Counting Axioms Do Not Polynomially Simulate

Counting Axioms Do Not Polynomially Simulate Counting Gates Russell Impagliazzo and Nathan Segerlind December 6, 2003

Abstract For every prime m ≥ 2, we give a family of tautologies that require super-polynomial size constant-depth Frege proofs from Countm axioms, and whose algebraic translations have constant-degree, polynomial size polynomial calculus refutations over Z m . This shows that constant-depth Frege systems with counting axioms modulo m do not polynomially simulate constant-depth Frege systems with counting gates modulo m. Our primary technical tool is a switching lemma that uses random substitutions of one variable for another in addition to random 0/1 restrictions.

1

Introduction

A major open propositional in propositional proof complexity is to prove superpolynomial proof size lower bounds for constant-depth Frege systems with counting gates. There have been exponential proof size lower bounds proven for constant-depth Frege systems with counting axioms [1, 2, 5, 4, 10, 22, 6], and an efficient simulation of constant-depth Frege systems with counting gates by constant-depth Frege systems with counting axioms would give the desired size lower bounds for the system with counting gates. We show that a polynomial simulation is not possible. The long-term goals for propositional proof complexity include proving lower bounds for Frege systems, textbook systems based on modus ponens, and abstract proof systems, arbitrary algorithmic methods for verifying tautologies [13]. Unfortunately, even Frege systems yet seem resistant to efforts for proving size lower bounds, so we narrow our attention to Frege proofs in which the lines of the proofs are restricted to a weak circuit class. This approach has been fruitful for constant-depth Frege systems: the method of random restrictions used in circuit complexity to prove lower bounds for bounded depth circuits with unbounded fan-in AND and OR gates [14, 15, 3] can also prove size lower bounds for proof systems which use such constant-depth formulas [19, 17]. In contrast, the approximation method, which has been successful for proving lower bounds for constant-depth circuits using modular gates [20], has not yet given any

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lower bounds for constant-depth Frege systems with modular gates [10]. Proving lower bounds for these Frege systems is hence an important and longstanding open problem in propositional proof complexity. Work on proving lower bounds for constant-depth Frege systems with modular gates has so far been successful only for (apparently) weaker subsystems: constant-depth Frege systems augmented with counting axioms [1, 2, 5, 4, 10, 22, 6], the Nullstellensatz system [4, 10, 6, 11, 8], and the polynomial calculus [12, 21, 16, 7, 9]. We address the question of whether constant-depth Frege systems with counting axioms are, in fact, weaker than constant-depth Frege systems with counting gates. We show that there are tautologies which are polynomially provable in constantdepth Frege with modular gates that are not provable with polynomial-size, constantdepth Frege with counting axioms. This rules out the possibility of proving a lower bound on Frege with modular counting gates via a polynomial simulation by Frege with counting axioms, for which we already have a lower bound. Moreover, the tautologies we prove the lower bound for have polynomial size proofs in the polynomial calculus, so we actually get the slightly stronger result of a separation between proof sizes for the polynomial calculus and constant-depth Frege systems with counting axioms. The analogous circuit result is proving a distinction between circuits with modular gates and sums of AC 0 circuits. One could do this with a combination of random restriction and degree arguments: Consider the OR of n modular sums, each on disjoint sets of n variables. Look at any sum of AC 0 functions that supposedly computes this function. A random restriction to the variables will, with high probability, reduce the above circuit to a sum of small decision trees, and hence, to a low degree polynomial over Z2 . However, the function will with high probability still have one variable unset from each block of n, and so will still have degree n, that of the OR. Thus, the circuit cannot compute the function. Our tautologies are in some sense analogous to the above function. We replace the OR by a tautology that is easy for the polynomial calculus but hard for Nullstellensatz, the linear induction tautologies: the negation of S1 , Sr → Sr+1 , 1 ≤ r ≤ M − 1, ¬SM . Linear induction tautologies are known to have constant degree, polynomial size proofs in the polynomial calculus and to require logarithmic degree to proof in Nullstellensatz, [11, 8]. Of course, constant-depth Frege systems have linear size proofs of the linear induction tautologies, so we must obfuscate the Sr ’s so that they are not definable by constant-depth formulas. We replace each Sr by the modular sum of disjoint sets of N variables, {xr,i | 1 ≤ i ≤ N }: Sr is true if and only if an even number of the xr,i ’s are satisfied. We wish to use the fact that it is difficult with AC 0 formulas to compute Sr from xr,i . A delicate point is to express the constraints Sr → Sr+1 without making Sr implicitly definable as such a formula, as the standard uses of intermediate variables would. P Instead, we think of Sr → Sr+1 as the mod 2 equation: P P x xr+1,j . One way to express this would be to x = S = S S = r,i r,i i i i+1 i j i have auxiliary variables representing a perfect matching on the satisfied monomials of P P i,j xr,i xr+1,j and j xr,j . In the polynomial calculus (and hence in constant-depth Frege systems with counting axioms) we can define each Sr from xr,i , prove the implications using the matchings, and then apply the proof for the induction principles to the Sr ’s. The overall structure of the lower bound proof is analogous to our circuit bound sketch, and very similar to that of other lower bounds for constant-depth Frege with

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counting axioms, in particular [6]. First, given an alleged polynomial-size constant depth proof, we construct a representation of the subformulas of the proof so that each formula is associated with a small depth decision tree. This is done with the iterated application of a switching lemma. Next, the representation is used to create a lowdegree Nullstellensatz refutation of the algebraic translation of our tautology. Finally, we proof a degree lower bound for the algebraic system by giving a constant degree reduction of the linear induction tautologies to our tautologies and appealing to the known degree lower bounds for Nullstellensatz refutations of linear induction [11, 8]. The technical novelty and difficulty in our paper is in the switching lemma. In the process of collapsing formulas into short decision trees, the proofs of [19, 17, 6] apply random 0/1 partial assignments to the variables which reduce the tautologies to smaller tautologies of the same family. For the induction on sums principles, no 0/1 partial assignment that does not set every underlying variable in each row can collapse a DNF into a short decision tree with high probability. For this reason, we simplify the tautologies by not only setting many variables to 0s and 1s, but also substituting one variable for another.

1.1

Outline of the Paper

The initial sections make formal definitions that are used throughout this paper. In sections 2 and 3, we define the proof systems we will reason about and introduce tautologies that we use for the separation. Section 4 shows that the principles can be refuted with small polynomial calculus proofs. The principle tasks in the paper are the proof of the switching lemma and its use to construct a shallow interpretation for the formulas in the proof. Sections 5 through 10 address these topics. Section 5 introduces a modification of the induction on sums principles that are needed for the proof of the switching lemma. Sections 6 and 7 describe the basics on restrictions and decisions. Our switching lemma does not works for 0/1 restrictions, and must substitute one variable for another. This process, which we call simplification, is introduced in section 8. The switching lemma proper, theorem 17, is proved in section 9. Finally, the switching lemma is used to construct a shallow interpretation of the formulas in the proof (called a “k-evaluation”) in section 10. Section 11 uses the k-evaluation to construct a low-degree Nullstellensatz refutation of the algebraic induction on sums principles. Section 12 uses a reduction from the linear induction principles to prove a degree lower bound for the algebraic induction on sums principles. Section 13 puts everything together to prove the lower bound, and section 14 concludes the paper with a brief discussion on the limits of our techniques and proposed future work.

2

The Proof Systems

Propositional proof systems are usually viewed as deriving tautologies by applying inference rules to a set of axioms. However, it can be useful to take the dual view that such proof systems establish that a set of hypotheses is unsatisfiable by deriving FALSE from the hypotheses and axioms. Such systems are called refutation systems. The Nullstellensatz and polynomial calculus systems demonstrate that sets of polynomials

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have no common solution, and are inherently refutation systems. Frege systems are traditionally viewed as deriving tautologies, but for ease of comparison, we treat them as refutation systems. Furthermore, we will be discussing propositional formulas and polynomials in the same set of variables. This is justified by identifying the logical constant FALSE with the field element 0 and the logical constant TRUE with the field element 1.

2.1

Constant-Depth Frege Systems

A Frege system is a sound, implicationally complete propositional proof system over a finite set of connectives with a finite number of axiom schema and inference rules. By the methods of Cook and Reckhow [13], any two Frege systems simulate one another up to a polynomial factor in size and a linear factor in depth. For concreteness, the reader can keep in mind the following W Frege system whose connectives are NOT gates, ¬, and unbounded fan-in OR gates, , and whose inferenceW rules W are: (1) Axioms AW∨ ¬A, (2) (¬A)∨C Y A W (X∪Y W ) (3) Cut A∨B B∨C (4) Merging WX∨ Weakening A∨B (X∪Y ) (5) Unmerging X∨ Y . Let H be a set of formulas. A derivation from H is a sequence of formulas f1 , . . . , fm so that for each i ∈ [m], either fi is a substitution instance of an axiom, fi is an element of H, or there exist j, k < i so that fi follows from fj and fk by the application of an inference rule to fj and fk . The formulas in H are called the hypotheses of the derivation. For a given formula F , a proof of F is a derivation from the empty set of hypotheses whose final formula is F . For fixed set of hypotheses H, a refutation of H is a derivation from H whose final formula is FALSE. The size of a derivation is the total number of symbols appearing in it. We say that a family of tautologies τn , each of size s(n), has polynomial size constant-depth Frege proofs (refutations) if there are constants c and d so that for all n, there is a proof (refutation) of τn so that each formula in the proof has depth at most d, and the proof (refutation) has size O (sc (n)).

2.2

Counting Axioms Modulo m

Constant-depth Frege with counting axioms modulo m is the extension of constantdepth Frege systems that has axioms that state for integers m, N , m ≥ 2 and N 6≡m 0, it is impossible to partition a set of N elements into pieces of size m. Definition 2.1 Let m > 1 and N 6≡m 0 be given. Let V be a set of N elements. For m each e ∈ [V ] , let there be a variable xe .

CountV m =





_  ^  ¬xe  ∨ 

v∈V

e∈[V ]m e3v

_

(xe ∧ xf )

e,f ∈[V ]m e⊥f

Frege with counting modulo m derivations are Frege derivations that allow the use ] of substitution instances of Count[N m (with N 6≡m 0) as axioms.

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2.3

Nullstellensatz Refutations

One way to prove that a system of polynomials Pk f1 , . . . , fk has no common roots is to give a list of polynomials p1 , . . . , pk so that i=1 pi fi = 1. Because we are interested in translations of propositional formulas, we add the polynomials x2 − x as hypotheses to guarantee all roots are zero-one roots. Definition 2.2 For a system of polynomials f1 , . . . , fk in variables x1 , . . . , xn over a field F , a Nullstellensatz refutation is a list of polynomials p1 , . . . , pk , r1 , . . . , rn satisfying the following equation: k X

pi fi +

i=1

n X j=1


rj x2j − xj = 1

For a polynomial q, Nullstellensatz derivation of q from f1 , . . . , fk is a list of polynomials p1 , . . . , pk , r1 , . . . , rn satisfying the following equation: k X

pi fi +

i=1

n X j=1


rj x2j − xj = q

The degree
of the refutation (derivation) is the maximum degree of the polynomials pi fi , rj x2j − xj . We define the size of a Nullstellensatz refutation (derivation) to be the number of monomials appearing in p1 , . . . , pk and f1 , . . . , fk . Hilbert’s weak Nullstellensatz guarantees that over a field, all unsatisfiable systems of polynomials have Nullstellensatz refutations [18]. We can define Nullstellensatz refutations over any ring, but such systems are no longer complete. In this paper, we work with Nullstellensatz refutations of polynomials over Zm , and for the sake of generality, we make no assumptions on m unless otherwise stated.

2.4

Polynomial Calculus

Definition 2.3 Let f1 , . . . , fk be polynomials over a field F . A polynomial calculus refutation of f1 , . . . , fk over F is a sequence of polynomials g1 , . . . , gm so that, gm = 1, and for each i ∈ [m], either gi is fl for some l ∈ [k], gi is xl 2 − xl for some l ∈ [n], gi is agj + bgl for some j, l < i, a, b ∈ F , or gi is xl gj for some j < i, l ∈ [n]. The size of a polynomial calculus refutation is the total number of monomials appearing in the polynomials of the refutation. The degree of a polynomial calculus refutation is the maximum degree of a polynomial that appears in the refutation.

