A Lower Bound for Polynomial Calculus with Extension Rule
aa r X i v : . [ c s . CC ] O c t A Lower Bound for Polynomial Calculus with Extension Rule
Yaroslav Alekseev ∗ Abstract
In this paper we study an extension of the Polynomial Calculus proof system where we canintroduce new variables and take a square root. We prove that an instance of the subset-sumprinciple, the bit-value principle x + 2 x + . . . n − x n = 0 ( BVP n ), requires refutations ofexponential bit size over Q in this system.Part and Tzameret [18] proved an exponential lower bound on the size of Res - Lin (Resolutionover linear equations [21]) refutations of
BVP n . We show that our system p-simulates Res - Lin andthus we get an alternative exponential lower bound for the size of
Res - Lin refutations of
BVP n . In essence, the study of propositional proof complexity started with the work of Cook and Reckhow[10]. The first superpolynomial bound on the proof size was proved in a pioneering work of Tseitin[25] for regular resolution. Since then, many proof systems have been studied, some of them arelogic-style (working with disjunctions, conjunctions and other Boolean operations) and some of themare algebraic (working with arbitrary polynomials).In this work, we consider extensions of two systems, an algebraic one and a logic-style one.
Algebraic proof systems.
Lower bounds for algebraic systems started with an exponential lowerbound for the
Nullstellensatz [3] system. The main system considered in this paper is based onthe
Polynomial Calculus system [7], which is a dynamic version of
Nullstellensatz . Many exponentiallower bounds are known for the size of
Polynomial Calculus proofs for tautologies like the PigeonholePrinciple [15, 22] and Tseitin tautologies [4]. While most results concern the representation of Booleanvalues by 0 and 1, there are also exponential lower bounds over the {− , +1 } basis [24].Many extensions of Polynomial Calculus and
Nullstellensatz have been considered before. Buss etal. [5] showed that there is a tight connection between the lengths of constant-depth Frege proofswith
M OD p gates and the length of Nullstellensatz refutations using extension axioms. Impagliazzo,Mouli and Pitassi [14] showed that a depth-3 extension of
Polynomial Calculus called
ΣΠΣ - PC p-simulates CP ∗ (an inequalities-based system, Cutting Planes [6, 11] with coefficients written in unary)over Q . Also, they showed that a stronger extension of Polynomial Calculus , called
Depth - k - PC , p-simulates Cutting Planes and another inequalities-based system
Sum - of - Squares ; the simulations canbe conducted over F p m for arbitrary prime number p if m is sufficiently large.Also very strong extensions were considered: Grigoriev and Hirsch [12] considered algebraic sys-tems over formulas. Grochow and Pitassi [13] introduced the Ideal Proof System, IPS , which can beconsidered as the version of
Nullstellensatz where all polynomials are written as algebraic circuits (seealso [19, 20] for earlier versions of this system). ∗ Steklov Institute of Mathematics at St. Petersburg, St. Petersburg, Russia, and Chebyshev Laboratory at St. Pe-tersburg State University ogic-style systems.
While exponential lower bounds for low-depth proof systems (both algebraicand logical ones) are known for decades, the situation with higher depth proof systems is much worse.The present knowledge is limited to exponential bounds for constant-depth Frege systems over deMorgan basis (that is, without xor’s or equivalences) [1, 3, 5]. In particular, no truly exponentiallower bounds are known for the size of refutations of formulas in CNF in (dag-like) systems that workover disjunctions of equations or inequalities (see [17] as the first paper defining these systems andcontaining partial results).
Res - Lin (defined in [21]), working with disjunctions of linear equations, isthe second system considered in our paper, and it can be viewed as a generalization of Resolution.Part and Tzameret [18] proved an exponential lower bound for (dag-like)
Res - Lin refutations over Q for the bit-value principle BVP n . Although this is the first exponential lower bound for this system,the instance does not constitute a translation of a formula in CNF. Itsykson and Sokolov [16] consideranother extension of the resolution proof system that operates with disjunctions of linear equalitiesover F named Res ( ⊕ ) and proved an exponential lower bound on the size of tree-like Res ( ⊕ ) -proofs. We extend
Polynomial Calculus with two additional rules. One rule allows to take a square root (itwas introduced by Grigoriev and Hirsch [12] in the context of transforming refutation proofs of non-Boolean formulas into derivation proofs; our motivation to take square roots is to consider an algebraicsystem that is at least as strong as
Res - Lin even for non-Boolean formulas, see below). Another ruleis an algebraic version of Tseitin’s extension rule, which allows to introduce new variables. We willdenote our generalization of
Polynomial Calculus as Ext - PC √ .In this work we give a positive answer to the question raised in [14] asking for a technique forproving size lower bounds on Polynomial Calculus without proving any degree lower bounds. Alsowe give an answer to another question raised in [14] by proving an exponential lower bound for thesystem with an extension rule even stronger than that in ΣΠΣ - PC , which is another extension ofPolynomial Calculus presented in the aforementioned work.We consider the following subset-sum instance, called Binary Value Principle ( BVP n ) [2, 18]: x + 2 x + . . . n − x n = 0 , and prove exponential lower bound for the size of Ext - PC √ Q refutations of BVP n . Note that BinaryValue Principle does not correspond to the translation of any CNF formula and thus the questionabout proving size lower bound on the refutation of formulas in CNF without proving degree lowerbounds remains open . Theorem 1.1.
