{X}^{\prime}\left(\left(\exists\mathbf{X}F_{3}(\mathbf{X},\mathbf{Y})\wedge F_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow F_{3}(\mathbf{X}^{\prime},\mathbf{Y})\right)$. This proves condition (b) of $F_{1}\preceq_{syn}F_{3}$. 2. 2.

Suppose $\bigwedge_{y_{j}\in\mathbf{Y}}\left(F|_{y_{j}=0}\Leftrightarrow F|_{y_{j}=1}\right)$ and $\pi\models F(\mathbf{X},\mathbf{Y})$. Then $F$ is semantically independent of $\mathbf{Y}$ and $\forall\mathbf{Y}F({\pi}\!\!\downarrow\!\!{\small{\mathbf{X}}},\mathbf{Y})=1$ holds. Therefore, $\forall\mathbf{Y}\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})=1$. Since $\forall\mathbf{Y}\exists\mathbf{X}\mathsf{form}({{\pi}\!\!\downarrow\!\!{\small{\mathbf{X}}}})=1$ trivially, it follows that $\forall\mathbf{Y}\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\Rightarrow\exists\mathbf{X}^{\prime}\mathsf{form}({{pi}\!\!\downarrow\!\!{\small{\mathbf{X}}}})\right)$. Therefore condition (a) of Definition 4 is satisfied. Condition (b) of Definition 4 follows from the observation that since $\pi\models F$ and $F$ is semantically independent of $\mathbf{Y}$, we have $\forall\mathbf{Y}\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})=1$ and $\forall\mathbf{Y}\forall\mathbf{X}\mathsf{form}({{\pi}\!\!\downarrow\!\!{\small{\mathbf{X}}}})\Rightarrow F(\mathbf{X},\mathbf{Y})$. 3. 3.

Suppose $\bigwedge_{x_{i}\in\mathbf{X}}\left(F|_{x_{i}=0}\Leftrightarrow F|_{x_{i}=1}\right)$. Then $F$ is semantically independent of $\mathbf{X}$. Substituting $1$ for $\mathsf{\widetilde{F}}$ in condition (a) of Definition 4, we get a tautology. Similarly, substituting $1$ for $\mathsf{\widetilde{F}}$ in condition (b) of Definition 4, we get $\forall\mathbf{Y}\forall\mathbf{X}^{\prime}\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\Rightarrow F(\mathbf{X}^{\prime},\mathbf{Y})\right)$. Since $F$ is semantically independent of $\mathbf{X}$, the above formula is also a tautology. Hence condition (b) of Definition 4 is also satisfied. 4. 4.

If $F$ is positive unate in $x_{i}$, then $F|_{x_{i}=0}\Rightarrow F|_{x_{i}=1}$. It follows that $F\Leftrightarrow(\neg x_{i}\wedge F|_{x_{i}=0})\vee(x\wedge F|_{x_{i}=1})\Rightarrow F|_{x_{i}=1}$. Therefore, $\forall\mathbf{Y}\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\Rightarrow\exists\mathbf{X}^{\prime}(x_{i}^{\prime}\wedge F(\mathbf{X}^{\prime},\mathbf{Y})|_{x_{i}^{\prime}=1})\right)$. This proves condition (a) of Definition 4. To show that condition (b) of the definition also holds, note that $x_{i}^{\prime}\wedge F(\mathbf{X}^{\prime},\mathbf{Y})|_{x_{i}^{\prime}=1}\Rightarrow F(\mathbf{X}^{\prime},\mathbf{Y})$ is trivially a tautology. The proof for the case when $F$ is negative unate in $x_{i}$ is analogous to the one above. 5. 5.

Suppose $\mathsf{\widetilde{F}}_{1}\preceq_{syn}F_{1}$ and $\mathsf{\widetilde{F}}_{2}\preceq_{syn}F_{2}$.