3

Induction on Sums Principles

Suppose that we have M rows of N boolean variables. There is no assignment to these variables so that the modular sum of the first row is 0, the sum of the final row is 1, and for each non-final row r, if the sum of row r is zero, then the sum of row r + 1 is also zero. This is the idea behind the unsatisfiable sets of clauses we call the induction on sums principles.

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Let M andSN be positive integers. Let R1 , . . . , RM be disjoint sets of N elements, M for each i ∈ r=1 we have a boolean variable Xi . The variables {Xi | i ∈ Rr } are identified as the “the r’th row of Boolean variables”. Because constant depth circuits require exponential size to compute modular sums, it takes some care to represent the principles as an unsatisfiable set of clauses. It is helpful to view the constraints as a system of M + 1 many quadratic equations modulo m. P Equation 0: 0 Pi∈R1 Xi = P P Equation r, for 1 ≤ r < M : ( i∈Rr Xi )( j∈Rr+1 Xj ) − j∈Rr+1 Xj = 0 P Equation M : i∈RM Xi − 1 = 0

To give a set of propositional clauses which are satisfied only when these equations are satisfied, we add extension variables and constraints expressing “in each equation, the number of satisfied monomials satisfied is even.” For shorthand, we define multisets Ur corresponding to the monomials in equation r. U0 = {{i} | i ∈ R1 }, for r ∈ {1, . . . , M − 1}, Ur = {{i, j} | i ∈ Rr , j ∈ Rr+1 } ∪ (m − 1) × {{j} | j ∈ Rr+1 }, and UM = {{i} | i ∈ RM } ∪ {∅}. (The notation (m − 1) × S denotes the multi-set with m − 1 copies of each element from S.) For each equation, we add a set of “partitioning variables”, Ye for e ∈ [Ui ]m . The induction on sums principles state that for each equation, these variables define a twopartition on the satisfied monomials. Because the particular index sets play no role, except for their sizes, we refer to all such principles as the M by N induction on sums principle. Definition 3.1 Let M and N be positive integers, and let R1 , . . . , RM and U0 , . . . , Um be given as in the preceding paragraphs. The M by N induction on sums principle, IS(M, N), is the following set of clauses: For each r, 0 ≤ r ≤ M , for each I ∈ Ur , for each I ∈ Ur , e ∈ [Ur ]m , i ∈ I, m for each e, f ∈ [Ur ] , e ⊥ f ,

W

i∈I ¬Xi ¬Ye ∨ Xi

∨

¬Ye ∨ ¬Yf

W

e3I

Ye

We will also need to treat this set of a clauses as a set of polynomials. Let AIS m (M, N) denote the following set of polynomials: For each r, 0 ≤ r ≤ M , for each I ∈ Ur , for each I ∈ Ur , e ∈ [Ur ]m , i ∈ I, m for each e, f ∈ [Ur ] , e ⊥ f ,

4

Q

i∈I Xi Ye (Xi −

Y e Yf

P

1)

e3I

Ye − 1




An Upper Bound for the Polynomial Calculus

Theorem 1 The AISm (M, N ) system of polynomials has a degree 3, size O(M N 3 ) polynomial calculus refutation. P Proof: The polynomial calculus derivation iteratively i∈Rr Xi for each row r. Finally, P P i∈RM Xi − 1 and subtract it from i∈RM Xi to derive 1. But first we use the axioms

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to derive polynomials stating that the number of monomials satisfied in each equation is even. Q For each r and each I ∈ Ur , e ∈ [Ur ]m with e 3 I, let qI,e = Ye i∈I Xi − Ye . When |I| = 1, qI,e = Xi Ye − Ye belongs to AISm (M, N ), and when I = {i, j}, there is a degree 3 derivation of qI,e from AISm (M, N ): Xi (Xj Ye − Ye ) + Xi Ye − Ye = Xi Xj Ye − Ye = qI,e P Q For each r and each I ∈ Ur , let pI = e3I Ye − i∈I Xi . These polynomials have degree ≤ 3 derivations from AISm (M, N ) and the qI,e ’s:
P
P
Q Q P Q i∈I Xi e3I Ye − 1 − e3I qI,e = Pe3I Ye Q i∈I Xi − qI,e − i∈I Xi = Y − X = p e i I e3I i∈I

For each equation r, 0 ≤ r ≤ M , when we negate the sum of all monomials, P we disP cover that the sum of the satisfied monomials is even. The final identity I∈Ur e3I Ye = 0 is true because each variable Ye appears exactly twice in the sum.
P P Q P P − I∈Ur pI = PI∈Ur Q e3I i∈I PXi − Pe3I Ye = PI∈Ur Qi∈I Xi − I∈Ur e3I Ye = I∈Ur i∈I Xi P We derive U0 = {{i} | i ∈ i∈Rr Xi for each Pnow iteratively Q P Pr, 1 ≤ r ≤ M . Because P R1 }, I∈U0 i∈I Xi = i∈R1 Xi . To derive j∈Rr+1 Xj from i∈Rr Xi , for r < M , we proceed as follows:  P P
Q P i∈Rr Xi i∈I Xi − j∈Rr+1 Xj I∈Ur P P P P P = i∈Rr j∈Rr+1 Xi Xj + j∈Rr+1 Xj − i∈Rr j∈Rr+1 Xi Xj P = j∈Rr+1 Xj P Q P P Finally, notice that I∈UM i∈I Xi = i∈RM Xi + 1. We subtract i∈RM Xi from this to obtain 1.

5

Modified Induction on Sums Principles

To prove the size lower bounds for constant-depth Frege with counting axioms proofs of the induction on sums principles, we will not work directly with the induction on sums principles but with a related variant. There are two difficulties that make the induction on sums principles unwieldy to use. First, a restriction of the IS m (M, N ) principle usually does not yield an instance of some ISm (M 0 , N 0 ) principle for any M 0 and N 0 . This kind of closure simplifies matters when working with decision trees. The second, and more substantial, difficulty is encountered in the proof of the switching lemma. The idea is to restrict the partition variables, Ye , so that each e can contain at most one monomial. This is done by adding a set of “extra-points” whose size is divisible by m, and placing an m-partition on the satisfied monomials and the extra points with the restriction that each edge used contains at most one monomial. Such a partition would imply that the equation is 0 modulo m because each satisfied monomial be grouped with m − 1 of extra points, and the total number of extra points is 0 modulo m.

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Definition 5.1 Fix a positive integer M . Let R1 , . . . , RM be disjoint sets of size N . Let U0 = {{i} | i ∈ R1 }, for r ∈ {1, . . . , M − 1}, let Ur = {{i, j} | i ∈ Rr , j ∈ Rr+1 } ∪ (m − 1) × {{j} | j ∈ Rr+1 }, and let UM = {{i} | i ∈ RM } ∪ {∅}. For each r, and let Vr be a set of distinct points so that |Vr | = m|Ur |. For each r, let Er = {e ∈ [Ur ∪ Vr ]m | |e ∩ Ur | ≤ 1}. SM SM For each i ∈ r=1 Rr , there is a propositional variable Xi , and for each e ∈ r=0 Er , there is a propositional variable Ye . The M by N modified induction on sums principle, MISm (M, N), is: For each r, 0 ≤ r ≤ M , for each I ∈ Ur , for each e with I ∈ e and i ∈ I, for each p ∈ Vr , for each e ⊥ f ,

W

i∈I ¬Xi ¬Y W e ∨ Xi

∨

e3p Ye ¬Ye ∨ ¬Yf

W

e3I

Ye

To ease our constructions of decision trees and k-evaluations, we generalize the definition so that a restriction of MISm is also an instance of MISm on a different index set. These principles need a substantial amount of bookkeeping to describe their index sets. The family of index sets describing an instance of a modified induction on sums principle is called a universe. Definition S 5.2 Fix a positive integer SM M . Let R1 , . . . , RM , S1 , . . . SM be disjoint sets. Let R = M R , and let S = r=1 r r=1 Sr . Let M0 = {{i} | i ∈ R1 ∪ S1 }, for r ∈ {1, . . . , M − 1}, let Mr = {{i, j} | i ∈ Rr ∪ Sr , j ∈ Rr+1 ∪ Sr+1 } ∪ (m − 1) × {{j} | j ∈ Rr+1 ∪ Sr+1 }, and let MM = {{i} | i ∈ RM ∪ SM } ∪ {∅}. For each r, let Ur be a subset of Mr so that (I ∈ Mr \ Ur ) → I ⊆ S (i.e., the monomials not in Ur have been satisfied). For each r, and let Vr be a set of distinct points so that |Vr | ≡m |Mr \ Ur |. The tuple (R1 , . . . , RM , S1 , . . . , SM , U0 , . . . , UM , V0 , . . . , VM ) is called a universe. Definition 5.3 Let U = (R1 , . . . , RM , S1 , . . . , SM , U0 , . . . , UM , V0 , . . . , VM ) be a uniSM verse. For each i ∈ R, let there be a propositional variable X i , and for each e ∈ r=0 Er , let there be a propositional variable Ye . The modified induction on sums principle for U, MISm (U), is the following set of clauses: For each r, 0 ≤ r ≤ M , for each I ∈ Ur , for each e, I ∈ e ∩ Ur , i ∈ I ∩ R, for each p ∈ Vr , for each e ⊥ f ,

W

i∈I∩R ¬Xi ¬Y W e ∨ Xi

e3p Ye ¬Ye ∨ ¬Yf

∨

W

e3I

Ye

Let AMISm (U) denote the following set of polynomials For each r, 0 ≤ r ≤ M , For each I ∈ Ur , for each e, I ∈ e ∩ Ur , i ∈ I ∩ R, For p ∈ Vr , For each e ⊥ f ,


P Xi e3I Ye − 1 Y Pe (1 − Xi ) e3p Ye − 1 Y e Yf Q

8

i∈I∩R

The MISm (M, N ) principles are an instance of the MISm (U) principles with an appropriate choice of universe. Definition 5.4 Let U = (R1 , . . . , RM , S1 , . . . , SM , U0 , . . . , UM , V0 , . . . , VM ) be a universe. We say that U is an (M, N ) universe if for each r, 1 ≤ r ≤ M , Sr = ∅, U0 = {{i} | i ∈ R1 }, |V0 | = m|U0 |, for r ∈ {1, . . . , M − 1}, let Ur = {{i, j} | i ∈ Rr , j ∈ Rr+1 } ∪ (m − 1) × {{j} | j ∈ Rr+1 }, |Vr | = m|Ur |, UM = {{i} | i ∈ RM } ∪ {∅}, and |VM | = m|UM |. Because the sets S1 , . . . , SM are empty, we will often write (M, N ) universes as (R1 , . . . , RM , U0 , . . . , UM , V0 , . . . , VM ). Definition 5.5 Let M and N be positive integers. AISm (M, N) denotes any system of polynomials AISm (U) where U is an (M, N ) universe. Definition 5.6 Let U = (R1 , . . . , RM , S1 , . . . , SM , U0 , . . . , UM , V0 , . . . , VM ) be a universe. The length of U, l(U), is M and the width of U, w(U), is defined as follows: w(U) = min ({|Rr | | 1 ≤ r ≤ M } ∪ {bVr /mc | 0 ≤ r ≤ M })