Any
Ext - PC √ Q refutation of BVP n requires size Ω( n ) . The technique we use for proving this lower bound is similar to the technique for proving condi-tional
IPS lower bound in [2]. However, since
Ext - PC proof system is weaker than Ideal Proof System ,we get an unconditional lower bound. The main idea of conditional lower bound in [2] is to provecomplexity lower bound on the free term in the end of
IPS -refutation of
BVP n over Z and then showthat IPS Z simulates IPS Q . One difference is that instead of concentrating on the complexity of com-puting the free term of the proof, we concentrate on prime numbers being mentioned in the proof(and thus appearing as factors of the free term).Then we consider Res - Lin and show that
Ext - PC √ Q simulates Res - Lin and thus get an alternativelower bound for
Res - Lin . Corollary 1.2 (Informal) . Any
Res - Lin refutation of
BVP n requires size Ω( n ) . In Section 2 we recall the definition of Polynomial Calculus ( PC ) and give the definitions of PolynomialCalculus with square root ( PC √ ) and Extended Polynomial Calculus with square root ( Ext - PC √ ).In Section 3 we prove exponential lower bound on the size of Ext - PC √ Q refutations of BVP n . Westart with considering derivations with integer coefficients ( Ext - PC √ Z ) and show that the free term inthe end of such refutation of BVP n is not just large but also is divisible by all primes less then n (seeTheorem 3.1). Then, in Theorem 3.3, we convert proofs over Q into proofs over Z without changingthe set of primes mentioned in the proof and thus get an Ext - PC √ Q lower bound.In Section 4 we show that Ext - PC √ Q simulates Res - Lin and thus we get an alternative lower boundfor the size of
Res - Lin refutations of
BVP n . In this paper we are going to work with polynomials over integers or rationals. We define the size ofa polynomial roughly as the total length of the bit representation of its coefficients:
Definition 1 (Size of a polynomial) . Let f be an arbitrary integer or rational polynomial in variables { x , . . . , x n } . • If f ∈ Z [ x , . . . , x n ] then Size ( f ) = P ⌈ log | a i |⌉ where a i are the coefficients of f . • If f ∈ Q [ x , . . . , x n ] then Size ( f ) = P ⌈ log | p i |⌉ + ⌈ log | q i |⌉ where p i ∈ Z , q i ∈ N and p i q i are thecoefficients of f . Definition 2 (Polynomial Calculus) . Let
Γ = { P , . . . , P m } ⊂ F [ x , . . . , x n ] be a set of polynomialsin variables { x , . . . , x n } over a field F such that the system of equations P = 0 , . . . , P m = 0 hasno solution. A Polynomial Calculus refutation of Γ is a sequence of polynomials R , . . . , R s where R s = 1 and for every l in { , . . . , s } , R l ∈ Γ or is obtained through one of the following derivationrules for j, k < l • R l = αR j + βR k for α, β ∈ F • R l = x i R k The size of the refutation is P sl =1 Size ( R l ) . The degree of the refutation is max l deg ( R l ) . Now we consider a variant of Polynomial Calculus proof system with additional square rootderivation rule (see [12]). Moreover, we extend our definition from fields to rings . Definition 3 (Polynomial Calculus with square root) . Let
Γ = { P , . . . , P m } ⊂ R [ x , . . . , x n ] be aset of polynomials in variables { x , . . . , x n } over a ring R such that the system of equations P =0 , . . . , P m = 0 has no solution. A PC √ R refutation of Γ is a sequence of polynomials R , . . . , R s where R s = M for some constant M ∈ R, M = 0 and for every l in { , . . . , s } , R l ∈ Γ or is obtained throughone of the following derivation rules for j, k < l • R l = αR j + βR k for α, β ∈ R R l = x i R k for some i ∈ { , . . . , n } • R l = R k for some i ∈ { , . . . , n } The size of the refutation is P sl =1 Size ( R l ) , where Size ( R l ) is the size of the polynomial R l . Thedegree of the refutation is max l deg ( R l ) . Note.
We will consider Q or Z as the ring R . For both of those rings, if we consider Boolean case,where axioms x i − x i = 0 added, our system will be complete, which means that for every unsatisfiableover { , } assignment system { f i ( ~x ) = 0 } there is a PC √ R refutation. Also, note that if R is a domainand P = 0 for some P ∈ R [ ~x ] , then P = 0 . We now define a variant of PC √ R , Ext - PC √ R where the proof system is additionally allowed tointroduce new variables y i corresponding to arbitrary polynomials in the original variables x i . Definition 4 (Extended Polynomial Calculus with square root) . Let
Γ = { P , . . . , P m } ⊂ R [ x , . . . , x n ] be a set of polynomials in variables { x , . . . , x n } over a ring R such that the system of equations P = 0 , . . . , P m = 0 has no solution. A Ext - PC √ R refutation of Γ is a PC √ R refutation of a set Γ ′ = { P , . . . , P m , y − Q ( x , . . . , x n ) , y − Q ( x , . . . , x n , y ) , . . . , y m − Q m ( x , . . . , x n , y , . . . , y m − ) } where Q i ∈ R [ ~x, y , . . . , y i − ] are arbitrary polynomials.The size of the Ext - PC √ R refutation is equal to the size of the PC √ R refutation of Γ ′ . In order to prove lower bound for the
Ext - PC √ Q proof system, we consider the following subset-suminstance [2, 18]: Definition 5 (Binary Value Principle
BVP n ) . The binary value principle over the variables x , . . . , x n , BVP n for short, is the following unsatisfiable system of linear equations: x + 2 x + . . . n − x n + 1 = 0 ,x − x = 0 , x − x = 0 , . . . , x n − x n = 0 . Theorem 3.1.