1. (a)

Since $\exists\mathbf{X}\left(F_{1}(\mathbf{X},\mathbf{Y})\vee F_{2}(\mathbf{X},\mathbf{Y})\right)\Leftrightarrow\left(\exists\mathbf{X}F_{1}(\mathbf{X},\mathbf{Y})\vee\exists\mathbf{X}F_{2}(\mathbf{X},\mathbf{Y})\right)$ and $\exists\mathbf{X}\left(\mathsf{\widetilde{F}}_{1}(\mathbf{X},\mathbf{Y})\vee\mathsf{\widetilde{F}}_{2}(\mathbf{X},\mathbf{Y})\right)\Leftrightarrow\left(\exists\mathbf{X}\mathsf{\widetilde{F}}_{1}(\mathbf{X},\mathbf{Y})\vee\exists\mathbf{X}\mathsf{\widetilde{F}}_{2}(\mathbf{X},\mathbf{Y})\right)$, condition (a) of Definition 4 follows immediately. To see why condition (b) of the definition holds, notice that $\exists\mathbf{X}\left(F_{1}(\mathbf{X},\mathbf{Y})\vee F_{2}(\mathbf{X},\mathbf{Y})\right)\wedge\left(\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\vee\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow\left(\exists\mathbf{X}F_{1}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\vee\left(\exists\mathbf{X}F_{2}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)$. By condition (b) for $\mathsf{\widetilde{F}}_{1}\preceq_{syn}F_{1}$ and $\mathsf{\widetilde{F}}_{2}\preceq_{syn}F_{2}$, it follows that $\left(\exists\mathbf{X}F_{1}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow F_{1}(\mathbf{X}^{\prime},\mathbf{Y})$ and $\left(\exists\mathbf{X}F_{2}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow F_{2}(\mathbf{X}^{\prime},\mathbf{Y})$. Hence, by transitivity of $\Rightarrow$, we get $\exists\mathbf{X}\left(F_{1}(\mathbf{X},\mathbf{Y})\vee F_{2}(\mathbf{X},\mathbf{Y})\right)\wedge\left(\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\vee\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow\left(F_{1}(\mathbf{X}^{\prime},\mathbf{Y})\vee F_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)$. Since this holds for all $\mathbf{Y}$ and $\mathbf{X}^{\prime}$, condition (b) of Definition 4 is satisfied. 2. (b)

Since the output supports of $F_{1}$ and $F_{2}$, and similarly of $\mathsf{\widetilde{F}}_{1}$ and $\mathsf{\widetilde{F}}_{2}$, are disjoint, we have $\exists\mathbf{X}\left(F_{1}(\mathbf{X},\mathbf{Y})\wedge F_{2}(\mathbf{X},\mathbf{Y})\right)\Leftrightarrow\left(\exists\mathbf{X}F_{1}(\mathbf{X},\mathbf{Y})\wedge\exists\mathbf{X}F_{2}(\mathbf{X},\mathbf{Y})\right)$, and $\exists\mathbf{X}\left(\mathsf{\widetilde{F}}_{1}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{2}(\mathbf{X},\mathbf{Y})\right)\Leftrightarrow\left(\exists\mathbf{X}\mathsf{\widetilde{F}}_{1}(\mathbf{X},\mathbf{Y})\wedge\exists\mathbf{X}\mathsf{\widetilde{F}}_{2}(\mathbf{X},\mathbf{Y})\right)$. Therefore, condition (a) of Definition 4 follows immediately.

To see why condition (b) of the definition holds, notice that $\exists\mathbf{X}\left(F_{1}(\mathbf{X},\mathbf{Y})\wedge F_{2}(\mathbf{X},\mathbf{Y})\right)\wedge\left(\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow\left(\exists\mathbf{X}F_{1}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\wedge\left(\exists\mathbf{X}F_{2}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)$. By condition (b) for $\mathsf{\widetilde{F}}_{1}\preceq_{syn}F_{1}$ and $\mathsf{\widetilde{F}}_{2}\preceq_{syn}F_{2}$, it follows that $\left(\exists\mathbf{X}F_{1}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow F_{1}(\mathbf{X}^{\prime},\mathbf{Y})$ and $\left(\exists\mathbf{X}F_{2}(\mathbf{X},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow F_{2}(\mathbf{X}^{\prime},\mathbf{Y})$. Hence, by transitivity of $\Rightarrow$, we get $\exists\mathbf{X}\left(F_{1}(\mathbf{X},\mathbf{Y})\wedge F_{2}(\mathbf{X},\mathbf{Y})\right)\wedge\left(\mathsf{\widetilde{F}}_{1}(\mathbf{X}^{\prime},\mathbf{Y})\wedge\mathsf{\widetilde{F}}_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)\Rightarrow\left(F_{1}(\mathbf{X}^{\prime},\mathbf{Y})\wedge F_{2}(\mathbf{X}^{\prime},\mathbf{Y})\right)$. Since this holds for all $\mathbf{Y}$ and $\mathbf{X}^{\prime}$, condition (b) of Definition 4 is satisfied.