5.1

Reducing ISm to MISm

We will prove size lower bounds for the MISm (M, N ) principles, but this will suffice to prove size lower bounds for ISm (M, N ) as well. Theorem 2 There exists substitution Σ so that for each clause H ∈ ISm (M, (m + 1)N + m(m − 1)), either Σ(H) = 1, Σ(H) = X ∨ ¬X, or Σ(H) ∈ MISm (M, N ). Proof: Let Q1 , . . . , QM be index sets of size 2N and let T0 , . . . , Tm be the sets of monomials for ISm (M, 2N ). partition each Rr into two sets, Rr and Sr with |Rr | = N and |Sr | = mN +m(m−1). 1 1 be an m-partition of = {{i, j} ∈ Tr | j ∈ Sr+1 }. Let Pr,i For each i ∈ Rr , let Hr,i 1 0 2 2 Hr,i . For each j ∈ Rr+1 , let Hr,j = {{i, j} ∈ Tr | i ∈ Sr }, and let Pr,j be an m-partition S 2 . For r, 0 ≤ r ≤ M , let U = {I ∈ T | I ⊆ R }, and let Vr = {I ∈ Tr | of HS r r r,j r r I ⊆ r Sr }. Let U be the universe (R1 , . . . , RM , ∅, . . . , ∅, U0 , . . . , UM , V0 , . . . , VM ); this is not an (M, N ) universe, but it is close and we will deal with it at the end. Let Σ be the following substitution:  Ye if e ∈ [Ur ∪ Vr ]m and |e ∩ Ur | ≤ 1     X if e ∈ P 1 1 if i ∈ Sr i r,i Σ(Xi ) = Σ(Ye ) = 2 Xi if e ∈ {I, J} ∈ Pr,i Xi if i ∈ Rr    0 otherwise W W Consider an axiom of the form i∈I ¬Xi ∨ e3I Ye , with I ∈ Tr . Consider the case when I ⊆ S. In this case, I ∈ Vr . For each i ∈ I, Σ(Xi ) = 1 and for each e 3 I,WΣ(Ye ) = YeWif e ∈ [Ur ∪WVr ]m and |e ∩ Ur | ≤ 1, and Σ(Ye ) = 0 otherwise. Therefore, Σ( i∈I ¬Xi ∨ e3I Ye ) = e3I Ye . Consider the case when I ⊆ R. In this case, I ∈ Ur . For each i ∈ I, Σ(Xi ) = Xi m and for each e 3 I, Σ(YeW ) = Ye if e W∈ [Ur ∪ Vr ]W with e ∩W Ur = {I}, and otherwise Σ(Ye ) = 0. Therefore, Σ( i∈I ¬Xi ∨ e3I Ye ) = i∈I ¬Xi ∨ e3I Ye . If I = {i, j} with i ∈ Rr and j ∈ Sr+1 , then Σ(Xi ) = Xi and there is exactly one e ∈ [Tr ]m so that Σ(Ye ) = Xi , for all other e, Σ(Ye ) = 0. Therefore, Therefore,

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W W Σ( i∈I ¬Xi ∨ e3I Ye ) = ¬Xi ∨ Xi . The case for I = {i, j} with i ∈ Sr and j ∈ Sr+1 is handled similarly. m For each axiom ¬Ye ∨ ¬Yf with e, f ∈ [Ur ] , e ⊥ f , it is easily checked by a similar case analysis that Σ(¬Ye ∨ ¬Yf ) = 1 or Σ(¬Ye ∨ ¬Yf ) ∈ MISm (U). The only issue remaining is that U is not an (M, N ) universe. Notice that |U0 | = N , |V0 | = mN + m(m − 1), |UM | = N + (m − 1), |VM | = mN + m(m − 1), and for 1 ≤ r ≤ M − 1, |Ur | = (m − 1)N + N 2 and |Vr | = (m − 1)mN + (mN + m(m − 1))2 . We group together some of the extraneous points of Vr by setting some Ye ’s to 1 and thus obtaining Vr0 ⊆ Vr with |Vr0 | = m|Ur |. Corollary 3 If there is a size s, depth d refutation of ISm (M, (m + 1)N + m(m − 1)) then there is a size s, depth d refutation of MIS(M, N ).

6

Restrictions and Partial Assignments

Definition 6.1 Let U be a universe. The following pairs of literals in Vars(U) are inconsistent: ¬Xi and Ye , when i ∈ I and I ∈ e, Xi and ¬Xi , Ye and Yf , when e ⊥ f , and Ye and ¬Ye . For a set of literals S, if there exist l1 , l2 ∈ S so that l1 and l2 are inconsistent, then S is said to be inconsistent. Definition 6.2 Let U be a universe. Let S be a consistent set of literals from Vars(U). For a formula F , the restriction of F by S, F S , is defined in the usual way, replacing a literal l by 1 is l ∈ S and replacing l by 0 if some l 0 ∈ S is inconsistent with l. Compound expressions are simplified when explicitly satisfied or falsified. Here is an easy lemma we give without proof. Lemma 4 Let U be a universe, and let S, T, V be subsets of Vars(U). 1. (S T ) V = S (T ∪V ) 2. If S and T are consistent and S is closed then S ∪ T = S ∪ (T S ). Definition 6.3 Let U be a universe. Let B be a set of literals in Vars(U). We say that B is closed if for every Ye ∈ B, we have that for every I ∈ e ∩ Ur , i ∈ I, Xi ∈ B. The motivation for the more complicated formulation of MIS m with the universes is that we want the restriction of an MISm principle to be another instance of an MISm principle. In actuality, we only need this hold for restrictions that do not negate an edge variable. Definition 6.4 Let U = (R1 , . . . , RM , S1 , . . . , SM , U0 , . . . , UM , V0 , . . . , VM ) be a uniSM SM verse. Let i ∈ r=1 Rr be given, and let e ∈ r=0 , so that if there is I ∈ e ∩ Ur for some r, then I ⊆ Sr ∪ Sr+1 . 1. The restriction of U by Xi , U Xi , is defined as follows: for 1 ≤ r ≤ M , i 6∈ Rr , let Rr0 = Rr , and Sr0 = Sr , for the r so that i ∈ Rr , let Rr0 = Rr \ {i} and Sr0 = Sr ∪ {i}. U Xi = (R0 1 , . . . , R0 M , S 0 1 , . . . , S 0 M , U0 , . . . , UM , V0 , . . . , VM )

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2. The restriction of U by ¬Xi , U ¬Xi , is defined as follows: for 1 ≤ r ≤ M , i 6∈ Rr , let Rr0 = Rr , and Sr0 = Sr , for the r so that i ∈ Rr , let Rr0 = Rr \ {i} and Sr = Sr . For r, 0 ≤ r ≤ M , let Ur0 = Ur \ {I ∈ Ur | i ∈ I}. U ¬Xi = (R0 1 , . . . , R0 M , S1 , . . . , SM , U 0 0 , . . . , U 0 M , V0 , . . . , VM ) 3. The restriction of U by Ye , U Ye , is defined as follows: For r, 0 ≤ r ≤ M , so that e 6∈ Er , let Ur0 = Ur and Vr0 = Vr , for the r so that e ∈ Er , let Vr0 = Vr \ e and let Ur0 = Ur \ e. U Ye = (R1 , . . . , RM , S1 , . . . , SM , U 0 0 , . . . , U 0 M , V 0 0 , . . . , V 0 M ) Let π be a consistent, closed set of literals containing only positive instances of Y e variables. The restriction of U by π, U π is defined by iteratively restricting by Xi ∈ π, ¬Xi ∈ π, and finally by Ye ∈ π. Lemma 5 Let U be a universe, and let π be a consistent, closed subset of literals containing only positive instances of Ye variables. MISm (U) π = MISm (U π ). Lemma 6 Let U be a universe, and let π be a consistent, closed subset of literals containing only positive instances of Ye variables. l(U π ) = l(U) and w(U π ) ≥ w(U) − |π|.

7

Decision Trees

A decision tree is a way of computing functions in which at every internal node of the tree, a variable (or small set of variables) and the answers to the queries determine a path to a leaf which is labeled with the value of the function on all assignments extending that sequence of answers. At each internal node of a decision tree, a query is made either of an underlying variable Xi , asking whether the variable is true, or to a monomial in Ur , asking “Is this monomial satisfied, and if so, what partition edge covers this monomial? Or, is the monomial not satisfied?”, or to an extra point in Vr , asking “What partition edge covers the extra point?” Edges are labeled with answers to these queries. The branches leaving a node correspond to all of the answers consistent with the previous answers on the branch leading up to that node. Definition 7.1 Let U = (R1 , . . . , RM , S1 , . . . , SM , V0 , . . . , VM ) be a universe. The SM queries of U, Q(U), are value queries, Value(I), for each I ∈ r=0 Vr , extra point matching queries, Match(p), for each r ∈ {0, . . . M } and p ∈ Vr , monomial matching queries, Match(I), for each r ∈ {0, . . . M } and I ∈ Ur . Associated with each query is a set of answers. An answer is a closed set of literals telling what happens to a point or monomial. Answers appear as labels for branches in decision trees. Definition 7.2 Let U = (R1 , . . . , RM , S1 , . . . , SM , V0 , . . . , VM ) be a universe. Let R = SM U r=1 Rr . Let Q a be query of U. The answer to Q in U, AN S (Q), is defined as follows:

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SM 1. For i ∈ r=1 Rr , AN S U (Value({i})) = {{Xi }, {¬Xi }} 2. For I ∈ Ur , 0 ≤ r ≤ M , AN S U (Match(I)) = {{Xii | i ∈ I ∩ R} |  ∈ {−1, 1}I∩R, ∃j ∈ I ∩ R, j = −1} ∪{{Ye } ∪ {Xk | k ∈ I ∩ R} | e ∈ Er , I ∈ e} 3. For p ∈ Vr , 0 ≤ r ≤ M , AN S U (Match(p)) = {{Ye } ∪ {Xk | k ∈ I ∩ R, I ∈ e ∩ Ur } | e ∈ Er , p ∈ e}

Definition 7.3 Let U = (R1 , . . . , RM , S1 , . . . , SM , V0 , . . . , VM ) be a universe. BLits(U) S is the set containing exactly the following literals: Xi , for each i ∈ M r=1 Rr , ¬Xi , for SM each i ∈ r=1 Rr , and Ye , for each 0 ≤ r ≤ M, e ∈ Er . Lemma 7 Let U be a universe and let Q be a query. If π is a consistent, closed subset of BLits(U) so that |π| < w(U) then there exists L ∈ AN S U (Q) so that L is consistent with π.

Lemma 8 Let U be a universe and Q be a query. If π is consistent, closed subset of BLits(U) then AN S U π (Q) = {A π | A ∈ AN S U (Q), A π 6= 0}. Definition 7.4 Let U be a universe. Fix a set of values, V . A decision tree (over U) is a rooted tree T that satisfies the following recursive definition: T can be a single node, labeled with a value from V . If T has height > 0, then the root is labeled with a query Q ∈ Q(U). For each L ∈ AN S U (Q), there is an arc labeled L, underneath which is a decision tree over U L . The height of a decision tree is its height as a tree. Definition 7.5 Let U be a universe and let k be a positive integer. T (U, k) is the set of decision trees in U of height ≤ k. Definition 7.6 Let T be a decision tree. For each path p in T from the root to a leaf, let Bp be the partial assignment obtained by taking the union of the edge labels along p. Because the distinct answers for any query are mutually inconsistent, the mapping p 7→ Bp is injective. Definition 7.7 Let T be a decision tree. Br(T) is defined as Br(T) = {Bp | p is a path in T from the root to a leaf } For each v ∈ V , Brv (T) = {Bp | p is a path in T from the root to a leaf with label v } In particular, if T is a single node labeled with v, then Brv (T ) = {1}, and for all w ∈ V \ {v}, Brw (T ) = ∅. We will often view Br(T ) as a set of terms, with each branch corresponding to W the conjunction of the literals it contains. Using this interpretation, Brv (T ) is the disjunction of all branches that lead to a leaf labeled v, and for a collection of trees T i , W i ∈ I, i∈I Brv (Ti ) denotes the disjunction of branches that lead to a leaf labeled v in some tree Ti .