Any
Ext - PC √ Z refutation of BVP n requires size Ω(2 n ) . Moreover, the absolute valueof the constant in the end of our Ext - PC √ Z refutation consists of at least C · n bits for some constant C > . Also, the constant in the end of our Ext - PC √ Z refutation is divisible by every prime numberless than n .Proof. Assume that { R , . . . , R t } is the Ext - PC √ Z refutation of BVP n . Then we know that { R , . . . , R t } is PC √ Z refutation of some set Γ ′ = { G ( ~x ) , F ( ~x ) , . . . , F n ( ~x ) , y − Q ( ~x ) , . . . y m − Q m ( ~x, y , . . . , y m − ) } where G ( ~x ) = 1 + P i = ni =1 ( i − x i , F i ( ~x ) = x i − x i and Q i ∈ Z [ ~x, y , . . . , y i − ] .By the definition of Ext - PC √ Z refutation we know that there exists an integer constant M = 0 suchthat F t = M . Claim 3.2. M is divisible by every prime number less than n . roof of claim : Consider arbitrary integer number ≤ k < n and its binary representation b , . . . , b n .Let k + 1 be prime . Then G ( b , . . . , b n ) = k + 1 , F i ( b , . . . , b n ) = b i − b i = 0 . Also consider integers c , . . . , c m such that c i = Q i ( b , . . . , b n , c , c , . . . , c i − ) . Now we will prove by induction that everyinteger number R i ( b , . . . , b n , c , . . . , c m ) is divisible by k + 1 and thus M is divisible by every primenumber less than n . Base case: if i = 1 , then R i = G ( b , . . . , b n , c , . . . , c m ) = k +1 or R i = F i ( b , . . . , b n , c , . . . , c m ) =0 or R i ( b , . . . , b n , c , . . . , c m ) = c i − Q i ( b , . . . , b n , c , . . . , c i − ) = 0 which means that R i is divisibleby k + 1 . Induction step: suppose we know that R j is divisible by k + 1 for any j ≤ i . Now we will showit for R i +1 . There are four cases:1. If R i +1 ∈ Γ ′ , then this case is equivalent to the base case and R i +1 ( b , . . . , b n , c , . . . , c m ) isdivisible by k + 1 .2. If R i +1 = αR j + βR s for α, β ∈ Z and j, s ≤ i , then R i +1 ( b , . . . , b n , c , . . . , c m ) is divisible by k + 1 because R j ( b , . . . , b n , c , . . . , c m ) and R s ( b , . . . , b n , c , . . . , c m ) are divisible by k + 1 and α and β are integers.3. If R i +1 = x j R s or R i +1 = y j R s , then R i +1 ( b , . . . , b n , c , . . . , c m ) is divisible by k + 1 because R s ( b , . . . , b n , c , . . . , c m ) is divisible by k + 1 and b i and c i are integers.4. If R i +1 = R s , then we know that R s ( b , . . . , b n , c , . . . , c m ) is divisible by k + 1 . Suppose R i +1 ( b , . . . , b n , c , . . . , c m ) is not divisible by k + 1 . Then R i +1 ( b , . . . , b n , c , . . . , c m ) is not di-visible by k +1 since k +1 is prime . But R i +1 ( b , . . . , b n , c , . . . , c m ) = R s ( b , . . . , b n , c , . . . , c m ) which leads us to a contradiction.Since every R i ( b , . . . , b n , c , . . . , c m ) is divisible by k +1 , we know that M = R s ( b , . . . , b n , c , . . . , c m ) is divisible by every k + 1 less than n , and in particular M is divisible by every prime number lessthan n . Claim
So we know that M is divisible by the product of all prime numbers less than n . Then we knowthat | M | > ( π (2 n ))! where π (2 n ) is the number of all prime numbers less than n . By the primenumber theorem π (2 n ) > C n n . By Stirling’s approximation we get | M | > (cid:18) C n n (cid:19) ! > C ′ · (cid:18) C n n (cid:19) C nn > C ′′ (cid:16) n (cid:17) C nn > C ′′ (2 n C ) which means that M consists of at least C · n bits and therefore any Ext - PC √ Z refutation of BVP n requires size Ω(2 n ) .In order to prove a lower bound over Q , we need to convert an Ext - PC √ Q proof into Ext - PC √ Z proof. Theorem 3.3.