∎

See 6

Proof.

To prove part (1), notice that $F\Rightarrow{\mathsf{Fun}}_{\mathbf{T}}$. Hence, whenever $F(\mathbf{X},\mathbf{Y})$ is satisfied, each of the functional definitions in ${\mathsf{Fun}}_{\mathbf{T}}$ are also satisfied. Therefore, condition (a) of Definition 4 is satisfied. For condition (b) of Definition 4, notice that for every value of $\mathbf{Y}$, only when the value of $\mathbf{X}^{\prime}$ is as given by ${\mathsf{Fun}}_{\mathbf{T}}(\mathbf{X}^{\prime},\mathbf{Y})$, does ${\mathsf{\widetilde{F}}}(\mathbf{X}^{\prime},\mathbf{Y})$ evaluate to $1$. For these values of $\mathbf{X}^{\prime}$, if $\mathbf{Y}$ is such that $\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})$ holds, then $F(\mathbf{X}^{\prime},\mathbf{Y})$ must also hold since $\mathbf{X}=\mathbf{T}$ and $F\Rightarrow{\mathsf{Fun}}_{\mathbf{T}}$.

To prove part (2), consider $\theta_{F,\mathbf{T},x_{i},0}$ to be a tautology; the proof for the case of $\theta_{F,\mathbf{T},x_{i},1}$ being a tautology is analogous. We show below that (a) $\forall\mathbf{Y}\left(\exists\mathbf{X}F(\mathbf{X},\mathbf{Y})\Rightarrow\exists\mathbf{X}^{\prime}(x_{i}^{\prime}\wedge F(\mathbf{X}^{\prime},\mathbf{Y})|_{x_{i}^{\prime}=1})\right)$, and (b) $\forall\mathbf{Y}\forall\mathbf{X}\left((x_{i}\wedge F(\mathbf{X},\mathbf{Y})|_{x_{i}=1})\Rightarrow F(\mathbf{X},\mathbf{Y})\right)$. Let $\sigma$ be an arbitrary element in $2^{|\mathbf{Y}|}$. To see why (a) holds, suppose $F(\mathbf{X},\sigma)=1$. If $x_{i}=1$, we set $\mathbf{X}^{\prime}=\mathbf{X}$ and it follows that $(x_{i}\wedge F(\mathbf{X}^{\prime},\sigma)|_{x_{i}=1})=1$. If $x_{i}=0$, we set $x_{j}^{\prime}=x_{j}$ for every $x_{j}\in\mathbf{X}\setminus(\mathbf{T}\cup\{x_{i}\})$, set $x_{i}^{\prime}=1$ and set the value of every $x_{j}^{\prime}$ for $x_{j}\in\mathbf{T}$ according its functional definition in ${\mathsf{Fun}}_{\mathbf{T}}(\mathbf{X}^{\prime},\mathbf{Y})$. Since $\theta_{F,\mathbf{T},x_{i},0}$ is a tautology, it follows that $(x_{i}^{\prime}\wedge F(\mathbf{X}^{\prime},\sigma)|_{x_{i}^{\prime}=1})=1$. To see why (b) holds, suppose $(x_{i}\wedge F(\mathbf{X},\mathbf{Y})|_{x_{i}=1})=1$. It follows trivially that $x_{i}$ must be set to $1$, and $F(\mathbf{X},\mathbf{Y})=1$. ∎

See 7

Proof.