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Definition 7.8 Let F be a DNF in Vars(U). Let T be a decision tree over U. We say that T strongly represents for F if each leaf of T is labeled with a 0 or 1, for each σ ∈ Br1 (T ), there exists a term C of F so that C π ⊆ σ, and for each σ ∈ Br0 (T ), for every term C of F , C σ = 0. Lemma 9 Let T be a decision tree over U of height h. Let π be a closed partial assignment. If 2Ht(T ) + |π| < w(U), then there is a branch in T consistent with π. Proof: The proof is by induction on the height of T . The statement is trivial for trees consisting of a single node. Assume that the lemma holds for all trees of height h. Let T be a tree of height h + 1 over U and let π be a partial assignment so that 2h + 2 + |π| < w(U). Let Q be the query on the root of T , and for each L ∈ AN S U (Q), let TL be the subtree immediately underneath the root, underneath the arc labeled L. Because |π| < w(U), we may apply lemma 7 and choose L ∈ AN S U (Q) which is consistent with π. TL is a decision tree over U L , and has height at most h. Moreover, 2h+||π|| < w(U)−2 ≤ w(U L ), so by the induction hypothesis, we can find β ∈ Br(TL ) consistent with π. L ∪ β is the desired branch of T . Definition 7.9 Let U be a universe. For each v ∈ Vars(U), the query associated with v, Qv , is defined as follows: If v = Xi then Qv = Value(i). If v = Ye , then Qv = Match(p), where p is the lexicographically first element of e.

8

Simplifications

The switching lemma requires us to not only apply a random restriction, but also to make substitutions of literals for variables. The process of applying such transformations to decision trees is called simplification. We use simplifications in a matter similar to the way restrictions are used in other proofs of lower bounds for constant-depth Whereas it is clear how to define a 0/1 restriction of a decision tree in a manner that guarantees several nice properties, doing so with substitutions requires more care. Given a mapping on literals, we will extend it in the natural way to a mapping on formulas and a mapping on sets of literals. Definition 8.1 Let U and V be universes. A simplification is a mapping Σ which maps Lits(U) to Lits(V) ∪ {0, 1} and Q(U) ∪ {null} to Q(V) ∪ {null} with the following properties: 1. 2. 3. 4.

If l1 , l2 ∈ Lits(U) are inconsistent, then Σ(l1 ) and Σ(l2 ) are inconsistent. For all queries Q ∈ Q(U), if Σ(Q) 6= null, then Σ(AN S U (Q)) = AN S V (Σ(Q)) For each v ∈ Vars(U), if Σ(Qv ) 6= null, then Σ(Qv ) = QΣ(v) . For every hypothesis H ∈ MISm (U), Σρ (H) = 1, Σρ (H) = w ∨ ¬w (for some variable w), or Σρ (H) ∈ MISm (V).

Property (1) allows us to strengthen property (2). Because distinct answers to a query are inconsistent, at most one answer to Q can simplify to an answer of Σ(Q). Lemma 10 Let Σ be a simplification from U to V, and let Q be a query of U so that Σ(Q) 6= null. For each B ∈ AN S V (Σρ (Q)), there is exactly one A ∈ AN S U (Q) so that Σρ (A) = B.

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Lemma 11 Let Σ be a simplification from U to V, and let A be a closed, consistent set of literals from BLits(U). Σ is also a simplification from U A to V Σ(A) . Lemma 12 Let Σ1 be a simplification from U to V, and let Σ2 be a simplification from V to W. Then Σ2 ◦ Σ1 is a simplification from U to W. Definition 8.2 Let Σ be a simplification from U to V. Let T be a decision tree over U. The simplification of T by Σ, Σ(T), is defined recursively as follows: If T is a single node, then Σ(T ) = T . If T has multiple nodes, then Σ(T ) is the tree constructed as follows: Let Q be the label on the root of T , and for each A ∈ AN S U (Q), let TA denote the subtree of T immediately underneath the root, underneath the edge labeled A If there exists A ∈ AN S U (Q) so that Σ(A) = 1 then Σ(T ) = Σ(TA ) Otherwise, the root of Σ(T ) is labeled with Σ(Q) For each A ∈ Σ(AN S U (Q)) place an arc labeled Σ(A) underneath the root and underneath this arc, place a copy of Σ(TA ) Lemma 13 Let U and V be universes, and let Σ be a simplification from U to V. Let T be a decision tree over U. If T is a decision tree of height h over U then Σ(T ) is a decision tree of height ≤ h over V. Proof: We prove this by induction on h. If h = 0, then T is a single node labeled with a value, and Σ(T ) is that same node. This is a decision tree over V. Assume that the lemma holds for all trees of height ≤ h. Let T be a decision tree of height h + 1 over U. Let Q be the query on the root of T . There are two cases: there exists A ∈ AN S U (Q) so that Σ(A) = 1, and when for all A ∈ AN S U (Q), Σ(A) 6= 1. In the case when there exists A ∈ AN S U (Q) so that Σ(A) = 1. Let A be the unique such element of AN S U (Q). By definition, Σ(T ) = Σ(TA ). By definition, TA is a decision tree over U A and by lemma 11, Σ is a simplification from U A to V Σ(A) = V. Therefore, by the induction hypothesis, Σ(TA ) is a decision tree over V. In the case when for all A ∈ AN S U (Q), Σ(A) = 0, Σ(Q) 6= null. By definition, the label on the root of Σ(T ) is Σ(Q), and for each A ∈ AN S Q (U) so that Σ(A) 6= 0, there is an arc labeled Σ(A) leading to Σ(TA ). For each such A, because Σ is a simplification from U A to V Σ(A) , by the induction, Σ(TA ) is a decision tree over V Σ(A) . Because Σ(AN S U (Q)) = AN S V (Σ(Q)), we have that Σ(T ) is a decision tree over V. Lemma 14 Let Σ be a simplification from U to V. Let T be a decision tree over U with leaf labels from some set V , Σ(Brv (T )) = Brv (Σ(T )). Proof: The proof is by induction on the height of T . If T is a single node, then the lemma trivially holds. Suppose that the lemma holds for all trees of height ≤ h. Let T be a decision tree of height h + 1 over U. Let Q be the query on the root of T . By the induction hypothesis and lemma 11, for each A ∈ AN S Q , Σ(Brv (TA )) = Brv (Σ(TA )).

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There are two cases: there exists A ∈ AN S U (Q) so that Σ(A) = 1, and for all A ∈ AN S U (Q), Σ(A) 6= 1. Consider the case when there exists A ∈ AN S U (Q), so that Σ(A) = 1. In this case, Σ(T ) = Σ(TA ). Because distinct answers to a query are inconsistent, for every A0 ∈ AN S U (Q) π , if A0 6= A, then Σ(A) = 0. Therefore, Σ(Brv (T )) = Σ(Brv (TA )) and thus Brv (Σ(T )) = Brv (Σ(TA )) = Σ(Brv (TA )) = Σ(Brv (T )). Now consider the case when for all A ∈ AN S U (Q), Σ(A) 6= 1. Σ(Brv (T )) = {Σ(A ∪ B) | A ∈ AN S U (Q), Σ(A) 6= 0, B ∈ Brv (TA )} = {Σ(A) ∪ B 0 | A ∈ AN S U (Q), Σ(A) 6= 0, B 0 ∈ Σ(Brv (TA ))} = {Σ(A) ∪ B 0 | A ∈ AN S U (Q), Σ(A) = 6 0, B 0 ∈ Brv (Σ(TA ))} = Brv (Σ(T ))

Corollary 15 Let Σ be a simplification from U to V. Let T be a decision tree in U with leaf labels from {0, 1}. Let T c denote the tree obtained by inverting the leaf labels of T . 1. If Br(T ) = Br0 (T ) then Br(Σ(T )) = Br0 (Σ(T )). 2. If Br(T ) = Br1 (T ) then Br(Σ(T )) = Br1 (Σ(T )). 3. Σ(T c ) = (Σ(T ))c W W 4. Σ( Br1 (T )) = Br1 (Σ(T ))

Lemma 16 Let T be a decision tree which strongly represents a DNF F = strongly represents Σ(F ).

W

Ci , Σ(T )

Proof: It is easily checked that if Σ(F ) is constant, then Σ(T ) is the appropriate constant tree. By lemma 14, for any leaf label value v, and for each σ ∈ Brv (Σ(T )), there exists τ ∈ Brv (T ) so that Σ(τ ) is non-constant and Σ(τ ) = σ. Let σ ∈ Br1 (T ) be given. Choose τ ∈ Br1 (T ) so that Σ(τ ) = σ and choose a term C of F so that C ⊆ τ . We have Σ(C) ⊆ Σ(τ ) = σ as needed. Let σ ∈ Br0 (T ) be given. Choose τ ∈ Br0 (T ) so that Σ(τ ) = σ. Let C be a term of F . Because T strongly represents F , C τ = 0. Because Σ is a simplification, Σ(C) Σ(τ ) = 0, and therefore Σ(C) σ = 0.

9

The Switching Lemma

Theorem 17 Let r, c be positive constants. There exist constants  > 0 and h so that for all M , and all N sufficiently large, for every (M, N ) universe U and every collection R of at most N c many r-DNFs, there exists an (M, N  ) universe U 0 and simplification Σ : U → U 0 so that for every F ∈ R, Σ(F ) is strongly represented by a decision tree of height at most h.

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The standard tool for proving lower bounds for constant-depth circuits and constantdepth Frege systems is to apply random restrictions to the formulas which collapse them into short decision trees. This approach does not suffice to prove size lower bounds for refutations the MISm principles. We require a process which we call “random simplification” that first applies a random restriction and then substitutes some literals for other variables. In subsection 9.1 we show why this approach does not suffice for our tautologies. The random simplifications are described in subsection 9.2. The crucial result of this section is that the simplification process is indeed a simplification as described in section 8, lemma 20. Subsection 9.3 contains the proof of our switching lemma, in the guise of theorem 21. We combine theorem 20 and theorem 21 to prove theorem 17.

9.1

The Inadequacy of Restrictions Alone

Restrictions alone do not seem sufficient to simplify formulas while preserving the ISm and MISm . This is because for a monomial Xi Xj with i ∈ Rr , j ∈ Rr+1 , it is possible that Xi is set to 1 and Xj is left unset. Because the monomial may in the future be set to 0 or 1, depending on how Xj is set, we can neither match it to an extra point nor say that it is unmatched. If the random restriction sets a reasonable numbers one and leaves a reasonable number of variables unset in each row, this event happens with reasonably large probability. W Fix i ∈ Rr and consider a disjunction j∈Rr+1 Ye , for a fixed i ∈ Rr . In the event e3{i,j}

that Xi is left unset, none of the matching variables in the disjunction are set to 1. The height of the decision tree needed to represent this DNF is proportional The decision tree to represent the restriction of this DNF requires height proportional to the number of unset variables in row Rr+1 (a matching query must be made for each {i, j}) and this is too large for our purposes.

9.2

Presimplifications

What we do is apply a random restriction and then substitute one variable for another to avoid the difficulties described in the preceding section. In particular, for monomials {i, j} with ρ(Xi ) = 1 and ρ(Xj ) = ∗, we reserve a particular p ∈ Vr and substitute Xj for the variable M{i,j},p and 0 for every other M{i,j},q . Notice that this reduces the disjunction of subsection 9.1 to the single variable Xj which is strongly represented by a decision tree of height 1. Definition 9.1 Let U be an (M, N ) universe, and let ρ be a partial assignment to Vars(U). Let r ∈ {1, . . . , M − 1}, i ∈ Rr , j ∈ Rr+1 be given. We say that the monomial {i, j} is a half-star of ρ if ρ(Xi ) = 1 and ρ(Xj ) = ∗, or, ρ(Xi ) = ∗ and ρ(Xj ) = 1. For each half-star I = {i, j}, some e 3 I is reserved to cover I should XI become 1. The monomial can be covered by no other edge. Of course, if the monomial XI later becomes a 0, then something must be done with the extra points in the reserved edge. Notice that the monomial {i, j} becomes 0 only when Xj is set to 0. In this case, all the other half-stars {i0 , j} become 0 and the extra points of the edges for these half-stars become free as well. We simply place a predetermined partition on all of these

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extra-points and use it when Xj is set to 0: for f belonging to this partition, ¬Xj is substituted for Yf . Definition 9.2 Let U be an (M, N ) universe. Let L ≤ N be given so that K = N −L is 2 an integer divisible by m. An L-presimplification for U is a consistent, closed partial assignment ρ satisfying the following properties: 1. In each row, ρ sets K ones: ∀r ∈ {1, . . . , M }, |{i ∈ Rr | Xi ∈ ρ}| = K. 2. In each row, ρ sets K zeroes: ∀r ∈ {1, . . . , M }, |{i ∈ Rr | ¬Xi ∈ ρ}| = K. 3. In equations 1 through M − 1, 2mKN − mK 2 + m(m − 1)K many extra points are covered: ∀r ∈ {1, . . . , M − 1}, |{p ∈ Vr | ∃e ∈ [Vr ]m , p ∈ e, Ye ∈ ρ}| = 2mKN − mK 2 + m(m − 1)K. 4. In equations 0 and M , (m + 1)K extra points are covered: ∀r ∈ {0, M }, |{p ∈ V r | 2 ∃e ∈ [Vr ] , p ∈ e, Ye ∈ ρ}| = K. 5. In equations 0 through M − 1, every satisfied monomial is matched. ∀r ∈ {0, . . . , M − 1}, ∀I ∈ Ur , (∀i ∈ I, Xi ∈ ρ) ⇒ ∃e 3 I, Ye ∈ ρ 6. In equation M , all satisfied monomials but m − 1 are matched. |{I ∈ UM | ∃e 3 I, Ye ∈ ρ}| = K 7. Potential matches are reserved for the half-stars: For each r ∈ {1, . . . , M − 1}, i ∈ Rr , j ∈ Rr+1 so that ρ(Xi ) = 1 and ρ(Xj ) = ∗, i,j there is ei,j ρ ∈ Er with {i, j} ∈ eρ so that • ∀f, {i, j} ∈ f, f 6= ei,j ρ , ρ(Yf ) = 0 • ρ(Yei,j )=∗ ρ For each r ∈ {1, . . . , M − 1}, j ∈ Rr+1 so that ρ(Xj ) = ∗, let Sρj = Let Eρj be a fixed m-partition on the points of Sρj .