Any
Ext - PC √ Q refutation of BVP n requires size Ω(2 n ) .Proof. Assume that { R , . . . , R t } is the Ext - PC √ Q refutation of Γ of the size S . Then we know that { R , . . . , R t } is a PC √ Q refutation of some set Γ ′ = Γ ∪ { y − Q ( ~x ) , . . . , y m − Q m ( ~x, y , . . . , y m − ) } where Q i ∈ Q [ ~x, ~y ] . Also, we know that R t = M for some M ∈ Q .Consider integers M , . . . , M m where M i is equal to the product of denominators of all coeffi-cients of polynomial Q i . Also consider all polynomials R j ( ~x, ~y ) which was derived by using linearcombination rule which means that R j = αR i + βR k . Then we consider all constants α and β { γ , γ , . . . , γ f } ⊂ Q . Now consider the set of all denominators of the constants in { γ , γ , . . . , γ f } and denote this set as { δ , δ , . . . , δ l } ⊂ N .Also consider the products of all denominators of coefficients of polynomials { R , . . . , R t } . Wewill denote the set of those integers as { L , . . . , L t } ⊂ N .Now we will construct the Ext - PC √ Z refutation of Γ such that the constant in the end of this proof isequal to M c · M c · · · M c m m · δ c m +1 · · · δ c m + l l · L c m + l +1 · · · L c m + l + t t · M where { c , c , · · · , c m + l + t } ⊂ N ∪{ } .Firstly, we will translate polynomials Q i into some integer polynomials Q ′ i . Consider Q ′ ( ~x ) = M · Q ( ~x ) where M is equal to the product of denominators of all coefficients of polynomial Q .Then Q ′ ∈ Z [ ~x ] and T = M . Then consider Q ′ ( ~x, y ′ ) = T · Q ( ~x, y ′ T ) where T is equal to T α · M where α is an arbitrary non-negative integer such that Q ′ ∈ Z [ ~x, y ′ ] . Then for every i we consider Q ′ i ( ~x, y ′ , . . . , y ′ i − ) = T i · Q i ( ~x, y ′ T , . . . , y ′ i − T i − ) where T i = T α i · T α i · · · T α ii − i − · M i where α i , . . . , α ii − are arbitrary integers such that Q ′ i ∈ Z [ ~x, y ′ , . . . , y ′ i − ] . Note that we are not interested in the sizeof the integers α ij so they could be arbitrary large.Now we will construct PC √ Q refutation { R ′ , . . . , R ′ s } of the set Γ ′′ = Γ ∪ { y ′ − Q ′ ( ~x ) , . . . y ′ m − Q ′ m ( ~x, y ′ , . . . , y ′ m − ) } of the following form: this refutation duplicates the original refutation { R , . . . , R t } in all cases except when the polynomial R i was derived by multiplying by some variable y j from somepolynomial R k . In this case we will multiply corresponding polynomial by y ′ j and then multiply it by T j . Formally, we will prove the following claim: Claim 3.4.
There is an PC √ Q refutation { R ′ , . . . , R ′ s } of the set Γ ′′ = Γ ∪ { y ′ − Q ′ ( ~x ) , . . . y ′ m − Q ′ m ( ~x, y ′ , . . . , y ′ m − ) } for which the following properties holds: • For every polynomial R ′ i ( ~x, y ′ , . . . , y ′ m ) one of the following equations holds: R ′ i ( ~x, y · T , . . . , y m · T m ) = R j ( ~x, y , . . . , y m ) for some j or R ′ i ( ~x, y · T , . . . , y m · T m ) = T k · R j ( ~x, y , . . . , y m ) forsome k and j . • If R ′ i ( ~x, y ′ , . . . , y ′ m ) was derived from R ′ j ( ~x, y ′ , . . . , y ′ m ) and R ′ k ( ~x, y , . . . , y m ) by taking linearcombination with rational constants α and β (which means that R ′ i = αR ′ j + βR ′ k ), then α = T f and β = 0 for some f or there is some polynomial R h ( ~x, y ′ , . . . , y ′ m ) which was derived fromsome polynomials R k and R l by using linear combination with constants α and β .Proof of claim : The proof is an easy (but lengthy) inductive argument and is given in the Appendix. Claim
Now we will show that Γ ′′ has a PC √ Z refutation in which the constant in the end is equal to M c · M c · · · M c m m · δ c m +1 · · · δ c m + l l · L c m + l +1 · · · L c m + l + t t · M. In order to do this we will fix a PC √ Q refutation { R ′ , . . . , R ′ s } of Γ ′′ with the properties from theClaim 3.4 and construct a PC √ Z refutation of Γ ′′ by induction. Moreover, we will construct a PC √ Z refutation { R ′′ , . . . , R ′′ f } in which every polynomial R ′′ i is equal to M d · M d · · · M d m m · δ d m +1 · · · δ d m + l l · L d m + l +1 · · · L d m + l + t t · R ′ i for some non-negative integers d , . . . , d m + l + t and some polynomial R ′ i .Informally, we are going to multiply each line in our PC √ Q refutation by some constant in orderto get correct PC √ Z refutation. But since we can’t divide polynomials in our PC √ Z refutation byany constant, we will duplicate original PC √ Q refutation multiplied by some constant of the form6 d · M d · · · M d m m · δ d m +1 · · · δ d m + l l · L d m + l +1 · · · L d m + l + t t every time we would like to simulate derivationin the original proof. Induction statement:
Let { R ′ , . . . , R ′ i } be a PC √ Q derivation from Γ ′′ with the properties from theClaim 3.4. Then there exists a PC √ Z derivation { R ′′ , . . . , R ′′ f } from Γ ′′ such that• f ≤ i .• There is some constant F i = M b · M b · · · M b m m · δ b m +1 · · · δ b m + l l · L b m + l +1 · · · L b m + l + t t ∈ N suchthat F i · R ′ = R ′′ f − i +1 , F i · R ′ = R ′′ f − i +2 , . . . , F i · R ′ i = R ′′ f Base case: If i = 1 then R ′ i ∈ Γ ′′ . Then we can take R ′′ = R ′ i . Induction step:
Suppose we have already constructed the PC √ Z refutation { R ′′ , R ′′ , . . . , R ′′ f } for whichthe induction statement is true. Then there are four cases depending on the way the R ′ i +1 is derived. Case 1 : If R ′ i +1 ∈ Γ ′′ then F i +1 = F i and R ′′ f +1 = R ′ i +1 , R ′′ f +2 = F i +1 · R ′ , R ′′ f +3 = F i +1 · R ′ , . . . , R ′′ f + i +1 = F i +1 · R ′ i , , R ′′ f + i +2 = F i +1 · R ′ i +1 Case 2 : If R ′ i +1 = x j R ′ l or R ′ i +1 = y ′ j R ′ l then F i +1 = F i , R ′′ f +1 = F i +1 · R ′ , R ′′ f +2 = F i +1 · R ′ , . . . , R ′′ f + i = F i +1 · R ′ i and R ′′ f + i +1 = x j R ′′ f − i + l = F i +1 · R ′ i +1 or R ′′ f + i +1 = y j R ′′ f − i + l = F i +1 · R ′′ i +1 . Case 3 : If R i +1 = αR j + βR k where α = p q and β = p q where { p , q , p , q } ⊂ Z . Then we cantake F i +1 = q q F i , R ′′ f +1 = q q · R ′′ f − i +1 = F i +1 · R ′ , R ′′ f +2 = q q · R ′′ f − i +2 = F i +1 · R ′ , . . . , R ′′ f + i = q q · R ′′ f = F i +1 R ′ i and R ′′ f + i +1 = p q · R ′′ f − i + j + p q · R ′′ f − i + k = M i +1 R ′ i +1 . From the Claim 3.4 we know that α = T k for some k and β = 0 , or q and q are equal to some δ k and δ r . From the induction statement weknow that F i = M b · M b · · · M b m m · δ b m +1 · · · δ b m + l l · L b m + l +1 · · · L b m + l + t t . Then, since T k = M r k · · · M r mk m , we know that F i +1 = M b ′ · M b ′ · · · M b ′ m m · δ b ′ m +1 · · · δ b ′ m + l l · L b ′ m + l +1 · · · L b ′ m + l + t t , and the induction statement stays true. Case 4 : Suppose R ′ i +1 = R ′ j . We know that R ′ i +1 ( x , . . . , x n , y ′ , . . . , y ′ m ) = R k ( x , . . . , x n , y ′ T , . . . , y ′ m T m ) or R ′ i +1 ( x , . . . , x n , y ′ , . . . , y ′ m ) = T h · R k ( x , . . . , x n , y ′ T , . . . , y ′ m T m ) for some h . Then we can take M ′ = L k · T α · T α · · · T α m m = L k · M α ′ · M α ′ · · · M α ′ m m for some non-negative integers α , . . . , α m ,such that M ′ · R ′ i +1 is an integer polynomial. We know that such integers α , . . . , α m exist since L k is the product of all denominators of coefficients of polynomial R k .Then we can take F i +1 = M ′ · F i . It’s obvious that F i +1 · R ′ i +1 is an integer polynomial. Then wecan make the following PC √ Z derivation: R ′′ f +1 = F i ( M ′ ) · R ′′ f − i + j = ( F i M ′ ) · R ′ j ,R ′ f +2 = M ′ · R ′ f − i +1 = F i +1 · R , R ′ f +3 = M ′ · R ′ f − i +2 = F i +1 · R , . . . , R ′ f + i +1 = M ′ · R ′ f = F i +1 R i . R ′′ f + i +2 = F i M ′ · R ′ i +1 and since R ′′ f +1 = ( F i M ′ ) · R ′ j we know that ( R ′′ f + i +2 ) = R ′′ f +1 and we get a correct PC √ Z derivation.Since M ′ = L p · M α ′ · M α ′ · · · M α ′ m m we know that F i +1 = M b ′ · M b ′ · · · M b ′ m m · δ b ′ m +1 · · · δ b ′ m + l f · L b ′ m + l +1 · · · L b ′ m + l + t t , and the induction statement stays true.So now we have a Ext - PC √ Z refutation of Γ such that the constant in the end of this refutationis equal to M c · M c · · · M c m m · δ c m +1 · · · δ c m + l l · L c m + l +1 · · · L c m + l + t t · M . Suppose that M = p ′ q ′ where p ∈ Z and q ∈ N . Then, from Theorem 3.1 we know that M c · M c · · · M c m m · δ c m +1 · · · δ c m + l f · L c m + l +1 · · · L c m + l + t t · p ′ is divisible by every prime number less than n . Since M , . . . , M m , δ , . . . , δ l , L , . . . , L t are positive integers we know that M · M · · · M m · δ · · · δ l · L · · · L t · p ′ is divisible byevery prime number less than n . Also we know that log ⌈ M ⌉ + · · · + log ⌈ M m ⌉ + log ⌈ δ ⌉ + · · · + log ⌈ δ l ⌉ + log ⌈ L ⌉ + · · · + log ⌈ L t ⌉ + log ⌈ p ⌉ ≤ O ( Size ( S )) because all constants M , . . . , M m , L , . . . , L t are products of denominators in the lines of our refuta-tion { R , . . . , R t } and constants δ , . . . , δ l are denominators of rationals in linear combinations usedin our derivation.On the other hand, we know that M · M · · · M m · δ · · · δ l · L · · · L t · p ′ ≥ Ω( n ) since our product is divisible by every prime number less than n . Then we know that S ≥ Ω( n ) .8 Connection between
Res - Lin , Ext - PC √ Q and Ext - PC Q Following [21], we define
Res - Lin proof system.