Observe that for any system of acyclic f-defs $(\mathbf{T},{\mathsf{Fun}}_{\mathbf{T}})$ in $F$, since $F(\mathbf{X},\mathbf{Y})\Rightarrow{\mathsf{Fun}}_{\mathbf{T}}$, the formula $\theta_{F,\mathbf{T},x_{i},a}$ is a tautology iff $F(\mathbf{X},\mathbf{Y})|_{x_{i}=a}\Rightarrow\exists\mathbf{T}\,F(\mathbf{X},\mathbf{Y})|_{x_{i}=1-a}$ is a tautology. It is now easy to see that if $\mathbf{T}^{\prime}\subseteq\mathbf{T}\subseteq\mathbf{X}$ and $\theta_{F,\mathbf{T}^{\prime},x_{i},a}$ is valid, then $\theta_{F,\mathbf{T},x_{i},a}$ is valid as well. ∎

See 8

Proof of Theorem 8.

The reverse direction is proved by first applying Theorem 1(ii) to $\mathsf{\widetilde{F}}$, and then noting that since $\mathsf{\widetilde{F}}\preceq_{syn}F$, every Skolem function vector for $\mathbf{X}$ in $\mathsf{\widetilde{F}}$ is also a Skolem function vector for $\mathbf{X}$ in $F$. For the forward direction, let $\mathbf{\Psi}(\mathbf{Y})$ be a Skolem function vector for $\mathbf{X}$ in $F$ such that the size of an AND/OR/NOT gate circuit representation of $\mathbf{\Psi}$ (denoted $|\mathbf{\Psi}|$) is polynomial in $|F|$. As mentioned in Section II, every such circuit can be converted to NNF in time $\mathcal{O}(|\mathbf{\Psi}|)$. Hence the NNF representation of $\mathbf{\Psi}$ is of size at most polynomial in $|F|$. Therefore, w.l.o.g we consider $\mathbf{\Psi}$ to be in NNF. Now consider the specification $\mathsf{\widetilde{F}}(\mathbf{X},\mathbf{Y})\equiv\bigwedge_{i=1}^{n}\left((x_{i}\wedge\psi_{i}(\mathbf{Y}))\vee(\neg x_{i}\vee\neg\psi_{i}(\mathbf{Y}))\right)$. Since no paths from $x_{i}$ and $\neg{x_{i}}$ ($x_{i}\in\mathbf{X}$) meet at an $\wedge$-labeled node in the circuit representation of $\mathsf{\widetilde{F}}$, it follows that $\mathsf{\widetilde{F}}(\mathbf{X},\mathbf{Y})$ is in SynNNF. Furthermore, by construction of $\mathsf{\widetilde{F}}$, every Skolem function vector for $\mathbf{X}$ in $\mathsf{\widetilde{F}}$ is necessarily component-wise semantically equivalent to $\mathbf{\Psi}$, which is itself a Skolem function vector for $\mathbf{X}$ in $F$. Therefore, conditions (a) and (b) in Definition 4 are satisfied by $\mathsf{\widetilde{F}}$, and hence $\mathsf{\widetilde{F}}\preceq_{syn}F$. ∎

A-D Proof from Section V

See 9

Proof.

To see that $\mathsf{C2Syn}$ always terminates, notice that every time the recursion level $\ell$ in Algorithm 2 increases, the set of output variables in the remaining set of clauses reduces by $1$. Hence, the maximum value of $\ell$ can only be $|\mathbf{X}|$, and the recursion always terminates. To see why FDRefine (Algorithm 1) terminates, notice that every time $\mathbf{T}^{\prime}$ changes, its size increases by at least $1$, and hence $\mathbf{T}^{\prime}$ can change at most $|\mathbf{X}|$ times. Similarly, every time $\mathcal{S}^{\prime}$ changes, at least one variable is added to $\mathbf{T}^{\prime}$, and hence $\mathcal{S}^{\prime}$ cannot change more than $|\mathbf{X}|$ times.

To see that the returned specification refines $\varphi_{{\mathcal{S}}}$, notice that each of the return statements in Algorithm 2 (i.e., lines $3$, $6$, $8$, $12$, $17$ and $30$) uses one of the properties of refinement already discussed in Section IV. Specifically, the correctness of line $3$ is trivial. The correctness of lines $6$ and $8$ use Propositions 5(2) and 5(2). The correctness of lines $12$ and $17$ use Lemma 6(1). The correctness of line $30$ uses Proposition 5(5). ∎