S

i∈Rr ρ(Xi )=1

ei,j ρ .

• If e ∈ Eρj then ρ(Ye ) = ∗. • ∀e, if e ∩ Sρj 6= ∅ and e 6∈ Eρj then ρ(Ye ) = 0. For each r ∈ {1, . . . , M − 1}, i ∈ Rr , j ∈ Rr+1 so that ρ(Xi ) = ∗ and ρ(Xj ) = 1, there is fρi,j ∈ Vr so that • ∀e, {i, j} ∈ e, e 6= fρi,j , rho(Ye ) = 0. • ρ(Yfρi,j ) = ∗ For each r ∈ {1, . . . , M − 1}, i ∈ Rr so that ρ(Xi ) = ∗, let Tρi = Fρi be a fixed m-partition on the points of Tρi .

S

j∈Rr+1 ρ(Xj )=1

fρi,j . Let

• If f ∈ Fρi , then ρ(Yf ) = ∗. • ∀f , if f ∩ Tρi 6= ∅ and f 6∈ Fρi then ρ(Yf ) = 0. 8. If l is a literal inconsistent with a variable v and v ∈ ρ then ¬l ∈ ρ. 9. No other literals belong to ρ. Lemma 18 Let U be an (M, N ) universe, and let π be a consistent, closed set of literals with |π| < N . There exists a (N − |π|)-simplification ρ so that π ⊆ ρ.

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Definition 9.3 Let U be an (M, N ) universe, and let ρ be an L-presimplification for U. An e-extension of ρ is an (L − 2e)-presimplification κ so that ρ ⊆ κ. We call this an e-extension rather than a 2e extension because it adds e ones in each row and e zeroes in each row. Definition 9.4 Let M and N be positive integers. Let U = (R1 , . . . , RM , U0 , . . . , UM , V0 , . . . , VM ) be an (M, N ) universe. For each r, 1 ≤ r ≤ M , we define Σρ (Rr ) := {i ∈ Rr | ρ(Xi ) = ∗}. For each r, 0 ≤ r ≤ M , we define Σρ (Ur ) Σρ (Vr ) Σρ (U)

:= {I ∈ Ur | ∀i ∈ I, ρ(Xi ) = ∗} i,j := {p ∈ Vr | ∀i ∈ Rr , ∀j ∈ Rr+1 , p 6∈ ei,j ρ , p 6∈ fρ , ∀e 3 p, ρ(Ye ) 6= 1} := (Σρ (R1 ), . . . , Σρ (RM ), Σρ (U0 ), . . . , Σρ (UM ), Σρ (V0 ), . . . , Σρ (VM ))

Lemma 19 Let U = (R1 , . . . , RM , U0 , . . . , UM , V0 , . . . , VM ) be an (M, N ) universe and let ρ be an L-presimplification for U. Σρ (U) is an (M, L) universe. Definition 9.5 Let U be an (M, N ) universe, and let ρ be a presimplification for U. For each query Q ∈ Q(U), we define the simplification of Q by ρ, Σρ (Q), as follows: If Q = Value(i) then if Σρ (Xi ) = Xi then Σρ (Q) = Value(i) otherwise Σρ (Q) = null if Q = Match({i}) then if Σρ (Xi ) = Xi then Σρ (Q) = Match(i) otherwise Σρ (Q) = null If Q = Match({i, j}), then if Σρ (Xi ) = Xi and Σρ (Xj ) = Xj then Σρ (Q) = Match({i, j}) if Σρ (Xi ) = 1 and Σρ (Xj ) = Xj then Σρ (Q) = Value(j) otherwise, Σρ (Q) = null If Q = Match(p), then if p ∈ ei,j ρ then Σρ (Q) = Value(j) if p ∈ fρi,j then Σρ (Q) = Value(i) if p ∈ Σρ (Vr ) for some r, then Σρ (Q) = Match(p) otherwise Σρ (Q) = null Lemma 20 Let U be an (M, N ) universe and let ρ be a presimplification for U. Σ ρ is a simplification from U to Σρ (U). Proof: We show that the three criteria of the definition of simplification are met: 1. If l1 , l2 ∈ Vars(U) are inconsistent, then either (i) Σρ (l1 ) = 0 or Σρ (l2 ) = 0, or (ii) Σρ (l1 ) and Σρ (l2 ) are inconsistent. This is non-trivial only when Σρ (l1 ) 6= l1 or Σρ (l2 ) 6= l2 . Without loss of generality, there are two cases to consider, l1 = MI,p , with Σρ (MI,p ) = Xi , and l1 = Ye , with Σρ (l1 ) = ¬Xi . Consider the case when l1 = Ye and Σρ (Ye ) = Xi with i ∈ I ∈ e. This happens when e is the edge reserved for the monomial {i, j} with ρ(Xi ) = ∗ and ρ(Xj ) = 1. There are three possibilities for what l2 can be: ¬Xi , ¬Xj , and Yf with f ⊥ e. If

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l2 = ¬Xi , then Σρ (l2 ) = ¬Xi . If l2 = ¬Xj then Σρ (l2 ) = 0. If l2 = Yf with f ⊥ e, then either Σρ (l2 ) = 0 or Σρ (l2 ) = ¬Xi . Consider the case when l1 = Ye and Σρ (Ye ) = ¬Xi . This happens when e ∈ Eρi or e ∈ Fρi , and e 3 {i, j} with ρ(Xj ) = 1. The possibilities for what l2 can be are ¬Xi , ¬Xj , and Yf with f ⊥ e. If l2 = ¬Xi then Σρ (l2 ) = ¬Xi . If l2 = ¬Xj then Σρ (l2 ) = 0. If l2 = Yf with e ⊥ f then either Σρ (Yf ) = Xi or Σρ (Yf ) = 0. 2. For all queries Q ∈ Q(U), if Σρ (Q) 6= null, then Σ(AN S U (Q)) = AN S Σρ (U ) (Σρ (Q)). Consider a query of the form Q = Match({i, j}). The other queries are handled with similar case analyses and we omit them to save space. Recall that AN S U (Match({i, j})) = {{¬Xi , ¬Xj }, {¬Xi , Xj }, {Xi , ¬Xj }} ∪{{Ye , Xi , Xj } | e 3 I} Consider the case when Σρ (Xi ) = Xi and Σρ (Xj ) = Xj . In this case Σρ (Q) = Match({i, j}). Clearly, for each e ∈ [Σρ (Ur )∪Σρ (Vr )]m , with {i, j} ∈ e, Σρ ({Ye , Xi , Xj }) = {Ye , Xi , Xj }, and for each e 6∈ [Σρ (Ur ) ∪ Σρ (Vr )]m , {i, j} ∈ e, Σρ (Ye , Xi , Xj }) = 0. Also, for A = {¬Xi , ¬Xj }, {¬Xi , Xj }, or {Xi , ¬Xj }, we trivially have that Σρ (A) = A. Therefore, Σ(AN S U (Q)) = AN S Σρ (U ) (Σρ (Q)). Consider the case when Σρ (Xi ) = 1 and Σρ (Xj ) = ∗. In this, case Σρ ({¬Xi , ¬Xj }) = 0, Σρ ({¬Xi , Xj }) = 0, and Σρ ({Xi , ¬Xj }) = {¬Xj }. For each e 3 I, Σρ (Ye ) = 0 if e is not the edge reserved for {i, j} and Σρ (Ye ) = Xj if it is. Therefore, Σρ (AN S U (Q)) = {{Xj }, {¬Xj }} = AN S Σρ (U ) (Value(j)) 3. For each v ∈ Vars(U), if Σρ (Qv ) 6= null, then Σρ (Qv ) = QΣρ (v) . This is trivial whenever Σρ (v) = v. The only other cases are when Σρ (Ye ) = Xi and Σρ (Yf ) = ¬Xi . In the former case, Σρ (Match(I)) = Value(i) = QXi and in the latter case Σρ (Match(p)) = Value(i) = QXi . 4. For every hypothesis H ∈ MISm (U), Σρ (H) = 1, Σρ (H) = w ∨ ¬w (for some variable w), or Σρ (H) ∈ MISm (Σρ (U)). The proof is a case analysis for each type of W clause in MIS m (U). To save space, we give the proof only for axioms of the form e3p Ye , for some p ∈ Vr , 0 ≤ r ≤ M . The other axioms are handled similarly. In the case that there exists e 3 Iso that Σρ (Ye ) = 1, Σρ (A) = 1. In the case when p is belongs to an edge e reserved for some half-star, {i, j} with Σρ (Xj ) = Xj and Σρ (Xi ) = 1, we have that Σρ (Ye ) = Xj . Also, there is exactly one W edge f 3 p with Σρ (Yf ) = ¬Xj ; for all other g 3 p, Σρ (Yg ) = 0. Therefore, Σρ ( e3p = Xj ∨ ¬Xj . The remaining case is when p ∈ Σρ (Vr ). In this case, for e ∈ [Σρ (Ur ) ∪ Σρ (Vr )]m , |e ∪ Σρ (Ur )| ≤ 1, Σρ (Ye ) = Ye . For e ∈ [Σρ (Ur ) ∪ Σρ (Vr )]m with e ∩ (Vr \ Σρ (Vr )) 6= ∅, we have that Σρ (Ye ) = 0. Similarly,We containsWa half-star or a satisfied monomial, then Σρ (Ye ) = 0. Therefore, Σρ ( e3p Ye ) = e3p Ye .

