NP-Hard to Linearly Approximate. University of California, San Diego. Computer Science Department. August 3, Abstract

Minimum Propositional Proof Length is NP-Hard to Linearly Approximate Michael Alekhnovich Faculty of Mechanics & Mathematics Moscow State University, Russia michael@mail.dnttm.ru Shlomo Moran y Department of Computer Science Technion, Israel Institute of Technology Haifa, Israel 32000 moran@cs.technion.ac.il Sam Buss Department of Mathematics University of California, San Diego sbuss@ucsd.edu Toniann Pitassi z Computer Science Department University of Arizona Tucson, AZ 85721 toni@cs.arizona.edu August 3, 1998 Abstract We prove that the problem of determining the minimum propositional proof length is NP-hard to approximate within any constant factor. These results are very robust in that they hold for almost all natural proof systems, including: Frege systems, extended Frege systems, resolution, Horn resolution, the sequent calculus, the cut-free sequent calculus, as well as the polynomial calculus. Also, if NP is not in QP = DTIME(n logo(1) n ), then it is impossible to approximate minimum propositional proof length within a factor of 2 log(1?") n for any " > 0. Our hardness of approximation results usually apply to proof length measured either by number of symbols or by number of inferences, for tree-like or dag-like proofs. We introduce the Monotone Minimum (Circuit) Satisfying Assignment problem and prove the same hardness results for Monotone Minimum (Circuit) Satisfying Assignment. Supported in part by NSF grant DMS-9503247 and grant INT-9600919/ME-103 from NSF and MSMT (Czech Republic) y Research supported by the Bernard Elkin Chair for Computer Science and by US-Israel grant 95-00238 z Supported in part by NSF grant CCR-9457782, US-Israel BSF grant 95-00238, and grant INT-9600919/ME-103 from NSF and MSMT (Czech Republic) 1

1 Introduction This paper proves lower bounds on the hardness of nding short propositional proofs of a given tautology and on the hardness of nding short resolution refutations. When considering Frege proof systems, which are textbook-style proof systems for propositional logic, the problem can be stated precisely as the following optimization problem: Minimum Length Frege Proof: Instance: A propositional formula ' which is a tautology. Solution: A Frege proof P of '. Objective function: The number of symbols in the proof P. For a xed Frege system F, let min F (') denote the minimum number of symbols in an F-proof of '. An algorithm M is said to approximate the Minimum Length Frege Proof problem within factor, if for all tautologies ', M(') produces a Frege proof of ' of length min F ('). (Here, may be a constant or may be a function of the length of '.) We are interested only in polynomial time algorithms for solving this problem. However, there is a potential pitfall here since the shortest proof of a propositional formula could be substantially longer than the formula itself, 1 and in this situation, an algorithm with runtime bounded by a polynomial of the length of the input could not possibly produce a proof of the formula. In addition, it seems reasonable that a \feasible" algorithm which is searching for a proof of a given length ` should be allowed runtime polynomial in `, even if the formula to be proved is substantially shorter than `. Therefore we shall only discuss algorithms that are polynomial time in the length of the shortest proof (or refutation) of the input. Note that an alternative approach would be to consider a similar problem, Minimum Length Equivalent Frege Proof, an instance of which is a Frege proof of some tautology ', and the corresponding solutions are (preferably shorter) proofs of '. While our results are all stated in terms of nding a short proof to a given tautology, they hold also for that latter version where the instance is a proof rather than a formula. A yet dierent approach could be studying algorithms which output the size (i.e., number of symbols) of a short proof of the input formula, rather than the proof itself. In this case it is possible for an algorithm to have run time bounded by a polynomial of the length of the input formula, even if 1 Is is known that N P 6= con P implies that some tautologies require superpolynomially long Frege proofs. 2

the size of the shortest proof is exponential in the size of the formula. In the nal section of this paper, we show that strong non-approximability results can be obtained for algorithms with run time bounded by a polynomial of the length of the formula for a variety of proof systems. A related minimization problem concerns nding the shortest Frege proof when proof length is measured in terms of the number of steps, or lines, in the proof: Minimum Step-Length Frege Proof: Instance: A propositional formula ' which is a tautology. Solution: A Frege proof P of '. Objective function: The number of steps in the proof P. Resolution is a propositional proof system which is popular as a foundation for automated theorem provers. Since one is interested in nding resolution refutations quickly it is interesting to consider the following problem: Minimum Length Resolution Refutation Instance: An unsatisable set? of clauses. Solution: A resolution refutation R of?. Objective function: The number of inferences (steps) in R. The main results of this paper state that a variety of minimum propositional proof length problems, including the Minimum Length Frege Proof, the Minimum Step-Length Frege Proof and the Minimum Length Resolution Refutation problems, cannot be approximated to within a constant factor by any polynomial time algorithm unless P = NP. Furthermore, for these proof systems and for every constant, the Minimum Length Proof problems cannot be approximated to within a factor of (1? ) ln n unless NP DT IME(n O(log log n) ) or to within a factor of 2 log(1?") n unless NP QP, where QP, quasi-polynomial time, is dened to equal DTIME(2 (log n)o(1) ). Our results apply to all Frege systems, to all extended Frege systems, to resolution, to Horn clause resolution, to the sequent calculus, and to the cut-free sequent calculus; in addition, they apply whether proofs are measured in terms of symbols or in terms of steps (inferences), and they usually apply to either dag-like or tree-like versions of all these systems. We let F k ' mean that ' has an F-proof of k symbols. One of the rst prior results about the hardness of nding optimal length of Frege 3