Definition 6. A disjunction of linear equations is of the following general form: ( a (1)1 x + . . . + a (1) n x n = a (1)0 ) ∨ · · · ∨ ( a ( t )1 x + . . . + a ( t ) n x n = a ( t )0 ) (1) where t ≥ and the coefficients a ji are integers (for all ≤ i ≤ n , ≤ j ≤ t ). The semantics of such adisjunction is the natural one: We say that an assignment of integral values to the variables x , . . . , x n satisfies (1) if and only if there exists j ∈ { , . . . , t } so that the equation a ( j )1 x + . . . + a ( j ) n x n = a ( j )0 holds under the given assignment.The size of the disjunction of linear equations is P ni =1 P tj =1 | a ( j ) i | if all coefficients are writ-ten in unary notation. If all coefficients are written in binary notation then the size is equal to P ni =1 P tj =1 ⌈ log | a ( j ) i |⌉ . Definition 7.
Let K := { K , . . . , K m } be a collection of disjunctions of linear equations. An Res - Lin proof from K of a disjunction of linear equations D is a finite sequence π = ( D , . . . , D l ) ofdisjunctions of linear equations, such that D l = D and for every i ∈ { , . . . , l } , either D i = K j forsome j ∈ { , . . . , m } , or D i is a Boolean axiom ( x h = 0) ∨ ( x h = 1) for some h ∈ { , . . . , n } , or D i was deduced by one of the following Res - Lin inference rules, using D j , D k for some j, k < i : • Resolution : Let
A, B be two, possibly empty, disjunctions of linear equations and let L , L be two linear equations. From A ∨ L and B ∨ L derive A ∨ B ∨ ( αL + βL ) where α, β ∈ Z . • Weakening : From a (possibly empty) disjunction of linear equations A derive A ∨ L , where L is an arbitrary linear equation over { x , . . . , x n } . • Simplification : From A ∨ ( k = 0) derive A , where A is a, possibly empty, disjunction of linearequations and k = 0 is a constant. • Contraction : From A ∨ L ∨ L derive A ∨ L , where A is a, possibly empty, disjunction of linearequations and L is some linear equation.Note that we assume that the order of equations in the disjunction is not significant, while we contractidentical equations, especially.An Res - Lin refutation of a collection of disjunctions of linear equations K is a proof of the emptydisjunction from K . The size of an Res - Lin proof π is the total size of all the disjunctions of linearequations in π .If all coefficients in our Res - Lin proof π are written in the unary notation then we denote thisproof an Res - Lin U derivation. Otherwise, if all coefficients are written in the binary notation thenwe denote this proof an Res - Lin B derivation. Note.
In the original
Res - Lin proof system duplicate linear equations can be discarded from the dis-junction. Instead, we will use contraction rule explicitly. It is easy to see that both these variantsof
Res - Lin system are equivalent.
Definition 8.
Let D be a disjunction of linear equations: ( a (1)1 x + . . . + a (1) n x n = a (1)0 ) ∨ · · · ∨ ( a ( t )1 x + . . . + a ( t ) n x n = a ( t )0 ) e denote by b D its translation into the following system of polynomial equations: y · y · · · y t = 0 y = a (1)1 x + . . . + a (1) n x n − a (1)0 , y = a (2)1 x + . . . + a (2) n x n − a (2)0 , . . . , y t = a ( t )1 x + . . . + a ( t ) n x n − a ( t )0 If D is the empty disjunction, we define b D to be the single polynomial equation 1 = 0. Now we will prove that
Ext - PC √ Q p-simulates Res - Lin B and ΣΠΣ - P C Q p-simulates Res - Lin U . Theorem 4.1.