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9.3

An Independent Set Style Switching Lemma

Definition 9.6 Let U be an (M, N ) universe, and let ρ a presimplification for U. The minimum height of a decision tree strongly representing Σρ (F ) is denoted as hρ (F). Theorem 21 Let r, c be positive constants. There exist constants  > 0 and h so that for all M , and all N sufficiently large, for every (M, N ) universe U, for a randomly chosen N  -presimplification for U, ρ, Prρ [hρ (F ) ≥ h] ≤ N −c Lemma 22 Let U be an (M, N ) universe. Let L ≤ N be given, and let K = 12 (N − L). Let F be an r-DNF in Vars(U), and let ρ be an L-presimplification for U. If ρ is selected uniformly among L-presimplifications, then  2 Cr s 4r sL Prρ [hρ (F ) ≥ 4m2 r4 s2 )] ≤ K Proof:(of theorem 21 from lemma 22) Simply set  = 1/2cr and s = 4c. For N large enough so that N − N  ≥ (4/5)N and (10r s )s ≤ N c , s  2  2 1/2 s  2 Cr s 4r s(N )Cr 8r sN 4r sL ≤ ≤ Prρ [hρ (F ) ≥ 4m2 r4 s2 )] ≤ 1  K N −N  2 (N −N )  2 s s (10r2 s) 10r s −c ≤ ≤ ≤ N N 2c N 1/2 The proof of theorem 22 is the direct combination of theorems 26 and 31, which are proved in subsubsection 9.3.2 and 9.3.3, respectively. fi = Xi , ¬X gi = ¬Xi , Definition 9.7 We define the skeleton of a literal as follows: X fe = {Xi | i ∈ I}. The skeleton of a for e ∈ [Vr ]m , f Ye = Ye , and for e 3 I, I ∈ Ur , Y set of literals is defined as follows: [ ˜l e= S l∈S

Definition 9.8 Let U be an (M, N ) universe, and let ρ be a presimplification for U. Let B be a consistent, closed set of literals from BLits(Σρ (U)). We define Bρ , the pullback of B along ρ as follows: B ρ = {l | Σρ (l) = 1 or Σρ (l) ∈ B} Definition 9.9 Let U = (R1 , . . . , RM , S1 , . . . , SM , U0 , . . . , UM , V0 , . . . , VM ) be a universe. Let B be a set of literals in Vars(U). We say that B matches its ones if for every r ∈ {0, . . . M }, and every I ∈ Ur , if {Xi | i ∈ I} ⊆ B then there exists Ye ∈ B with I ∈ e. Definition 9.10 Let U be an (M, N ) universe. Let F be a DNF in the literals BLits(U). Let ρ be a presimplification for U. Let T1 , . . . , Ts be terms from F . We say that T1 , . . . , Ts are ρ-consistent if there exists κ, so that ρ ∪ T1 ∪ . . . ∪ Ts ⊆ κ. Let B be a consistent, closed partial assignment in the variables BLits(Σρ (U)) that matches its ones. We say that T1 , . . . , Ts are B-independent if for all 1 ≤ i < j ≤ s, Tei ∩ T˜j ⊆ B ρ .

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We say that that T1 , . . . , Ts form a B-independent-set for F with respect to ρ if T1 , . . . , Ts are B-independent, T1 , . . . , Ts are ρ-consistent and Σρ (F ) B is non-constant. We will say that T1 , . . . , Ts form an independent set for F with respect to ρ is there exists a B so that T1 , . . . , Ts is a B-independent-set for F with respect to ρ. We say that that T1 , . . . , Ts is a maximal B-independent-set for F with respect to ρ if for every term T 0 in F that is not one of T1 , . . . , Ts , the set T, T1 , . . . , Ts is not a B-independent-set for F with respect to ρ. Definition 9.11 Let U be an (M, N ) universe, and let ρ be an L-presimplification for U. Let F be a DNF in BLits(U). We say that ρ is s-bad for F if there is an independent set for F with respect to ρ of size s. Let κ be an e-extension of ρ. We say that κ is an s-encoding for ρ with respect to F if there exists an independent set for F with respect to ρ, T1 , . . . , Ts , so that κ satisfies T1 ∪ . . . ∪ Ts . Lemma 23 If ρ is not s-bad for F , then for any consistent, closed set of literals B that matches its ones so that Σρ (F ) B is non-constant, there exists a maximal Bindependent set of size at most s. Proof: A single, non-constant, term is trivially B-independent, and by repeatedly adding terms, we can construct a maximal B-independent set for F with respect to ρ.

9.3.1

Needed Facts About Simplifications and Restrictions

Lemma 24 If F B ρ is constant then Σρ (F ) B is constant and F B ρ = Σρ (F ) B . Proof: Suppose that F B ρ = 1. For each l ∈ F , l ∈ B ρ so Σρ (l) = 1 or Σρ (l) ∈ B. Therefore, Σρ (F ) B = 1. Suppose that F B ρ = 0. Choose l1 ∈ F and l2 ∈ B ρ that are inconsistent. By definition, Σρ (l2 ) = 1 or Σρ (l2 ) ∈ B. Because Σρ maps inconsistent literals to inconsistent literals, Σρ (F ) B = 0. Lemma 25 Let U be a universe, let ρ be a restriction, and let V = Σρ (U). Let B be a consistent, closed partial assignment to Vars(V) and let T1 , T2 be terms in Vars(U) so ^ f1 ∩ T f2 6⊆ B ρ , then Σ^ that Σρ (T1 ) 6= 0 and Σρ (T2 ) 6= 0. If T ρ (T1 ) ∩ Σρ (T2 ) 6⊆ B.

f1 ∩ T f2 \ B ρ . By definition of Σρ , Σρ (l) is not a matching literal Proof: Choose l ∈ T ρ ^ MI,p , so Σρ (l) ∈ Σ^ ρ (T1 ) ∩ Σρ (T2 ). Moreover, because l 6∈ B , Σρ (l) 6∈ B and Σρ (l) 6= 1. Finally, Σρ (l) 6= 0 because Σρ (T1 ) 6= 0 and B is closed.

9.3.2

With High Probability, a Random Presimplification is Not Bad

Theorem 26 Let M , N , L, r and s be positive integers, so that 2rs < L ≤ N . Set K = 12 (N − L). For any (M, N ) universe, U, for any r-DNF F in the literals BLits(U), with ρ a uniformly drawn L-presimplification for U,  2 Cr s 4r sL Prρ [ρ is s-bad for F ] ≤ K

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Theorem 26 is proved through a sequence of three lemmas. It shown that it is very unlikely that any adversary can recover much of an L-presimplification from a randomly chosen extension, lemma 27, then it is shown that if a presimplification is bad for F then an extension probably encodes the independent set, lemma 28, and an encoding for a bad presimplification can be exploited to guess a substantial part of the extension, lemma 29. Therefore, the probability of the original extension being bad is small. We first prove the three lemmas and then we combine them to prove theorem 26. Lemma 27 Let U be an (M, N ) universe, and let L and e be integers so that e ≤ L ≤ N . Set K = N −L 2 . If ρ is a uniformly drawn L-presimplification for U. and κ is a uniformly drawn e-extension of ρ, then for any adversary A that returns a set of s literals:  e s Prρ,κ [A(κ) ⊆ κ ˜ \ ρ] ≤ K Proof: Notice that the distribution on pairs (ρ, κ) so that ρ is a random L-presimplification and κ is a random e-extension of ρ is the same distribution on pairs (ρ, κ) where κ is a randomly chosen (L − e)-presimplification and ρ is a randomly selected subpresimplification of κ. This is simply because both choose uniformly among such pairs, which in turn holds because the number of e-extensions of an L-presimplification depends only on L and the number of L-presimplifications which are extended by a given (L − e)-presimplification depends only on L and e. Fix an (L−e)-presimplification κ and let S = A(κ). Without loss of generality, there are no literals of the form MI,p in S because the adversary knows that such literals do not belong to κ ˜ . We now show that a randomly selected ρ ⊆ κ is disjoint from S with less than the stated probability. By the principle of deferred decisions, this proves the lemma. For each r, 1 ≤ r ≤ M , let ar = |{i ∈ Rr | Xi ∈ S}|, br = |{i ∈ Rr | ¬Xi ∈ S}|, and for r, 0 ≤ r ≤ M , let cr = |{e ∈ [Vr ]m | Ye ∈ S}|. In row r, we choose K ones. The probability that positions set are not specified by S is
  ar   K+e−ar e e ar (K + e − ar )!K!e! K ≤ ≤
= K+e (K + e)!K!(e − ar )! K + e − ar + 1 K K

Similarly, we must choose K zeroes in row r, and the probability that this is done
br without meeting S is at most Ke . For the extra points in equation 0, an assignment that does meet S chooses K/2 pairs from (K + e)/2 − c0 , and for the extra points in equation M , it chooses K/2 pairs from (K + e)/2 − cM . The probabilities of these c c events are at most (e/K) 0 and (e/K) M , respectively. In equation r, 1 ≤ r ≤ M − 1, 1 2 we choose 2 (2KN + K + K) pairs e ∈ [Vr ]2 from ρκ that are disjoint from S. Let P denote the number of vertices of Vr covered by an edge Ye ∈ ρ with e ∈ [Vr ]m , and let P 0 denote the number of vertices of Vr covered by an edge Ye ∈ ρ with e ∈ [Vr ]m . Note that P = 2mKN −mK 2 +m(m−1)K and P 0 = 2m(K +e)N −m(K +e)2 +m(m−1)(K +e), and therefore P 0 − P = 2meN − 2meK − me2 + m(m − 1)e. The probability of this happening is:

 c c P +(P 0 −P )−cr P 0 −cr 2meN − 2meK − me2 + m(m − 1)e r P0 − P r P P = = ≤

P0 P +(P 0 −P ) P 2mKN − mK 2 + m(m − 1)K P P

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 cr   e cr 2meN − meK + m(m − 1)e = ≤ 2mKN − mK 2 + m(m − 1)K K Therefore, the probability that a randomly selected presimplification will have its skeleton disjoint from S is at most (e/K)a+b+c = (e/K)s . 

Lemma 28 There exists a positive constant C so that for any positive integers M and N , any (M, N ) universe U, all e and L so that 2rs ≤ e ≤ L ≤ N , and all Lpresimplifications for U, ρ, and when κ is a uniformly selected e-extension of ρ: Prκ [κ is an s-encoding for ρ wrt F | ρ is s-bad for F ] ≥

1 LCrs

Proof: Let ρ be an L-presimplification for U so that there is a family of terms from F , T1 , . . . , Ts , that form an s independent set for F with respect to ρ. Notice that κ satisfies T1 ∪. . .∪Ts if and only if κ satisfies Σρ (T1 ∪. . .∪Ts ). Let κ0 = Σρ (T1 ∪. . . ∪Ts ). For each r, 1 ≤ r ≤ M , let ar = |{i | i ∈ Rr , Xi ∈ κ0 }|. For each r, 1 ≤ r ≤ M , let br = |{i | i ∈ Rr , ¬Xi ∈ κ0 }|. For each r, 0 ≤ r ≤ M , let cr = |{p | ∃e ∈ [Vr ]m , Ye ∈ κ0 }|. For each r, 0 ≤ r ≤ M , let dr = |{I | ∃Ye κ0 , I ∈ e}|.

L r Probability of getting the ones correct, in row r: L−a e−ar / e = (L − ar )!e!(L − (conditioned e)!/(e − ar )!(L − e)!L! ≥ 1/Lar . Probability of getting
zeroes correct
L−ethe br r . ≥ 1/(L − e) / upon getting the ones correct), in row r: L−e−b e e−br To bound the probability of getting the partition edges that are contained within Vr correct, we note that there are mL2 + m(m − 1)L many points in Vr , for 1 ≤ r ≤ M − 1,
2 many (mLand mL + m for equations 0 and M , respectively), and mL +m(m−1)L m edges. Therefore, the chance of the extension including the cr specified edges is clearly 1 ≥ LO(mc . r) Similarly, the probability of getting the edges containing the satisfied monomials is clearly ≥ 1/LO(mdr ) . PM PM PM PM Let a = r=1 ar , b = r=1 br , c = r=0 cr and d = r=0 dr . Therefore, the probability of κ satisfying κ0 is at least: M M M Y 1 Y 1 Y 1 ≥ LO(a+b+c+d) ar br O(cr +dr ) L L L r=1 r=1 r=0

Because a+b+c+d ≤ 3rs, for L sufficiently large, this probability is at least 1/L Crs for some constant C. Definition 9.12 Let U be a universe, let F be an r-DNF in BLits(U) and let s be an integer. Let AF,s (κ) be the following algorithm: Let κ1 = κ For i = 1 to s : Find the lexicographically first term of F which is satisfied by κi , call it Ti . If there is no such term, abort with output “error”. Randomly choose li ∈ Tei κi+1 ← κi−1 \ {l} if l = Xj then κi+1 ← κi+1 \ {Ye | j ∈ I, I ∈ e} Output {l1 , . . . , ls }

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Lemma 29 Let U be an (M, N ) universe, let F be an r-DNF in BLits(U), and let s, e and L be integers with e ≤ L ≤ N . Let ρ be an L-presimplification for U. and let κ be an e-extension of ρ.  s 1 Pr [AF,s (κ) ⊆ κ ˜ \ ρ | κ is an s-encoding of ρ] ≥ 2r Proof: Fix ρ and κ so that F contains an independent set of size s with respect to ρ, and κ satisfies every term of this independent set. Let B ⊆ BLits(V) be the core of the independent set. For each t, 0 ≤ t < s, let Et denote the event that for all i, 1 ≤ i ≤ t, li ∈ κ e \ B ρ . We will show that for each t < s, P r [lt+1 ∈ κ e \ B ρ | Et ] ≥ 1/2r. Let t < s be given and assume that event Et holds, that is, for all i, 1 ≤ i ≤ t, li ∈ κ e \ B ρ . Notice that each {li } ∪ {Ye | li ∈ f Ye } can meet at most one term of T 0 ⊆ B ρ and the independent set because for terms T , T 0 of the independent set, Te ∩ f ρ li 6∈ B . Therefore, there is some term of the independent set which is satisfied by κt+1 . Let Tt+1 be the term found at the t + 1’th iteration of AF,s (κ), with Tt+1 κt+1 = 1. Because Σρ (F ) B is non-constant, F B ρ is nonconstant, therefore there is a literal of Tt+1 not set by B ρ . However, Tt+1 κt = 1 and thus there exists l ∈ Tt+1 \ B ρ . Because ρ B ρ matches its ones, there exists l ∈ Tg t+1 \ B . Because Tt+1 is a term of size at most ρ g r, Tg t+1 has size at most 2r. Therefore, lt+1 ∈ Tt+1 \ B with probability at least 1/2r.

Now we can show that a random L-presimplification is unlikely to be s-bad for F . Proof: (of theorem 26) Let M, N, r, s, L be given with L > 2rs. Let ρ be a randomly selected L-presimplification and let κ be a randomly selected 2rs-extension of ρ. Lemmas 28 and 29 tell us that Prρ,κ [κ is an s-encoding of F wrt ρ | ρ is s-bad for F ] ≥ Prρ,κ [AF,s (κ) ⊆ κ ˜ \ ρ | κ is an s-encoding of F wrt ρ] ≥ Combining these two inequalities gives us Prρ,κ [AF,s (κ) ⊆ κ ˜ \ ρ | ρ is s-bad for F ] ≥

1 LCrs
1 s 2r


1 s 1 2r LCrs

By lemma 27, we have that
2rs s ≥ Prρ,κ [DECODE(κ) ⊆ κ ˜ \ ρ] K ≥ Prρ,κ
[DECODE(κ) ⊆ κ ˜ \ ρ | ρ is s-bad for F ] Prρ [ρ is s-bad for F ] s 1
s 1 ≥ Pr [ρ is s-bad for F ] ρ Cr 2r L

Therefore,

9.3.3



4r2 sLCr K

s

≥ Prρ [ρ is s-bad for F ]

Building Decision Trees Using Maximal Independent Sets

The construction works by making a small set of queries for each variable that appears in the independent set. Maximality guarantees that each term outside of the independent set is either falsified or shortened by the answers to these queries. First, we define the queries associated with each literal.

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Definition 9.13 The query for Xi is Value(i). The query for ¬Xi is Value(i). The queries for Ye , are Match(p), for each p ∈ e ∩ Vr , and Match(I), for each I ∈ e ∩ Ur . Lemma 30 Let V be a universe, let B be a consistent, closed subset of BLits(U) that matches its ones and let l1 , l2 ∈ BLits(V) be inconsistent literals, both of which are consistent with B. There is a query Q of l1 , so that for every answer A ∈ AN S V (Q), either A falsifies l2 or A satisfies some l 0 ∈ le2 \ B. The proof of lemma 30 is a straightforward case analysis on l1 and l2 .

Theorem 31 Let U be an (M, N ) universe, let F be an r-DNF in the literals BLits(U), and let ρ be a presimplification for U that is not s-bad for F . Then Σρ (F ) has decision tree of height at most 4m2 r4 s2 . Moreover, this decision tree strongly represents F . Proof: For a presimplification ρ which is not s-bad for F , and a closed partial assignment to Vars(U) which matches its ones, B, we build a decision tree strongly representing Σρ (F ) B recursively as follows. The decision tree for Σρ (F ), TΣρ (F ),B , is constructed as follows: If for some term C of F , Σρ (C) B = 1, then TΣρ (F ),B is a leaf labeled with 1. If for all terms C of F , Σρ (C) B = 0, then TΣρ (F ),B is a leaf labeled with 0. Otherwise Choose C1 , . . . , Ct to be a maximal B-independent set for F with respect to ρ St Let T0 be the decision tree which, makes the queries for i=1 Σρ (Ci ) B , and underneath each branch, makes a matching query for every monomial satisfied by that branch. TΣρ (F ),B is the tree obtained by taking a copy of T0 , and underneath each branch B 0 ∈ Br(T0 ), placing a copy of TΣρ (F ),B∪B 0 This construction can proceed for at most 2r iterations. We will show that for each term C of F that does not belong to the maximal independent set, either Σρ (C) ^ is falsified or the size of Σ ρ (C) decreases by one after the query phase. Because B ^ matches its ones, whenever Σ ρ (C) B is constant, Σρ (C) B is constant. Therefore, because terms have skeletons of size at most 2r, there are at most 2r iterations. Let C be a term of F so that Σρ (C) B is non-constant and C does not belong to the maximal B-independent set. In the case that there is a term Ci in the maximal ^ ^ e∩C fi 6⊆ B ρ . By lemma 25, Σ independent set so that C ρ (C) ∩ Σρ (Ci ) 6⊆ B. Choose ^ ^ ^ l∈Σ ρ (C) ∩ Σρ (Ci ) \ B. When the l is queried, Σρ (C) is falsified or the size of Σρ (C) decreases. Consider the case when there is no extension of ρ that satisfies C 1 ∪. . .∪Ct ∪C. Because each term has size at most r, and r(s + 1) 6k. If A = i∈[K] ¬ e3i Fe ∨ W e⊥f ¬ (¬Fe ∨ ¬Ff ) is an instance of a counting axiom in R that T-evaluates to 0, then for each i ∈ [K] and π ∈ Br1 (TW Fi ) there is exactly one e 3 i so that there exists σ ∈ Br1 (TFe ) with π ⊇ σ. m

Proof: For each e ∈ [K] , let Te = TFe , and for each i ∈ [K], let Ti = TWe3i Fe . By W lemma 41, each e3i Fe T-evaluates to 1, and each ¬Fe ∨ ¬Ff T-evaluates to 1. m We show that if e, f ∈ [K] with e ⊥ f then there are no mutually consistent branches in Br1 (Te ) and Br1 (Tf ). Suppose that σ1 ∈ Br1 (Te ) and σ2 ∈ Br1 (Tf ) were consistent. Because 2Ht(T
¬Fe ∨¬Ff ) + ||σ1 ∪ σ2 || ≤ 6k < w(W), by lemma 9, there must exist τ ∈ Br T¬Fe ∨¬F
f consistent with σ1 ∪ σ2 . Because ¬Fe ∨ ¬Ff T-evaluates to 1, τ ∈ Br1 T¬Fe ∨¬Ff . Because T¬Fe ∨¬Ff strongly represents Br1 (T¬Fe ) ∪ Br1 (T¬Ff ), τ is extends some β ∈ Br0 (TFe ) ∪ Br0 (TFf ). This is impossible because σ1 ∈ Br1 (Te ) and σ2 ∈ Br1 (Tf ) and σ1 ∪ σ2 is consistent with τ . Now we show that for each i ∈ [K] and π ∈ Br(Ti ) there is exactly one e 3 i so that there exists σ ∈ Br1 (Te ) with π ⊇ σ. Let i ∈ S [K] and π ∈ Br(Ti ) be given. Because Br(Ti ) = Br1 (Ti ), and Ti strongly represents e3i Br1 (Te ), there exists e 3 i and σ ∈ Br1 (Te ) so that π ⊇ σ. By the preceding paragraph, there cannot be another f 3 i with σ 0 ∈ Br1 (Tf ) so that π ⊇ σ 0 because that σ and σ 0 would be consistent. Therefore, for each π ∈ Br(Ti ), there exists exactly one e 3 i so that there exists σ ∈ Br1 (Te ) with π ⊇ σ. Lemma 43 Let U and W be universes, let R be a refutation of MISm (U), and let T be a k-evaluation of R in W with w(W) ≥ 6k + 2. If there exists an instance of a counting axiom in R that T-evaluates to 0, then there exists is a generic system {G i | i ∈ [K]} of height at most mk over W. Proof: Suppose thatWA is an instance
of aWcounting axiom in R so that A T-evaluates W m to 0. Say that A = i∈[K] ¬ e3i Fe ∨ e⊥f ¬ (¬Fe ∨ ¬Ff ). For each e ∈ [K] , let Te = TFe , and for each i ∈ [K], let Ti = TWe3i Fe . For each i ∈ [K], let Si be the tree obtained by relabeling each leaf l of Ti with the unique e 3 i so that there is a σ ∈ Br1 (Te ) consistent with the branch to that leaf. We now show that for i, j ∈ [K] and each β1 ∈ Br(Si ), β2 ∈ Br(Sj ), if β1 and β2 are consistent then their leaf labels are the same. Suppose that β1 ∈ Br(Si ), β2 ∈ Br(Sj ) and β1 ∪ β2 is consistent. Let e1 be the leaf label for β1 and let e2 be the leaf label for β2 and suppose for the sake of contradiction that e1 6= e2 . Apply lemma 42 and choose σ1 ∈ Br1 (Te1 ) with σ1 ⊆ β1 , σ2 ∈ Br1 (Te2 ) with σ2 ⊆ β2 . Because β1 ∪ β2 is consistent, σ1 ∪ σ2 are consistent. Choose π ∈ Br(T¬(¬Fe1 ∨¬Fe2 ) ) that is consistent with β1 ∪ β2 . Because ¬(¬Fe1 ∨ ¬Fe2 ) T-evaluates to 0, π ∈ Br1 (T¬Fe1 ∨¬Fe2 ). By definition, π satisfies some δ ∈ Br1 (T¬Fe1 ) ∪ Br1 (T¬Fe2 ). However, such a δ is inconsistent with σ1 ∪ σ2 and therefore π is inconsistent with σ1 ∪ σ2 ; contradiction.

31

For each i ∈ [K], let Gi be the tree obtained as follows: Start with the tree Si , and underneath each branch σ that leads to a leaf labeled e = {i, i1 , . . . , im−1 }, replace the leaf with a copy of Si1 σ . Underneath each branch of this tree, replace a restricted copy of i3 , and so forth. Notice that every leaf of Gi that is underneath this leaf of Si will lead to a leaf labeled e. Clearly, the Gi ’s are decision trees of height ≤ mk, and each leaf of Gi is labeled with some edge e 3 i. Finally, we have that for all e = {i1 , . . . , im }, all any ij , ik ∈ e,
S Bre (Gij ) = {σj ∪ l∈[1,m] σl σj ...σi−1 | σl ∈ Bre (Sil ), 1 ≤ l ≤ m} l6=j Sm = { l=1 S σl | σl ∈ Bre (Sil ), 1
≤ l ≤ m} = {σk ∪ l∈[1,m] σl σk ...σi−1 | σl ∈ Bre (Sil ), 1 ≤ l ≤ m} l6=k

= Bre (Gik )

Lemma 44 Let U and V be universes, let R be a refutation of MISm (U), and let T be a k-evaluation of R of V. If 8k + 2 ≤ w(V) then there exists an (M, N − 2k) universe W and a generic system of height at most mk in W. Proof: By lemma 40, we may choose an (M, N −2k) universe W with and a k-evaluation T0 of R in W so that there exists some instance of a counting axiom A in R so that A T0 -evaluates to 0. By lemma 43, this guarantees the existence of a generic system of height at most mk in W.

11.2

From Generic System to Nullstellensatz Refutation

Definition 11.2 Let S be a subset of BLits(W). The monomial of S, pS , is defined as follows. pS =

Y

(1 − Xi )

¬Xi ∈S

Y

Xi

Xi ∈S

Y

Ye

Ye ∈S

Let T be a decision tree. The polynomial of T, pT , is the sum of the monomials of its branches. pT =

X

pπ

π∈Br(T )

LemmaP45 Let Q ∈ Q(U) be given. From AMISm (W) there is a Nullstellensatz derivation of A∈AN S W (Q) pA = 1 of degree at most 3.

Proof: The proof is done by case analysis on P the query Q. When Q = Value(i), AN S W (Value(i)) = {{Xi }, {¬Xi }}. Trivially, A∈AN S W (Q) pA = Xi + 1 − Xi = 1.

When Q = Match(p), with p ∈ Vr , AN S W (p) = {{Ye }∪{Xk | k ∈ I∩R, I ∈ e∩Ur } | e ∈ Er , p ∈ e}. Notice that for each I ∈ Ur , e 3Q I, there is a P family of degree ≤ 1 I,e polynomials cI,e , q ∈ AMIS (W), so that Y − Y X = m e e i q i∈I∩R q∈AMISm (W) cq q.

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Therefore, (using the fact that for all e, |e ∩ Ur | ≤ 1 to reduce the second expression to 1), P  P P P P I,e e3p Ye − 1 e3p I∈e∩Ur A∈AN S W (Q) pA + q∈AMISm (W) cq q − 
P P Q P P Q = A∈AN S W (Q) Ye i∈I∩R Xi + e3p I∈e∩Ur Ye − Ye i∈I∩R Xi − Y − 1 e3p e =1 When Q = Match(I), with I ∈ Ur , AN S W (Match(I)) = {{Xii | i ∈ I ∩ R} |  ∈ {−1, 1}I∩R, ∃j ∈ I ∩ R, j = −1} ∪ {{Ye } ∪ {Xk | k ∈ I ∩ R} | e ∈ Er , I ∈ e} Let A0 = {{Xii | i ∈ I ∩ R} |  ∈ {−1, 1}I∩R, ∃j ∈ I ∩ R, j = −1} and A1 = {{Ye } ∪ {Xk | k ∈QI ∩ R} | e ∈ Er , I ∈ e}PIt is easily Q shown by induction on |I ∩ R| that P A∈A1 pA − i∈I∩R Xi is one of the hypotheses A∈A0 pA = 1− i∈I∩R Xi . Moreover, of AMIS P m (W). ; this immediately implies that there is a degree ≤ 3 derivation of 1 from A∈AN S W (Q) pA and AMISm (W). Q

i∈I∩R

Xi

P

e3I



Q P Q P Q Ye − 1 = e3I i∈I∩R Xi Ye − i∈I Xi = A∈A1 pA − i∈I Xi

Lemma 46 Let W be a universe, and let T be a decision tree in W. There exists a family of P polynomials, cq , for q ∈ AMISm (W), each of degree < 3Ht(T ) so that PT = 1 + q∈AMISm (W) cq q.

Proof: We prove this by induction on the height of T . The base case is when h = 0 and T has a single node. By definition, pT = 1. For the induction step, assume that the lemma works for all decision trees of height h. Let T be a decision tree of height h + 1 vertices. Choose v to be a an internal vertex in T of depth h. Let Tv denote the decision tree obtained by deleting the children of v. Let πv be the branch of T leading to v. Let Q be the query of node v. Let A0 = {A ∈ AN S(Q) | A πv = 0} and A1 = {A ∈ AN S(Q) | A πv 6= 0}. By lemma 8, AN S Wπv (Q) = A1 . ! X X p T = p Tv − p πv + p A p πv = p Tv + p πv pA − 1 A∈AN S Wπv (Q)

A∈A1

By lemma 45, we may P choose a family P of polynomials P rq , for q ∈ AMISm (W), each of degree ≤ 3 so that A∈A1 pA − 1 = A∈A0 pA + q∈AMISm (W) rq q. Therefore, 

p T = p Tv + p πv 

X

pA +

A∈A0

X

q∈AMISm (W)



rq q 

Because each A ∈ A is inconsistent with πv , there is some q ∈ AMISm (W) a degree deg(pπv pA ) − 2 monomial sA,q so that pπv pA = sA,q q. Therefore, ! ! X X sA,q + rq q p T = p Tv + A∈A 0 q∈AMISm (W)

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For each polynomial sA,q , deg(rA,q ) ≤ deg(pπv pA ) − 2 ≤ 3h + 3 − 2 = 3h + 1 < 3(h + 1). By the induction hypothesis,Pthere exists family of degree < 3h polynomials tq , q ∈ AMISm (W), so that pTv = 1 + tq q. Therefore, pT = 1 +

X

tq q +

q∈AMISm (W)

q∈AMISm (W)

=1+

X

X

q∈AMISm (W)

For each q ∈ AMISm (W), let cq = such cq is clearly of degree < 3(h + 1).

X

A∈A0

P

sA,q

!

q∈AMISm (W)

X

sA,q

A∈A0

+ rq + tq P

!

A∈A0

!

+ rq

!

q

q



sA,q + rq + tq . Each

Lemma 47 Let W be a universe, K be an odd integer. If there exists a generic system {Ti | i ∈ [K]} height ≤ h over W then AMISm (W) has a Nullstellensatz refutation of degree < 3h. Proof: For each i ∈ [K], an application of lemma 46 shows that there is a family of polynomials ciq , for q ∈ AMISm (W), each of degree less than 3h, so that pTi = P P P 1+ q∈AMISm (W) ciq q. Therefore, summing over i ∈ [K] yields i∈[K] pTi = i∈[K] (1+ P P P i i i∈[K] q∈AMISm (W) cq q) = 1 + q∈AMISm (W) cq q. Because the Ti ’s form a generic system, every Pmonomial, P appears an even number of times. PKbranch and therefore every Therefore, i=1 pTi = 0 and hence 1 + i∈[K] q∈AMISm (W) ciq q.

11.3

The Proof of Theorem 39

Proof:(of theorem 39) By lemma 44, there is an (M, N − 2k) universe W and a generic system of height at most mk in W. By lemma 47, this guarantees the existence of a degree < 3mk Nullstellensatz refutation of AMISm (W).

12 A Degree Lower Bound for Nullstellensatz Refutations of AMISm Theorem 48 Let M be an integer, and let W be a universe of length M and width ≥ 1. All Nullstellensatz refutations of AMISm (W) have degree Ω(log M ) . The proof is a reduction to the linear induction principles of length M , which are known to require degree Ω(log M ) Nullstellensatz refutations. Definition 12.1 Let M be a positive integer. The linear induction principle of length M, IND(M), is a system of polynomials in the variables t1 , . . . , tM with coefficients from Z2 . It contains exactly the polynomials t1 , tM − 1 and, for each i < M , ti ti+1 − ti+1

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Theorem 49 [11, 8] The IND(M ) system has Nullstellensatz refutations of degree O(log M ) over any field. Moreover, over any field the system requires degree Ω(log M ) Nullstellensatz refutations. The following definition is almost verbatim from [9], although we are concerned with Nullstellensatz derivations whereas they were interested in polynomial calculus derivations. Definition 12.2 Let P (~x) and Q(~y) be two sets of polynomials over a field F. We say that P is (d1 , d2 )-reducible to Q if: (1) for every yi , there is a degree d1 definition of yi in terms of ~x. That is, for every i, there exists a degree d1 polynomial ri where yi will be viewed as being defined by ri (~x); (2) there exists a degree d2 Nullstellensatz derivation of the polynomials Q(~r(~x)) from the polynomials P (~x); (3) there exists a degree d2 Nullstellensatz derivation of the polynomials ri2 (~x)−ri (~x) from the polynomials of P (~x). Lemma 50 Suppose that P (~x) is (d1 , d2 )-reducible to Q(~y). Then if there is a degree d3 Nullstellensatz refutation of Q(~y ) then there is a degree d1 d3 + d2 Nullstellensatz refutation of P (x). We omit the proof of the following lemma; the reduction is simply an application an (N − 1)-simplification. Lemma 51 Let W be an (M, N ) universe with N ≥ 1. AMISm (M, 1) is (1, 1)-reducible to AMISm (W) Lemma 52 IND(M ) is (1, 1)-reducible to AIS(M, 1). Proof: Observe that when U is an (M, 1) universe, we may rename the underlying indices so that Rr = {r}. Notice that |V0 | = m, and |Vr | = m2 for 1 ≤ r ≤ M . For each r, 1 ≤ r ≤ M , let Er ⊂ Er be an m-partition on Ur ∪ Vr , and let Fr ⊆ Er be an m-partition on Vr . Let e0 = V 0 . We will use the following definitions for the variables of ISm (M, 1). Let Ye0 := 1−t1 , for each r, 1 ≤ r ≤ M , Xr := tr , for equation r, 1 ≤ r ≤ M − 1, for e ∈ Er , Ye := tr+1 , for f ∈ Fr , Yf := 1 − tr+1 , for all other e, Ye := 0, and for equation M , for e ∈ EM , Ye := tM , for f ∈ FM , Yf := 1 − tM . Recall that AMISm (M, 1) is the following system of polynomials. For each r, 0 ≤ r ≤ M , For each I ∈ Ur , for each e, I ∈ e ∩ Ur , i ∈ I ∩ R, For p ∈ Vr , For each e ⊥ f ,

Q

i∈I∩R Xi Y e P(1 − Xi ) e3p

P

Ye − 1

Y e Yf

The polynomials for equation 0:
P X1 = t1 (0 − 1) = −t1 e3I Ye − 1 Y (1 − X ) = 0(1 − t ) = 0 1 1 Pf Y − 1 = (1 − t ) − 1 = −t1 1 e3p e

35

e3I

Ye − 1




The polynomials for equation r, with 1 ≤ r ≤ M − 1 (we omit equations that are identically 0 because a factor Ye is identically 0): P  Xr+1 = tr+1 (tr+1 − 1) = t2r+1 − tr+1 e3{r+1} Ye − 1 P 
Xr Xr+1 = tr tr+1 (tr+1 − 1) = tr t2r+1 − tr+1 e3{r,r+1} Ye − 1 Ye (1 − Xr+1 ) = tr+1 (1 − tr+1 ) = −(t2r+1 − tr+1 ) Y = tr+1 (1 − tr ) = −(tr+1 tr − tr+1 ) Pe (1 − Xr ) Y − 1 = t + 1 − t = 0 e r+1 r+1 e3p Ye Yf = tr+1 (1 − tr+1 ) = −(t2r+1 − tr+1 ) The polynomials for equation M (we omit equations that a factor Ye is identically 0):
P = e3∅ PYe − 1  = (tM − 1) Y − 1 = t (t − 1) = XM M M e3{M } e Y (1 − X ) = t = e M M (1 − tM ) P Y − 1 = t + 1 − t − 1 = e M M e3p Ye Yf = tM (1 − tM ) =

are identically 0 because tM − 1 t2M − tM −(t2M − tM ) 0 −(t2M − tM )

We now combine these elements to prove theorem 48. Proof:(of theorem 48) Let d be the minimum degree of a Nullstellensatz refutation of AMISm (W). By lemmas 51 and 50, there is a degree d + 1 Nullstellensatz refutation of AMISm (M, 1). By lemmas 52 and 50, there is a degree d + 2 refutation of IND(M ). By theorem 49, we must have that d = Ω(log M ).

13

Putting It All Together

Theorem 53 Let c, d be positive constants. For sufficiently large values of N , there is no depth d refutation of MISm (N, N ) of size ≤ N c . Proof: Let c, d be given and suppose that for every N and (N, N ) universe U there is a refutation of IS(U) of size ≤ N c . Apply theorem 32 to get constants k and  so that for sufficiently large N there is an a k-evaluation T for R over an (N, N  ) universe V Take N large enough that N  > 8k + 2, and apply theorem 39, so we have a degree 3mk Nullstellensatz refutation of AISm (W) for an (N, N  −2k) universe W. By theorem 48, AISm (W) requires degree Ω(log N ) to refute; contradiction.

14

Conclusion

Demonstrating a tautology which requires exponential size proofs in constant-depth Frege with counting axioms seems more difficult. This task should be within the capabilities of the framework presented in this paper, and we have a candidate tautology based on graph pebbling. However, the quality of our switching lemma, theorem 17, is far inferior to what is needed for an exponential bound.

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