proofs was the second author's result [10] that, for a particular choice of Frege system F 1 with the language ^, _, : and!, there is no polynomial time algorithm which, on input a tautology ' and a k > 0, can decide whether F k 1 ', unless P equals NP. This result however applies only to a particular Frege system, and not to general Frege systems. It also did not imply the hardness of approximating Minimum Length Frege Proofs to within a constant factor. A second related result, which follows from the results of Krajcek and Pudlak [18], is that if the RSA cryptographic protocol is secure, then there is no polynomial time algorithm for approximating the Minimum Step-Length Frege Proof problem to within a polynomial. Another closely related prior result is the connection between the (non)automatizability of Frege systems and the (non)feasibility of factoring integers that was recently discovered by Bonet-Pitassi-Raz [9]. A proof system T is said be automatizable provided there is an algorithm M and a polynomial p such that whenever T n ' holds, M(') produces some T -proof of ' in time p(n) (see [12]). Obviously the automatizability of Frege systems is closely related to the solution of the Minimum Length Frege Proof problem: if a proof system S is automatizable, then the minimum length proof problem for S can be approximated to within a polynomial factor. The two concepts are not equivalent since automatization is a property of the search problem, whereas the minimum length proof problem concerns the associated decision problem. Our theorems give a linear or quasi-linear lower bound on the automatizability of the Minimum Proof Length problem based on the assumption that P 6= NP or that NP 6 QP. It has recently been shown by Bonet-Pitassi-Raz [9] that Frege systems are not automatizable unless Integer Factorization is in P, and more recently, that bounded-depth Frege systems are also not automatizable under a similar hardness assumption [5]. These results give stronger non-approximability conclusions, but require assuming a much stronger complexity assumption. For resolution, the rst prior hardness result was Iwama-Miyano's proof in [16] that it is NP-hard to determine whether a set of clauses has a read-once refutation (which is necessarily of linear length). Subsequently, Iwama [15] proved that it is in NP-hard to nd shortest resolution refutations; unlike us, he did not obtain an approximation ratio bounded away from 1. In section 2, we introduce the Monotone Minimum Satisfying Assignment and Monotone Minimum Circuit Satisfying Assignment problems and discuss the relevant prior results about the hardness of approximating NP-optimization problems. Section 3 contains the main results about the hardness of approximating minimum length Frege proofs. Section 4 discusses 4

the main results about the hardness of approximating minimum length resolution refutations, and Section 5 shows how to apply the same proofs to obtain hardness of approximating minimum length Polynomial Calculus refutations. Section 6 briey discusses the hardness of approximating shortest sequent calculus proofs and cut-free sequent calculus proofs. The proofs in sections 3 through 6 depend critically on reducing the Minimum Length Frege Proof problem to the Minimum Set Cover and Minimal Label Cover problems, via the Monotone Minimum (Circuit) Satisfying Assignment problem. 2 Monotone Minimum Satisfying Assignment The section introduces the Monotone Minimum Satisfying Assignment problem and shows it is harder to approximate than the Minimum Set Cover problem and the Minimum Label Cover. (The latter is needed for proving hardness of approximation within a superlinear factor). The reader can nd a general introduction to and survey of the hardness of approximation and of probabilistically checkable proofs in [4] and [2]. Recall that an A-reduction, as dened by [17], is a polynomial-time Karp-reduction which preserves the non-approximating ratio to within a constant factor. Consider the following NP-optimization problems: Monotone Minimum Satisfying Assignment: Instance: A monotone formula '(x 1 ; : : :; x n ) over the basis f_; ^g. Solution: An assignment hv 1 ; : : :; v n i such that '(v 1 ; : : :; v n ) = >. Objective function: The number of v i 's which equal >. We henceforth let (') denote the value of the optimal solution for the Monotone Minimum Satisfying Assignment problem for ' i.e., the minimum number of variables v i which must be set True to force ' to have value True. We will also consider the Monotone Minimum Circuit Satisfying Assignment problem which is to nd the minimum number of variables which must be set True to force a given monotone circuit over the basis f^; _g evaluate True. It does not matter whether we consider circuits with bounded fanin or unbounded fanin since they can simulate each other. It is apparent that Monotone Minimum Circuit Satisfying Assignment is at least as hard as Monotone Minimum Satisfying Assignment. Recall the Minimum Hitting Set problem, which is: 5

Minimum Hitting Set: Instance: A nite collection S of nonempty subsets of a nite set U. Solution: A subset V of U that intersects every member of S. Objective function: The cardinality of V. It is easy to see that Monotone Minimum Satisfying Assignment is at least as hard as Minimum Hitting Set: namely Minimum Hitting Set can be reduced (via an A-reduction) to the special case of Monotone Minimum Satisfying Assignment where the propositional formula is in conjunctive normal form. Namely, given S and U, identify members of U with propositional variables and form a CNF formula which has, for each set in S, a conjunct containing exactly the members of that set. Lund and Yannakakis [19] noted that Minimum Hitting Set is equivalent to Minimum Set Cover (under A-reductions). Furthermore, it is known that the problem of approximating Minimum Set Cover to within any constant factor is not in polynomial time unless P = NP [6]. If one makes a stronger complexity assumption, then one can obtain a better non-approximability result for Minimum Set Cover; namely, Feige [14] has proved that Minimum Set Cover cannot be approximated to within a factor of (1? ) ln n unless NP DTIME(n O(log log n) ). In fact, we can get stronger results than the above reduction of Minimum Set Cover to Monotone Minimum Satisfying Assignment. There are two ways to see this: rstly, we can use a construction due to S. Arora [private communication] to reduce Monotone Minimum Satisfying Assignment to the Minimum Label Cover problem, or alternatively we can use a \self-improvement" property of the Monotone Minimum Satisfying Assignment problem to directly prove better non-approximation results. Both approaches prove that Monotone Minimum Satisfying Assignment (log n)1? cannot be approximated to within a factor of 2 unless NP QP. The advantage of the rst approach is that it gives a sharper result, namely, a reduction of Minimum Label Cover to Monotone Minimum Satisfying Assignment for 4 -formula. The second approach is more direct in that it avoids the use of Label Cover. We present both approaches next. 6

Minimum Label Cover: (see [2]) Instance: The input consists of: (i) a regular bipartite graph G = (U; V; E), (ii) an integer N in unary, and (iii) for each edge e 2 E, a partial function e : f1; : : :; Ng! f1; : : :; Ng such that 1 is in the range of e. The integers in f1; : : :; Ng are called labels. A labeling associates a nonempty set of labels with every vertex in U and V. A labeling covers an edge e = (u; v) (where u 2 U, v 2 V ) i for every label ` assigned to v, there is some label t assigned to u such that e (t) = `. Solution: A labeling which covers all edges. Objective function: The number of all labels assigned to vertices in U and V. A 4 -formula is a propositional formula which is written as an AND of OR's of AND's of OR's. Theorem 1 (S. Arora) There is an A-reduction from Minimum Label Cover to Monotone Minimum Satisfying Assignment such that the instances of Label Cover are mapped to 4 formulas. It was proved in [1] that Minimum Label Cover is not approximable within a 2 log(1?") n factor unless NP QP. An immediate corollary of Theorem 1 is that Monotone Minimum Satisfying Assignment enjoys the same hardness of approximation, even when restricted to 4 -formulas. Proof. Suppose we have an instance of Label Cover (G; N; f e g). Let m = juj and k = jv j. For 1 i m and 1 ` N, let uì be a propositional variable with the intended meaning that uì = > i ` is one of the labels assigned to vertex i 2 U. Likewise, for 1 j m and 1 ` N let v`j denote the condition that ` is one of the labels assigned to vertex j 2 V. We shall construct a formula ' involving the variables uì and v`j so that the minimal satisfying assignments for ' are precisely those truth assignments which correspond to minimal weight labelings which cover all edges. We dene ' to be k^ j=1 N_ `=1 0 @ v`j ^ ^ ij(i;j)2e _ tj e(t)=` The formula ' clearly is a monotone 4 -formula and has size polynomial in N; m; k. It is easy to verify that any minimum satisfying assignment for ' corresponds to a labeling which covers all edges and which has a minimum number of labels: to see this, one should note that any minimum u t i 1 A 7

size labeling, as well as an minimum satisfying assignment, will have exactly one label assigned to each vertex V. 2 We next present a second proof that Monotone Minimum Satisfying Assignment is hard to approximate to within a 2 log(1?") n factor; although it does not yield the hardness of the 4 -formula case, it may be independent interest since because of its use of \self-improvement" and of non-constant depth propositional formulas. Lemma 2 (Self-improvement property) For any formulas ' 1, ' 2 there is a formula ' 1 ' 2 such that (' 1 ' 2 ) = (' 1 ) (' 2 ) and j' 1 ' 2 j j' 1 j j' 2 j. Proof. Let ' 1 = ' 1 (x 1 ; : : :; x k ), ' 2 = ' 2 (x 1 ; : : :; x`). Let (' 1 ' 2 ) be a composition of ' 1 and ' 2 : (' 1 ' 2 )(x 1; : : :; 1 x`1 ; : : :; x1 k ; : : :; x`k ) = ' 1(' 2 (x 1; : : :; 1 x`1 ); : : :; ' 2(x 1 k ; : : :; x`k )) The proof follows. 2 Theorem 3 There is no polynomial time algorithm which approximates (log n)1? Monotone Minimum Satisfying Assignment within a factor of 2 for any > 0 unless NP QP. Proof. Suppose we have a formula which is an instance of Satisability. Let n = j j. By the reduction to Minimum Hitting Set, there exists a polynomial time reduction from Satisability to approximation of Monotone Minimum Satisfying Assignment within factor 2 [6], i.e., a function f : 7! '; j'j = n O(1) such that if is satisable then (f( )) < C 1 (n) and otherwise (f( )) > C 2 (n) with gap g(n) = C 2(n) C > 2. 1 (n) Applying self-improvement k-times to ', we increase the gap g to 2 k. This gives the formula F ( ) = f( ) f( ) ::: f( ) (k-times), which has length N = jf ( )j < j'j k. If is satisable than (F ( )) < C k(n) 1 otherwise (F ( )) > C k(n), the new gap is at least 2 2k. If we take k(n) = (log n) c, we get so, N = j'j k(n) n (log n)c ; and log n (log N) 1 c+1 ; g 2 k(n) 2 (log N ) Suppose we have an algorithm A which approximates Monotone Minimum Satisfying Assignment to within the factor 2 (log n)1?. Take c large enough 8 c c+1 :

so that that c=(c + 1) > 1?. The function F is computable in QP, and applying algorithm A to the formula F ( ) solves determines the satisability of. That completes the proof of Theorem 3. 2 To summarize the above material, we have the following theorem: Theorem 4 (a) If P 6= NP, then there is no polynomial time algorithm which can approximate Monotone Minimum Satisfying Assignment (and hence Monotone Minimum Circuit Satisfying Assignment) to within a constant factor. (b) If NP 6 DTIME(n O(log log n) ), then Monotone Minimum Satisfying Assignment (and Monotone Minimum Circuit Satisfying Assignment) cannot be approximated to within a factor of (1? ) ln n where n equals the number of distinct variables. (c) If NP 6 QP, then there is no polynomial time algorithm which can approximate Monotone Minimum Satisfying Assignment (or Monotone Minimum Circuit Satisfying Assignment) to within a factor of 2 log(1?") n. In the next sections we reduce the Monotone Minimum Circuit Satisfying Assignment problem rst to the Minimum Length Frege Proof problem and then to other problems on minimum proof length. This will establish the same hardness of approximation results for the Minimum Length Frege Proof problem and similarly for the other proof length optimization problems. Open question. Is it possible to improve the non-approximation factor for Monotone Minimum Satisfying Assignment or Monotone Minimum Circuit Satisfying Assignment, or prove their hardness using just P 6= N P as a complexity theory hypothesis? In fact the known NP-hardness of Monotone Minimum Satisfying Assignment concerns 2 (CNF) formulae and Quasi NP-hardness uses 4 formulae. But in general the formula or circuit can have unbounded depth and thus it a priori has richer expressive abilities. Hence there could be some chance to prove its hardness by some other way without improving the corresponding factor of Label Cover, perhaps by using some extension of the self-improvement property. 9

3 Main Results for Frege Systems Our rst main results are that it is hard to approximate the length of the shortest Frege proof of a given tautology. In sections 3.2 and 3.3, we will treat the case where proof length is measured in terms of the number of symbols in the proof; we leave the case where proof length is measured in terms of number of lines to section 3.4. Hardness Theorem 5 (a) If P 6= NP, then there is no polynomial time algorithm which can approximate Minimum Length Frege Proof to within a constant factor. (b) If NP 6 QP, then there is no polynomial time algorithm which can approximate Minimum Length Frege Proof to within a factor of 2 log(1?") n for any ". This theorem holds whether Frege proofs are taken to be tree-like or dag-like. The proof of Theorem 5 will give a reduction of the Monotone Minimum (Circuit) Satisfying Assignment problem to the Minimum Length Frege Proof problem. Thus any hardness results for the Monotone Minimum Satisfying Assignment or Monotone Minimum Circuit Satisfying Assignment problem immediately also apply to the Minimum Length Frege proof problem. Thus, for instance, Theorem 4(b) implies that for some constant > 0, the Minimum Length Frege Proof problem cannot be approximated to within a factor of log n unless NP DTIME(n O(log log n) ). Similar comments apply to the rest of our theorems on the hardness of approximating proof length. 3.1 Preliminaries Frege proof systems are proof systems for propositional logic. A Frege proof system F is specied by its language L and a nite set of inference and axiom schemes. The language L is a nite set of Boolean connectives, which is complete in the sense that any Boolean function can be represented by an L-formula. The permissible inferences are specied schematically as inferences A 1 A 2 A k B which indicates that for any substitution of formulas for variables, B may be inferred from the formulas A 1 ; : : :; A k. We allow k = 0 in the above 10

scheme, which corresponds to axioms. Finally, the Frege proof system must be implicationally complete, i.e., if A 1 ; : : :; A k j= B then there is a derivation of B from the assumptions A 1 ; : : :; A k using the inferences of F. We dene the size or length, jcj, of a formula C to equal the number of symbols in C, where each occurence of a variable or a connective is counted as a symbol. Likewise, if P is a Frege proof, then the symbol size of P, jp j, equals the number of symbols in P. If F is a Frege system, then F ` ' means that there is an F-proof of '. The symbol size of a Frege proof is the total number of symbols in the proof. The step-length (or length) of a Frege proof is the number of lines in the proof. F n ' means that there exists an F-proof P such that jp j n. Typical examples of Frege system include the `textbook systems' which use the language f^; _; :;!g and have a nite set of axiom schemes and have modus ponens as their only other rule of inference. Of course there are many possible such textbook systems since there are many choices for the axiom schemes; however, they are all essentially equivalent in terms of proof length. Indeed the following holds: Theorem 6 ([13, 20, 21]) If F 1 and F 2 are Frege systems with the same language, then they linearly simulate each other; i.e., for all ', if F n 1 ' then F O(n) 2 ', and vice-versa. For Frege proof systems in diering languages, it is known that any two Frege systems F 1 and F 2 p-simulate each other, i.e., that any F 1 -proof can be translated into an F 2 -proof in polynomial time and vice-versa; see [13, 20] for precise denitions and proofs of this. Extended Frege proof systems are propositional proof systems which allow the introduction of abbreviations of formulas on the y. It is conjectured that the minimal symbol size for extended Frege proofs can sometimes be exponentially smaller than the corresponding minimal Frege proof; however, this is still open. We next dene the notion of \active" formulas in a proof, which will be useful for proving lower bounds on the lengths of proofs. Recall that an inference in a proof must be a substitution instance of an axiom scheme, i.e., each inference must be of the form A 1 A k A k+1 Consider a particular occurrence of a formula C as a subformula of a formula A i in the inference (I). If the principal connective of C is present already in (I) 11

the formula A i, then we say C is active w.r.t. the inference (I). Otherwise, C occurs as a (not necessarily proper) subformula of x for some variable x, and C is not active w.r.t. inference (I). If a formula C has some occurrence in a proof P which is active with respect to some inference of P, then C is said to be active in P. (The terminology is potentially confusing: it is important to note that an active formula of P may never occur as a formula in the proof P, but instead only as a subformula of formulas in P.) The next theorem lets us obtain a lower bound on the length of P, jp j, in terms of the lengths of the active formulas of P. Theorem 7 (see [11]) Let F be a Frege proof system. There is a constant such that if P is a Frege proof and we let ' range over active formulaoccurrences in P, then X jp j j'j ' Proof. A formula can be viewed as a tree with nodes labeled by connectives from L. The depth of a formula is dened to equal the height of this tree, namely the maximum number of connectives along any branch of the tree. Let d equal the maximum depth of the depths of the formulas which occur in the inference schemes of F, and set = (1=d). Clearly, any active occurrence of a formula in P must have its principal connective at distance at d? 1 from the root of the formula's tree. Thus, any given single symbol a occurring in P can occur inside at most d active occurrences of subformulas. From this, we immediately get d jp j j'j and the theorem follows immediately. 2 As mentioned earlier, the step-length of a proof P is equal to the number of steps or inferences (counting axioms as nullary inferences) in the proof. There is a linear relationship between the number of formulas active in P and the step length of P ; namely, Theorem 8 Let F be a Frege proof system. Then there is a constant so that if P is a proof and m is the number of distinct active formulas in P, then the step-length of P is m. Proof. We let equal the maximum number of subformulas that can be active in any given formula in P. For instance, if every inference scheme 12 X '

has depth bounded by d and P r is the maximum arity of connectives in the d language of F, then = i=0 ri works. Clearly Theorem 8 is true with = 1=. 2 It is an interesting (albeit trivial) observation that if no formula is repeated in P, then the number of steps in P is also linearly upper bounded by the number of distinct formulas active in P. 3.2 Frege Systems with language f^; _; :;!g We shall rst prove our main theorem for Frege proof systems which have language ^, _, : and!. This will introduce and motivate the proof in the next section for general Frege proof systems. Thus, consider an arbitrary Frege proof system F over the above set of connectives. To prove Theorem 5 for F, it will suce to give a reduction of the Monotone Minimum Satisfying Assignment problem to the Minimum Length Frege Proof problem for F. We will start with a construction from [10]: Denition Let p 0 ; p 1 ; : : : be propositional variables. Let? be the contradiction p 0 ^ :p 0, and let i 0 be the tautology (:p i _ p i ). Dene ì to be (? _ (? _ (? _ i 0 ) )), where there are ` occurrences of?. Note that ji lj = O(`). The next two lemmas state that the proof of ì is big for large `, and that a derivation of ì must essentially derive rst all i k for all k < `. Lemma 9 ([10]) ì 2 i has a tree-like F-proof of length O(`2). This lemma is proved by noting that F can derive successively i 0, i 1,, etc., until ì is derived. Denition Let be a formula and consider a particular ì. We dene =(ì ) to be the formula obtained from by replacing every occurrence of ì as a subformula of with the formula?. Note, that if `1 < `2 then `2 i =( `1 i ) is not a tautology anymore. If P is a proof, then we dene P=(ì ) be a sequence of formulas obtained by replacing every in P with =(ì ). Note that P=(ì ) will not, in general, be a valid proof. Lemma 10 Let P be a proof of and suppose that the formula ì active in P. Then, =(ì ) is a tautology. is not 13

The proof of Lemma 10 is immediate by the fact that if ì is not active in P, then P=(ì ) is identical to P, except that it may use a dierent substitution of formulas for variables, and hence it is still a valid proof. 2 Finally we need a lemma that we will use to upper bound the lengths of Frege proofs which are monotone Boolean combinations of formulas of the form ì. Lemma 11 Let '(x 1 ; : : :; x k ) be a formula of size j'j = n over the basis f_; ^g. Suppose hv j i is a satisfying assignment and that I is the index set of all v i such that v i = >. Then, for every set of formulas 1 ; 2 ; : : :; k we have a tree-like (and hence dag-like) proof of ^ i2i of length O(M k 3 + M n 2 ) = O(M n 3 ), i!! '( 1 ; : : :; k ) Proof. By a straightforward induction on the size of '. For the base case note that V i i! i0 has a proof of length O(M k 2 ); for the induction step observe that in any Frege system over the language f^; _; :;!g, we have that fx! y; x! zg O(1) (x! (y ^ z)), and fx! yg O(1) (x! (y _ z)). 2 To prove Theorem 5 for F, it suces to prove: Theorem 12 There is an A-reduction from the Monotone Minimum Satisfying Assignment problem to the Minimum Length Frege Proof problem for F. This reduction works for both the dag-like and the tree-like forms of F. Proof. Recall that an A-reduction is a polynomial time Karp-reduction which preserves the non-approximating ratio to within a constant factor. Suppose we have a formula '(x 1 ; : : :; x k ) of length n = j'j over the basis f_; ^g, which is an instance of Monotone Minimum Satisfying Assignment. Take m suciently large compared to n, say m = n 4, and consider the following tautology: = '( m 1 ; m 2 ; : : :; m k ): (a) Suppose that the optimal solution to the Monotone Minimum Satisfying Assignment problem for ' has (') p 1. We construct a tree-like (and hence dag-like) F-proof of of length O(p 1 m 2 )+O(mn 2 ) = O(p 1 m 2 ) as follows: First construct a tree-like proof of V i2i m i! ' of length O(mn 3 ) 14

using Lemma 11, then prove each i m for i 2 I with V a proof of length O(p 1 m 2 ) by Lemma 9, and then prove the conjunction i2i i m of these formulas and applies modus ponens - which adds O(mn) symbols to the proof. (b) Suppose that (') p 2. Let P be some F-proof of. Let I be the index set of all i such that ì is active in P for all 0 ` m. For j =2 I, choose j r so that jr j is not active in P. By Lemma 10 we have that, after replacing all jr j with? for all j =2 I, the formula remains a tautology. Hence the formula ' is satised by the truth assignment corresponding to the characteristic function of I, hence jij p 2 by choice of p 2. Thus jp j = (p 2 m 2 ), By Theorem 7 and the fact that the total length of the formulas ì, for i 2 I; 0 ` m, is (p 2 m 2 ). (a) and (b) establish that ' 7! is an A-reduction; namely, it is easy to see that the constructed reduction transforms a suciently large constant gap g between hard and easy instances of Monotone Minimum Satisfying Assignment into a constant gap g " in Minimum Length Frege Proof, hence if Minimum Length Frege Proof is approximable with some constant then Monotone Minimum Satisfying Assignment is approximable too. This proves Theorem 12 as well as Theorem 5 for the case of Frege systems with language f:; ^; _;!g. 2 The following theorem shows that Frege proof systems, even tree-like ones, can deal in a certain sense even with monotone circuits. Theorem 13 There is an A-reduction from the Monotone Minimum Circuit Satisfying Assignment problem to the Minimum Length Frege Proof problem for F. This is true for both tree-like and dag-like Frege systems. Proof. We present a reduction which works for both dag-like and tree-like Frege systems. Suppose we have circuit C with inputs x 1 ; : : :; x k given as a set of instructions x i := x ji;1 OP x ji;2 for i 2 fk + 1; : : :; ng, where OP is ^ or _, where j i;1 ; j i;2 < i and where x n is the output node. Construct a tautology = n^ i=k+1 ((x ji;1 OP x ji;2 )! x i )!! x n ; where x 1 ; : : :; x k are replaced with m; : : :; m 1 k. As before it will be sucient to take m large enough, i.e., m = n 4. The lower bound for the proof of is absolutely identical to that of Theorem 12, namely every proof of has length at least m 2 where 15

= (C). Now we estimate an upper bound, by constructing a tree-like (and hence dag-like) proof of length O(m 2 + mn 3 ) = O(m 2 ). Suppose C(v 1 ; : : :; v k ) = >, and let I = fi : v i = >g. Suppose = jij variables have value >. We need to nd a tree-like proof of of length O(m 2 + m n 3 ). Let r 1 ; r 2 ; : : :; r s = n be the increasing sequence of indices such that x r1 ; x r2 ; : : :; x rs are made true by the truth assignment ~v. Then, it is straightforward to construct a tree-like Frege proof of the formulas!! ^ n^ `^ x i! ((x ji;1 OP x ji;2 )! x i )! x ri i2i i=k+1 which proceeds by proving these successively with ` = 1; 2; 3; : : :; s. Since the conjunctions all have O(n) inputs and since the formulas have symbollength O(m n), the proof has symbol-length O(m n 3 ). (If one is careful and uses balanced conjunctions [8], the proof V can have length O(m n 2 log n).) Finally there is a tree-like proof of i2i x i of length O( m 2 ) since I has cardinality. Thus, with a nal modus ponens we get?v n i=k+1 ((x j i;1 OP x ji;2 )! x i )! V` i=1 x r i, from which we get with additional O(m n 2 ) symbols. In total, we get a tree-like proof of of length O(m 2 + m n 3 ) = O(m 2 ). 2 3.3 General Frege proof systems We now take up the proof of Theorem 5 for general Frege proof systems which use an arbitrary (complete) nite language of propositional connectives. Fix some such proof system F, with language L. We wish to construct L-formulas `;L i which are analogues of the formulas ì. To do this with similar size bounds, we need the following lemma: Lemma 14 (Reckhow [20]) There are L-formulas NOT(x; z), AND(x; y; z), OR(x; y; z), and IMP(x; y; z) such that (1) NOT(x; z) contains one occurrence of x, and AND(x; y; z), OR(x; y; z), and IMP(x; y; z) contain exactly one occurrence of each of x and y. (2) The four formulas represent the Boolean functions :x, (x^y), (x_y) and (x! y); in particular, the truth values of the formulas are independent of the truth value of z. Proof. The side variable z acts merely as a placeholder whose truth value is irrelevant: in fact, if the language L contains a constant symbol, then i=1 16

the use of the variable z is unnecessary. We shall assume that the constant symbols > and? are included in L; this may be assumed without loss of generality since the symbols > and? may be replaced everywhere by L-formulas equivalent to the formulas (z _ :z) and (z ^ :z), respectively. Let N be an L-formula containing only the variable x which represents the propositional function :x; N exists since L is a complete set of connectives. If N contains n occurrences of x, we write N = N(x; x; : : :; x) with each occurrence of x separately indicated. Let > i denote a vector of i occurrences of >, and dene? i similarly. By choice of N, N(> n ) has value False and N(? n ) has value True. Therefore, there is some 0 i < n such that N(> i ;? n?i ) has value True and N(> i+1 ;? n?i?1 ) has value False. Thus, we can take NOT(x) to be the formula N(> i ; x;? n?i?1 ). To prove the remainder of the lemma, it will suce to nd an L-formula X(x; y) which has one occurrence of each of x and y, and which has appearing in its truth table either three values True and one value False, or three values False and one value True. (Note that conjunction, disjunction and implication are three of the eight propositional functions whose truth table has this property.) This will suce since AND, OR and IMP can be readily dened from such a formula X and from NOT. Let A be an L-formula containing only the variables x and y which represents the propositional function (x ^ y). Assume A = A(x; : : :; x; y; : : :; y) has m occurrences of x and n occurrences of y, each occurrence separately indicated. Dene A i;> to be the formula A(> i ;? m?i ; > n ) and A i;? to be A(> i ;? m?i ;? n ). Now A m;> and A m;? have dierent truth values, and A 0;> and A 0;? have the same truth value. Clearly, there is a value 0 i < m so that A i;> and A i;? have the same truth value, but so that A i+1;> and A i+1;? have dierent truth values. Fix such an i and let B = B(x; y; : : :; y) be the formula A(> i ; x;? m?i?1 ; y; : : :; y). Note that B has one occurrence of x and n occurrences of y, each indicated separately. Let B i (x) be the formula B(x; > i ;? n?i ). From the denition of B, the multiset of the four truth values of B 0 (>), B 0 (?), B n (>) and B n (?) contains either three Trues and one False, or one True and three Falses. Therefore, there is some value 0 j < n so that the multiset of the truth values of B j (>), B j (?), B j+1 (>) and B j+1 (?) enjoys the same property. Letting X(x; y) be the formula B(x; > j ; y;? n?j?1 ) gives the desired formula. 2 For simplicity of notation, we shall henceforth suppress mentioning the occurrences of the side variable z. Denition If ' is a formula over the basis f:; ^; _;!g, then its L- translation, ' L, is the L-formula obtained by replacing the connectives 17

:, ^, _, and! with the formulas NOT, AND, OR and IMP in the obvious way. Because of the condition that x and y occur at most once in the formulas NOT, AND, OR and IMP, the size j' L j of ' L is O(j'j). We write `;L i to denote (ì )L ; thus j `;L i j = O(`). The next lemma will be used to give upper bounds on the lengths of F-proofs and is the analogue of Lemmas 9 and 11. Lemma 15 (1) `;L i has an F-proof of length O(`2). (2) Let '(x 1 ; : : :; x k ) be the L-translation of an f^; _g-formula. Suppose hv j i is a satisfying assignment and that I = fi : v i = >g. Then, for every set of formulas 1 ; 2 ; : : :; k, there is a tree-like F-proof of ^ i2i i! L! '( 1 ; : : :; k ) of length O(M k 2 n + M n 2 ) = O(M n 3 ), where n = j'j and M is the maximal size of i. Theorem 16 There is an A-reduction from the Monotone Minimum Satisfying Assignment problem to the Minimum Length Frege Proof problem for F. This reduction works both for tree-like F-proofs and dag-like F-proofs. Proof. The proof is similar to the proof of Theorem 12. Suppose we have formula '(x 1 ; : : :; x k ) of size n = j'j over the basis f_; ^g, which is an instance of Monotone Minimum Satisfying Assignment. Take m suciently large compared to n, say m = n 4, and consider the tautology = ' L ( m 1 ; m 2 ; : : :; m k ): (a) Suppose that the optimal solution to the Monotone Minimum Satisfying Assignment problem for ' has (') p 1. We have j' L j = O(j'j) so, by Lemmas 9 and 15, has a (tree-like) F-proof of length O(p 1 m 2 ) + O(m n 3 ) = O(p 1 m 2 ). (b) Suppose that (') p 2. Let P be some F-proof of. Let I be the index set of all i such that `;L i is active in P for all 0 ` m. For j =2 I, jr ;L choose j r so that j is not active in P. By Lemma 10 we have that, after replacing all jr;l j with? for all j =2 I, the formula remains a tautology. 18

Hence the formula ' is satised by the truth assignment corresponding to the characteristic function of I, hence jij p 2. The total of the lengths of the active formulas `;L i, i 2 I, is greater than p 2 m 2 ; therefore, by Theorem 7, jp j = (p 2 m 2 ). (a) and (b) establish that ' 7! is an A-reduction. This proves Theorems 16 and 5. 2 One can show that this reduction can be applied to monotone circuits as well: Theorem 17 There is an A-reduction from the Monotone Minimum Circuit Satisfying Assignment problem to the Minimum Length Frege Proof problem for F. This is true for both the tree-like and dag-like versions of F. The proof of this theorem is analogous to that of Theorem 13. 3.4 Hardness of approximating step-length The previous section discussed the problem of minimizing the number of symbols in a Frege proof. We now prove the hardness of approximating the minimum step-length of Frege proofs. Hardness Theorem 18 (a) If P 6= NP, then there is no polynomial time algorithm which can approximate Minimum Step-Length Frege Proof to within a constant factor. (b) If NP 6 QP, then there is no polynomial time algorithm which can approximate Minimum Step-Length Frege Proof to within a factor of 2 log(1?") n for any ". This theorem holds whether Frege proofs are taken to be tree-like or dag-like. To prove Theorem 18, x a Frege proof system F with language L. For proving upper bounds on the step-length of F-proofs we need the following lemma: Lemma 19 (1) `;L i has an F-proof of step-length O(`). 19

(2) Let '(x 1 ; : : :; x k ) be the L-translation of an f^; _g-formula. Suppose hv j i is a satisfying assignment and that I = fi : v i = >g. Then, for every set of formulas 1 ; 2 ; : : :; k, there is a tree-like F-proof of ^ i2i i! of step-length O(k 2 + n) = O(n 2 ).! '( 1 ; : : :; k ) Proof. (1) Since F is a schematic system there is a constant upper bound on the number of steps necessary to prove (? _ ') L from ' L. Therefore a F-proof of `;L i with O(`) steps can be given by proving successively ' k;l i for k = 0; 1; : : :; `. (2) is proved similarly to Lemma 11. 2 To nish the proof of Theorem 18 suces to prove: Theorem 20 There is an A-reduction from the Monotone Minimum Satisfying Assignment problem to the Minimum Step-Length Frege Proof problem for F. This reduction works both for tree-like F-proofs and dag-like F-proofs. Proof. Suppose we have formula '(x 1 ; : : :; x k ) of size n = j'j over the basis f_; ^g, which is an instance of Monotone Minimum Satisfying Assignment. Take m suciently large compared to n, say m = n 3, and consider the tautology = ' L ( m 1 ; m 2 ; : : :; m k ): (a) Suppose that the optimal solution to the Monotone Minimum Satisfying Assignment problem for ' has (') p 1. We have j' L j = O(j'j), so by Lemma 19 has a tree-like F-proof of length O(p 1 m+n 2 ) = O(p 1 m). (b) Suppose that (') p 2. Let P be some F-proof of. Let I be the index set of all i such that `;L i is active in P for all 0 ` m. For j =2 I, jr ;L choose j r so that j is not active in P. By Lemma 10 we have that, after replacing all jr;l j with? for all j =2 I, the formula remains a tautology. Hence the formula ' is satised by the truth assignment corresponding to the characteristic function of I, hence jij p 2. The total number of active formulas `;L i, i 2 I, is greater than or equal to p 2 m; therefore, by Theorem 8, jp j = (p 2 m). (a) and (b) establish that ' 7! is an A-reduction. This proves Theorems 20 and 18. 2 One can show that this reduction can be applied to monotone circuits as well: 20

Theorem 21 There is an A-reduction from the Monotone Minimum Circuit Satisfying Assignment problem to the Minimum Step-Length Frege Proof problem for F. This is true for both the tree-like and dag-like versions of F. Proof. The proof of this theorem is similar to that of Theorem 13. In fact, the identical reduction works and the same F-proofs establish the upper and lower bounds; however, the actual bounds change since we are measuring proof length in terms of steps instead of symbols. For the lower bound, it is easy to see that Lemma 10 implies that any proof of L must have step-length ( m). For the upper bound, it is straightforward to verify that the (tree-like) proof of L derived from the proof of constructed in the proof of Lemma 13 has length O( m + n 2 ). 2 Remark. All of our hardness results for approximating step-length and symbol-length of Frege proofs also apply to extended Frege systems. To see this, it suces to note that all the upper and lower bounds on the length of Frege proofs which were obtained in the proofs of Theorems 12, 13, 16 and 20 also apply to extended Frege proofs. Of course it is obvious that the upper bounds apply since every Frege proof is an extended Frege proof. The lower bounds also apply, since Theorems 7 and 8 are also true for extended Frege systems ([11]). 4 The Hardness of Resolution Resolution is a well-known proof system and its extensions are widely used as a foundation for many theorem proving systems. Thus it is of particular interest that it is dicult to nd approximately shortest resolution refutations. We shall be concerned exclusively with the propositional resolution systems. Recall that a literal is either a variable p or the negation of a variable p. A clause is a nite set of literals and is interpreted as the disjunction of its members. A set of clauses? is interpreted as the conjunction of its member clauses; thus a set of clauses can be identied with a formula in conjunctive normal form. The resolution rule allows an inference of the form C [ fpg D [ fpg C [ D It is well-known that resolution is sound and complete as a refutation system; namely, a set? is unsatisable if and only if the empty clause can be derived using only resolution inferences from the clauses in?. 21

A Horn clause is a clause which contains at most one positive literal. A resolution refutation consists of a sequence of clauses, ending with the empty clause. We can measure the length or size of a refutation in terms of either its step-length or its symbol-length. The step-length of the refutation is just equal to the number of clauses in the refutation. The symbol-length is dened to equal the sum of the cardinalities of the clauses appearing in the refutation. (Unlike the situation for Frege proofs where proof length is traditionally dened in terms of symbol-length, there seems to be no xed convention on how to measure the length of resolution refutations: thus, we shall always explicitly include one of the modiers `step-' or `symbol-'.) A refutation can be either tree-like or dag-like: unless it is explicitly stated otherwise, refutations are considered to be dag-like. We shall obtain the best possible results in that our upper bounds on length will apply to tree-like refutations and our lower bounds on length will apply to dag-like refutations. In the introduction, we introduced the Minimum Length Resolution Refutation problem: henceforth we'll be more precise and talk about the Minimum Step-Length Resolution Refutation and the the Minimum Symbol-Length Resolution Refutation problems. Hardness Theorem 22 (a) If P 6= NP, then there is no polynomial time algorithm which can approximate Minimum Step-Length (Symbol-Length) Resolution Refutation to within a constant factor. (b) If NP 6 QP, then there is no polynomial time algorithm which can approximate Minimum Step-Length (Symbol-Length) Resolution Refutation to within a factor of 2 log(1?") n for any ". These hardness results apply to both dag-like and tree-like resolution. In addition, the hardness results apply to Horn resolution, where all the input clauses are Horn clauses. To prove the Hardness Theorem, we shall reduce the Monotone Minimum Circuit Satisfying Assignment problem to Minimum Length Resolution Refutation problems. Let ' be an instance of Monotone Minimum Circuit Satisfying Assignment; we dene a set of clauses? ' which will be an A-reduction to the Minimum Length Resolution Refutation problems. Enumerate the subcircuits of ' as ' 1 ; : : :; '`, where the input variables are rst in the enumeration and where each ' i is listed only after all of its own subcircuits are enumerated, and thus '` is '. Obviously the number 22

` of subcircuits is less than the number of symbols n in '. We introduce new propositional variables y 1 ; : : :; y`, and dene the set? ' to contain the following clauses: a. The clause fy`g is in? '. b. For each i `, if ' i is (' j ^ ' k ), then the clause fy j ; y k ; y i g is in? '. c. For each i `, if ' i is (' j _ ' k ), then the clauses fy j ; y i g and fy k ; y i g are in? '. The above clauses describe the evaluation of '; however, note that they say nothing about the truth of the input variables x 1 ; : : :; x p of '. For each variable x i of ', we introduce new variables x i;j for j = 1; 2; : : :; m, and further include in? ' the following clauses: d. For each i p, the clauses fx i;1 g and fx i;m ; y i g and fx i;j ; x i;j+1 g for j = 1; : : :; m? 1 are included in? '. These clauses are said to be associated with x i. That completes the denition of? '. Informally,? ' asserts that there exists a truth-evaluation to all subcircuits of such that: the truth evaluation given to the output circuit is 0, the truth evaluation given to all input variables is 1, and the truth evaluation is consistent with all intermediate gate values. Clearly, this is an unsatisable formula since is monotone. Lemma 23 Let ' be an instance of Monotone Minimum Circuit Satisfying Assignment and let equal the cardinality of the minimum satisfying assignment for '. (1)? ' has a dag-like refutation with symbol-length (and hence step-length) equal to O(m + n). (2)? ' has a tree-like refutation of symbol-length O(m + n 2 ) and step-length O(m + n). Proof. (a) Let I fx 1 ; : : :; x p g specify a satisfying assignment for ' of cardinality. The dag-like proof proceeds by rst deriving the clause fy i g for each x i 2 I. Each fy i g is derived in m steps using the clauses associated with x i ; this part of the refutation takes m steps. The refutation then derives clauses fy i g starting with smaller values of i and ending with fy`g (e.g., the subcircuits of ' are processed in a bottom-up order). This takes O(n) steps. One further resolution with the input clause fy`g completes the 23

refutation. Each clause in the refutation contains a constant number of (in fact, at most three) literals. Hence the symbol-length of the refutation is also O(m + n). (b) The above proof is clearly not tree-like. To form a tree-like refutation, we use a top-down procedure to generate the refutation. The rst phase of the refutation starts with the clause fy`g and derives successively clauses of the form fy k1 ; y k2 ; : : :; y kr g with k 1 > k 2 > > k r. Such a clause is resolved with one of the (at most two) clauses that contain y k1 positively. This continues until we have a clause which contains only literals y i corresponding to input x i of '. It is possible to do this so that the remaining clause is just fy i : x i 2 Ig. For the second phase of the refutation, derive the clauses fy i g, for x i 2 I, with m steps, and for the third phases, use resolutions to derive the empty clause. There are obviously O(n) steps in the rst and third phases of the derivation, so the whole refutation has O(m + n) steps. Furthermore, each clause in the rst and third phase contains at most n literals. The second phase contains m clauses each with a constant number of literals. Therefore, the symbol-length of the tree-like refutation is O(m + n 2 ). 2 Lemma 24 Let ' and be as above. Then any resolution refutation (dag-like or tree-like) must have step-length, and hence symbol-length, of at least m. Proof. Let R be a resolution refutation. An input variable x i is dened to be R-analyzed if every one of the (m + 1)-clauses associated with x i is used in the refutation R. Obviously it will suce to prove that at least input variables are R-analyzed. In fact, if I is dened to equal the set of R-analyzed variables, then I implies a satisfying assignment for '. This last fact is almost immediate. To prove it formally, we dene a truth assignment as follows: (1) assigns truth values to variables y i according to the value I assigns to ' i (2) assigns True to x i;k i each clause fx i;1 g and fx i;j ; x i;j+1 g for 1 j < k is used in R. If I doesn't satisfy ', then would satisfy all the clauses used in the refutation R, which is impossible. Therefore, I is a satisfying assignment for '. 2 As an immediate corollary to these two lemmas we get: Theorem 25 There is an A-reduction from the Monotone Minimum Circuit Satisfying Assignment problem to the Minimum Length Resolution Refutation problems. This reduction works for both tree-like and dag-like refutations and for both step-length and symbol-length. Furthermore, the reduction produces only sets of Horn clauses. 24