Let π = ( D , . . . , D l ) be an Res - Lin B proof sequence of D l from some collection ofinitial disjunctions of linear equations Q , . . . , Q m . Also consider L , . . . , L t — all affine forms thatwe have in all disjunctions in our Res - Lin B proof sequence.Then, there exists an PC √ Q proof of b D l from b Q ∪ . . . ∪ b Q m ∪ { y = L , y = L , . . . , y t = L t } ofsize at most O ( p ( Size ( π ))) for some polynomial p .Proof. We proceed by induction on the number of lines in π . Base case: An Res - Lin B axiom Q i is translated into c Q i and Res - Lin B Boolean axiom ( x i = 0) ∨ ( x i = 1) is translated into PC axiom x i − x i = 0 . Induction step:
Now we will simulate all
Res - Lin B derivation rules in the PC √ Q proof.• Resolution : Assume that D i = A ∨ B ∨ ( αL + βL ) where D j = A ∨ L and D k = B ∨ L .Then, we have already derived polynomial equations y j = ( a (1) j x + . . . + a (1) jn x n − a (1) j ) , . . . , y jt j = ( a ( t j ) j x + . . . + a ( t j ) jn x n − a ( t j ) j ) ,y k = ( a (1) k x + . . . + a (1) kn x n − a (1) k ) , . . . , y kt k = ( a ( t k ) k x + . . . + a ( t k ) kn x n − a ( t k ) k ) ,y j · y j · · · y jt j = 0 , y k · y k · · · y kt k = 0 where A = ( a (2) j x + . . . + a (2) jn x n = a (2) j ) ∨ · · · ∨ ( a ( t j ) j x + . . . + a ( t j ) jn x n = a ( t j ) j ) ,B = ( a (2) k x + . . . + a (2) kn x n = a (2) k ) ∨ · · · ∨ ( a ( t k ) k x + . . . + a ( t k ) kn x n = a ( t k ) k ) L = ( a (1) j x + . . . + a (1) jn x n = a (1) j ) , L = ( a (1) k x + . . . + a (1) kn x n = a (1) k ) . Then we can derive y j · y j · · · y jt j · y k · · · y kt k = 0 , y j · y j · · · y jt j · y k · · · y kt k = 0 andthus ( αy j + βy k ) · y j · · · y jt j · y k · · · y kt k = 0 . Then there is some variable y i for whichholds y i = α ( a (1) j x + . . . + a (1) jn x n − a (1) j ) + β ( a (1) k x + . . . + a (1) kn x n − a (1) k ) and we can derive y i = αy j + βy k . Then we can derive y i · y j · · · y jt j · y k · · · y kt k = 0 which is part of b D i .• Weakening : Assume that D i = D j ∨ L where L is a linear equation. Then, we have alreadyderived polynomial equations y j = ( a (1) j x + . . . + a (1) jn x n − a (1) j ) , . . . , y jt j = ( a ( t j ) j x + . . . + a ( t j ) jn x n − a ( t j ) j ) ,y j · y j · · · y jt j = 0 . We know that there is some variable y for which y = b x + . . . b n x n − b where L is a linearequation b x + . . . b n x n = b . From y j · y j · · · y jt j = 0 we can derive y · y j · y j · · · y jt j = 0 which is part of b D i . 10 Simplification : Suppose that D i = A and D j = A ∨ ( k = 0) where k ∈ Z , k = 0 . Then, wehave already derived polynomial equations y j = ( a (1) j x + . . . + a (1) jn x n − a (1) j ) , . . . , y jt j − = ( a ( t j − j x + . . . + a ( t j − jn x n − a ( t j − j ) , y jt j = k,y j · y j · · · y jt j = 0 . From equation y j · y j · · · y jt j = 0 we can derive equation y j · y j · · · y jt j − · k = 0 from whichwe can derive y j · y j · · · y jt j − = 0 which is part of b D i .• Contraction : Assume that D i = A ∨ L and D j ∨ L ∨ L where L is a linear equation. Then,we have already derived polynomial equations y j = ( a (1) j x + . . . + a (1) jn x n − a (1) j ) , . . . , y jt j − = y jt j = ( a ( t j ) j x + . . . + a ( t j ) jn x n − a ( t j ) j ) ,y j · y j · · · y jt j − · y jt j = 0 . Then we can derive y jt j − = y jt j and y j · y j · · · y jt j − · ( y jt j − ) = 0 . Using multiplication we canderive y j · y j · · · y jt j − · ( y jt j − ) = 0 from which we can derive the equation y j · y j · · · y jt j − = 0 by using the square root rule. This equation is the last part of b D i because other parts werederived earlier. Definition 9.
Let
Γ = { P , . . . , P m } ⊂ F [ x , . . . , x n ] be a set of polynomials in variables { x , . . . , x n } over a ring R such that the system of equations P = 0 , . . . , P m = 0 has no solution. A ΣΠΣ - P C Q refutation of Γ is a PC R refutation of a set Γ ′ = { P , . . . , P m , Q , . . . , Q m } where Q i are polynomialsof the form Q i = y i − ( a i + P j a ij x j ) for some constants a ij ∈ R .The size of the ΣΠΣ - P C Q refutation is equal to the size of the PC R refutation of Γ ′ . Theorem 4.2.
Let π = ( D , . . . , D l ) be an Res - Lin U proof sequence of D l , from some collection ofinitial disjunctions of linear equations Q , . . . , Q m . Then, there exists an ΣΠΣ - P C Q proof of b D l from b Q ∪ . . . ∪ b Q m of size at most O ( p ( Size ( π ))) for some polynomial p .Proof. To prove this theorem we will use the following lemma from [14]:
Lemma ([14]) . Let
Γ = { P , . . . , P a , Q , . . . , Q b , X, Y } be a set of polynomials such that P = x − ( x − , P = x − ( x − , . . . , P a = x a − ( x − a ) ,Q = y − ( y − , Q = y − ( y − , . . . , Q b = y b − ( y − b ) ,X = x · x · x · · · x a , Y = y · y · y · · · y b . Then we can derive Γ ′ from Γ in ΣΠΣ - P C Q with derivation of size poly ( ab ) where Γ ′ = { Z , Z , . . . , Z a + b , Z } and Z = z − ( x + y ) , Z = z − ( x + y + 1) , Z = z − ( x + y + 2) , . . . , Z a + b = z a + b − ( x + y + a + b ) ,Z = z · z · z · · · z a + b . Now we will prove the theorem by induction on lines in π . Base case: An Res - Lin B axiom Q i is translated into c Q i and Res - Lin B Boolean axiom ( x i = 0) ∨ ( x i =1) is translated into PC axiom x i − x i = 0 . Induction step:
Now we will simulate all
Res - Lin B derivation rules in the Ext - PC √ Q proof.11 Resolution , Weakening , Simplification rules simulation is the same as in Theorem 4.1.•
Contraction : Assume that D i = A ∨ L and D j ∨ L ∨ L where L is a linear equation. Then,we have already derived polynomial equations y j = ( a (1) j x + . . . + a (1) jn x n − a (1) j ) , . . . , y jt j − = y jt j = ( a ( t j ) j x + . . . + a ( t j ) jn x n − a ( t j ) j ) ,y j · y j · · · y jt j − · y jt j = 0 . Then we can derive y jt j − = y jt j and y j · y j · · · y jt j − · ( y jt j − ) = 0 . Using lemma we canintroduce new variables { z − M , . . . , z M } and derive z − M = y jt j − + M, , z − M +1 = y jt j − + M − , . . . , z = y jt j − , z M = y jt j − − M,z − M · z − M +1 · · · z M − · z M = 0 , where M = | a ( t j − ) j | + | a ( t j − ) j | + . . . + | a ( t j − ) jn | . Then we can substitute y jt j − k for each z k oneby one and get equation f ( y jt j − ) = 0 where f ( y jt j − ) = b · y jt j − + b · y jt j − + . . . + b M +1 · y M +1 jt j − is some polynomial from Z [ y jt j − ] and b = ( M !) · ( − M . Then we can derive the following equation by using multiplicationrule: y j · y j · · · y jt j − · f ( y jt j − ) = b · y j · y j · · · y jt j − · y jt j − ++ y j · y j · · · y jt j − · ( y jt j − ) · ( b + b · y jt j − + . . . + b M +1 · y M − jt j − ) = 0 . Now, using the equation y j · y j · · · y jt j − · ( y jt j − ) = 0 we can derive b · y j · y j · · · y jt j − · y jt j − =0 and since b = 0 we can derive y j · y j · · · y jt j − · y jt j − = 0 . This equation is the last part of b D i because other parts were derived earlier.Now we will show that our lower bound provides an interesting counterpart to a result from [18]. Theorem 4.3 ([18]) . Any
Res - Lin B refutation of x + . . . + 2 n x n = 0 is of the size Ω( n ) .Proof. From Theorem 3.3 we know that any
Ext - PC √ Q refutation of BVP n requires size Ω( n ) and thusfrom Theorem 4.1 we know that there is some polynomial p such that for any Res - Lin B refutation of BVP n of size S the equation p ( S ) ≥ C · C · n holds. Then we know that for some constant C theequation S ≥ C · n holds. Open Problems
1. Theorem 4.1 says that
Ext - PC √ Q p-simulates any Res - Lin B derivation. Is the square root rulenecessary, that is, can we p-simulate Res - Lin B refutation in the Ext - PC Q proof system?2. A major question is to prove an exponential lower bound on the size of ΣΠΣ - PC Q refutation ofa translation of a formula in CNF. 12 cknowledgement I would like to thank Edward A. Hirsch for guidance and useful discussions at various stages ofthis work. Also I wish to thank Dmitry Itsykson and Dmitry Sokolov for very helpful commentsconcerning this work.
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Appendix
Claim 3.4.
There is an PC √ Q refutation { R ′ , . . . , R ′ s } of the set Γ ′′ = Γ ∪ { y ′ − Q ′ ( ~x ) , . . . y ′ m − Q ′ m ( ~x, y ′ , . . . , y ′ m − ) } for which the following properties holds: • For every polynomial R ′ i ( ~x, y ′ , . . . , y ′ m ) one of the following equations holds: R ′ i ( ~x, y · T , . . . , y m · T m ) = R j ( ~x, y , . . . , y m ) for some j or R ′ i ( ~x, y · T , . . . , y m · T m ) = T k · R j ( ~x, y , . . . , y m ) forsome k and j . • If R ′ i ( ~x, y ′ , . . . , y ′ m ) was derived from R ′ j ( ~x, y ′ , . . . , y ′ m ) and R ′ k ( ~x, y , . . . , y m ) by taking linearcombination with rational constants α and β (which means that R ′ i = αR ′ j + βR ′ k ), then α = T f and β = 0 for some f or there is some polynomial R h ( ~x, y ′ , . . . , y ′ m ) which was derived fromsome polynomials R k and R l by using linear combination with constants α and β .Proof of claim : We will construct PC √ Q refutation { R ′ , R ′ , . . . , R ′ s } of the set Γ ′′ by induction. Induction statement:
Let { R , . . . , R i } be a PC √ Q derivation from Γ ′ . Then there exists a PC √ Q derivation { R ′ , . . . , R ′ p } from Γ ′′ such that• p ≤ i . 14 For every R j ( x , . . . , x n , y , . . . , y m ) there exists some R ′ k ( x , . . . , x n , y ′ , . . . , y ′ m ) such that R ′ k ( x , . . . , x n , T · y , . . . , T m · y m ) = R j ( x , . . . , x n , y , . . . , y m ) . • All the properties mentioned in the claim are true for our derivation { R ′ , . . . , R ′ p } . Base case: If i = 1 then R i ∈ Γ ′ . If R i ∈ Γ then we can take R ′ = R . Otherwise, if R i = y j − Q j ( ~x ) then we can take R ′ = y ′ j − Q ′ j ( ~x, y ′ , . . . , y ′ j − ) and R ′ = y ′ j − Q ′ j ( ~x,y ′ ,...,y ′ j − ) T j . Then it’s obvious that R ′ ( ~x, T · y , . . . , T m · y m ) = R ( ~x, y , . . . , y m ) . Induction step: